Chapter 7: Area

Understanding the question
The square has to be partitioned into 4 regions such that the area of every region is exactly one-fourth of the area of the whole square. The question asks how many different (distinct) ways this can be done.

Step 1 : Listing a few obvious examples

• Draw the two medians (one horizontal and one vertical). This gives four smaller squares.
• Draw the two diagonals. This gives four congruent right-angled isosceles triangles.
• Draw one diagonal and then through the centre draw the line perpendicular to it. This again gives four congruent right-angled triangles (but oriented differently from the previous case).
• Draw any line parallel to the base at a distance $$\tfrac14$$ of the side–length from it, and another parallel line at a distance $$\tfrac34$$ of the side–length from the base. Now draw a vertical line through the mid-point of the base. The square is cut into four equal rectangles.
• Take an L-shaped path that starts from the centre, goes to the middle of one side, turns 90° and goes to the middle of the adjacent side. Rotate this path successively through 90°, 180° and 270°. The four images enclose four congruent L-shapes whose areas are each $$\tfrac14$$ of the whole square.

So even by casual experimentation we already have several different partitions.

Step 2 : A way to generate endlessly many partitions
Choose any path that starts from the centre of the square and ends at the mid-point of one of its sides. (The path can be straight, zig-zag or curved; it just must lie completely inside the square.) Call this path $$P$$. Rotate $$P$$ about the centre through 90°, 180° and 270°. The four rotated copies of $$P$$ enclose four regions that are congruent to one another, hence have equal area. Because the original path $$P$$ can be varied in infinitely many ways, the resulting partitions are also infinitely many.

Step 3 : Why the number is uncountable (and therefore certainly not finite)
Even if we restrict ourselves to merely straight-line paths that join the centre to the midpoint of a particular side, the slope of that segment can vary smoothly through an interval of real numbers. Each distinct slope produces a distinct set of four congruent regions, so there are as many partitions as there are real numbers in an interval—an uncountably infinite set.

Conclusion
There is no longest list that contains all the distinct ways; whenever we have a list we can slightly alter one of the cuts and obtain a new, different partition with the same equal area property. Hence the correct count is not a finite whole number.

Let the original square be $$ABCD$$ with side $$s$$. Our aim is always to produce four congruent regions; since congruent figures have the same area, each region will then automatically have area $$\frac{s^{2}}{4}$$.

Method 1 – “+ sign (four little squares)

1. Mark the mid–points $$P$$ of $$AB$$ and $$Q$$ of $$CD$$. Join $$PQ$$ – a vertical line through the centre $$O$$.
2. In the same way join the mid–points $$R$$ of $$BC$$ and $$S$$ of $$AD$$ to get the horizontal line $$RS$$.
3. The two lines intersect at right angles and cut the original square into four small squares, each of side $$\frac{s}{2}$$.
   Area of each small square = $$\Bigl(\tfrac{s}{2}\Bigr)^{2}=\tfrac{s^{2}}{4}$$.

(Draw two straight lines through the centre – one parallel to a side and the other perpendicular to it.)

Method 2 – “×” sign (two diagonals)

1. Draw the two diagonals $$AC$$ and $$BD$$; they meet at the centre $$O$$.
2. The diagonals form four congruent isosceles right–angled triangles: for instance $$\triangle AOB$$ has base $$AB=s$$ and height $$AO=\tfrac{s}{2}$$ (because $$O$$ is the mid-point of the diagonal).
   Area $$=\tfrac12\,s\,\tfrac{s}{2}=\tfrac{s^{2}}{4}$$.

Method 3 – Four equal vertical strips

1. On sides $$AB$$ and $$CD$$ mark the points whose abscissae are $$\tfrac{s}{4},\;\tfrac{s}{2},\;\tfrac{3s}{4}$$ (that is, divide the side into four equal parts).
2. Through each of those three points draw a line parallel to $$AD$$.
   This produces four congruent rectangles, each of width $$\tfrac{s}{4}$$ and height $$s$$, so area $$=s\times\tfrac{s}{4}=\tfrac{s^{2}}{4}$$.

Method 4 – Any curve with quarter-turn symmetry

1. Draw any curve $$\gamma$$ that starts at the centre $$O$$ and ends at one point on $$AB$$ (for example the mid-point).
2. Rotate the square (or the curve) through $$90^{\circ},\,180^{\circ},\,270^{\circ}$$ about $$O$$; each rotation gives a new copy of the same curve.
3. The four copies of $$\gamma$$ together cut the square into four regions that are exact quarter-turn images of one another, hence congruent, and so each has area $$\tfrac{s^{2}}{4}$$.

(Because the construction uses rotational symmetry the precise shape of $$\gamma$$ does not matter.)

Method 5 – Quarter-circle arcs

1. Mark the mid-points $$P,Q,R,S$$ of the four sides in order.
2. With each corner as centre and radius $$\tfrac{s}{2}$$ draw an arc joining the mid-points of the two adjacent sides. Four equal quarter-circles are produced inside the square.
3. These four arcs again divide the square into four congruent “curved-edge” regions related by $$90^{\circ}$$ rotation, so each region’s area is $$\tfrac{s^{2}}{4}$$ even though the boundary is not straight.

Further ideas: Any set of four congruent pieces obtained from the square by a symmetry of order 4 (rotation or repeated translation) will automatically give a valid division into equal areas. Students are encouraged to invent their own patterns—zig-zags, wavy lines, shapes made of dots—so long as the whole square is filled and the four pieces remain congruent.

To colour a rectangle evenly we need rangoli powder for the entire area of the rectangle.

The area $$A$$ of a rectangle with length $$l$$ and breadth $$b$$ is given by

$$A = l \times b$$

Rectangle 1

Length $$l_1 = 7\,\text{cm}$$, breadth $$b_1 = 4\,\text{cm}$$

$$A_1 = l_1 \times b_1 = 7\,\text{cm} \times 4\,\text{cm} = 28\,\text{cm}^2$$

Rectangle 2

Length $$l_2 = 8\,\text{cm}$$, breadth $$b_2 = 3\,\text{cm}$$

$$A_2 = l_2 \times b_2 = 8\,\text{cm} \times 3\,\text{cm} = 24\,\text{cm}^2$$

Comparing the two areas,

$$28\,\text{cm}^2 > 24\,\text{cm}^2$$

Hence Rectangle 1 has the larger area.

Therefore Rectangle 1 will require more rangoli powder when it is coloured evenly.

Given: A rectangle of length $$7\,\mathrm{cm}$$ and breadth $$4\,\mathrm{cm}$$ is divided by a diagonal into two congruent triangles.

Step 1 – Find the area of the rectangle

The area of a rectangle is the product of its length and breadth:

$$\text{Area of rectangle}=l\times b$$

Substituting $$l=7\,\mathrm{cm}$$ and $$b=4\,\mathrm{cm}$$,

$$\text{Area of rectangle}=7\times4=28\,\mathrm{cm^2}$$

Step 2 – Relate the triangles to the rectangle

The diagonal splits the rectangle into two congruent (equal) right-angled triangles, so each triangle occupies exactly one-half of the rectangle’s area.

$$\text{Area of one triangle}=\dfrac{1}{2}\times\text{Area of rectangle}$$

$$\text{Area of one triangle}=\dfrac{1}{2}\times28=14\,\mathrm{cm^2}$$

Therefore, the area of each triangle is $$14\,\mathrm{cm^2}$$.

Step 1 : What is area?
Area is the amount of surface a plane region occupies. In mathematics we want a single number that tells “how much space” is covered, no matter what the outline looks like.

Step 2 : Why choose a unit square?
Because a square of side 1 unit

• can tile the plane without leaving gaps or overlaps (so every region can be completely filled by such squares or parts of them),
• has straight edges that make counting exact,
• gives a standard that everyone can agree on. One unit square is defined to have area $$1\;\text{square-unit}$$. Counting how many such squares (or fractional squares) exactly cover the region gives a measure that is independent of the region’s shape.

Step 3 : Why can’t perimeter do the same job?

• Perimeter is a one-dimensional quantity (only length). Doubling every length of a figure multiplies the perimeter by $$2$$, but its area by $$4$$. Thus the two quantities behave differently under scaling.
• Different regions can have the same perimeter yet quite different areas. For instance:
  

  Shape	Dimensions (units)	Perimeter $$P$$	Area $$A$$
  Rectangle I	$$5 \times 4$$	$$2(5+4)=18$$	$$5\times4=20$$
  Rectangle II	$$7 \times 2$$	$$2(7+2)=18$$	$$7\times2=14$$

  Both rectangles have the same boundary length $$18$$ units, but the first covers $$20$$ square units of surface while the second covers only $$14$$.
• Conversely, two regions can have the same area with different perimeters (e.g. a square of side $$4$$ and a long thin rectangle of sides $$1$$ and $$16$$ both have area $$16$$ sq. units, but their perimeters are $$16$$ units and $$34$$ units respectively).

Step 4 : Conclusion
Since perimeter does not uniquely determine the amount of surface covered, it cannot serve as a measure of area. Counting the number of unit squares (or fractional parts of them) that exactly cover the region gives a unique, reliable and additive measure. Hence we define area that way, not by the length of the boundary.

Restatement of the doubt
Many students feel that “more boundary means more surface” and therefore expect:

• Two figures with the same perimeter must have the same area.
• The figure with the larger perimeter must also have the larger area.

Both expectations are false. We settle the matter by constructing explicit numerical examples that Class 8 students can calculate entirely on their own.

1. Same perimeter but different areas

Take two rectangles whose perimeters are each $$16\text{ cm}$$.

• Rectangle I: length $$l_1 = 3\text{ cm}$$, breadth $$b_1 = 5\text{ cm}$$.
• Rectangle II (actually a square): side $$l_2 = b_2 = 4\text{ cm}$$.

Verify their common perimeter:

Rectangle I: $$\text{Perimeter} = 2(l_1+b_1)=2(3+5)=2\times8=16\text{ cm}$$
Rectangle II: $$\text{Perimeter} = 2(l_2+b_2)=2(4+4)=2\times8=16\text{ cm}$$

Now compare their areas:

$$\text{Area of Rectangle I}=l_1\times b_1 = 3\times5 = 15\;\text{cm}^2$$
$$\text{Area of Rectangle II}=l_2\times b_2 = 4\times4 = 16\;\text{cm}^2$$

Although the boundaries are equal, $$15\;\text{cm}^2 \neq 16\;\text{cm}^2$$. Hence equal perimeter does not force equal area.

2. Larger perimeter but smaller area

Keep the square from above (side $$10\text{ cm}$$) and compare it with a long thin rectangle.

• Square: side $$s = 10\text{ cm}$$.
• Rectangle: length $$L = 1\text{ cm}$$, breadth $$B = 20\text{ cm}$$.

Perimeters:

Square: $$\text{Perimeter} = 4s = 4\times10 = 40\text{ cm}$$
Rectangle: $$\text{Perimeter} = 2(L+B) = 2(1+20) = 2\times21 = 42\text{ cm}$$

The rectangle clearly has a larger perimeter (42 cm vs 40 cm). Now compute their areas:

Square: $$\text{Area} = s^2 = 10^2 = 100\;\text{cm}^2$$
Rectangle: $$\text{Area} = L\times B = 1\times20 = 20\;\text{cm}^2$$

Here the shape with the larger perimeter (42 cm) actually encloses a much smaller area (20 cm²) than the square (100 cm²). Thus a larger perimeter does not guarantee a larger area.

3. What we can safely say

Perimeter measures the length of the boundary; area measures the amount of surface enclosed. They are related but one does not determine the other. Without extra information about the exact shape, neither of the two statements in the question is logically valid.

Therefore,

• Figures can have the same perimeter yet different areas.
• A figure with a larger perimeter can have either a larger area, an equal area, or even a smaller area.

Hence the conclusions suggested in the question are incorrect.

We have to exhibit two rectangles such that the first one has the larger perimeter while the second one has the larger area. A concrete numerical choice and the full working are given below.

1. Choice of dimensions
   
   • Region 1 – rectangle with length $$l_1 = 2 \text{ cm}$$ and breadth $$b_1 = 9 \text{ cm}$$.
   • Region 2 – rectangle with length $$l_2 = 4 \text{ cm}$$ and breadth $$b_2 = 5 \text{ cm}$$.
2. Perimeter calculation
   For a rectangle, $$\text{Perimeter}=2(\text{length}+\text{breadth}).$$ Hence
   • Region 1: $$P_1 = 2(l_1+b_1)=2(2+9)=2\times11=22 \text{ cm}$$.
   • Region 2: $$P_2 = 2(l_2+b_2)=2(4+5)=2\times9=18 \text{ cm}$$.
   Thus $$P_1 = 22\,\text{cm} > P_2 = 18\,\text{cm}$$.
3. Area calculation
   For a rectangle, $$\text{Area}=\text{length}\times\text{breadth}.$$ Therefore
   • Region 1: $$A_1 = l_1\times b_1 = 2\times9 = 18 \text{ cm}^2$$.
   • Region 2: $$A_2 = l_2\times b_2 = 4\times5 = 20 \text{ cm}^2$$.
   Hence $$A_1 = 18\,\text{cm}^2 < A_2 = 20\,\text{cm}^2$$.
4. Verification
   Collecting the perimeter and area comparisons obtained above: \[P_1 = 22\,\text{cm}\;>\;18\,\text{cm} = P_2,\qquad A_1 = 18\,\text{cm}^2\;<\;20\,\text{cm}^2 = A_2.\] Hence the required inequalities hold simultaneously: \[\boxed{\;P_1 \;>\; P_2 \quad\text{and}\quad A_1 \;<\; A_2\;}\] Thus the chosen pair of rectangles is a valid example — Region 1 has the larger perimeter, while Region 2 has the larger area, exactly as required.

You can draw both rectangles on a grid sheet: the 2 cm × 9 cm rectangle will look longer and thinner with 22 boundary units, while the 4 cm × 5 cm rectangle will enclose more unit squares even though its perimeter is only 18 units.

Idea of the construction

Just as we saw for rectangles, even among other plane figures a shape that "goes round a lot" need not cover a large surface. A very convenient pair is a fat, almost–equilateral triangle and a long, thin right-angled triangle. Their drawings look quite different, so the contrast between boundary and interior region is immediately visible.

Step 1 : Fix the two triangles

• Triangle $$T_1$$  (fat) : an isosceles right-angled triangle whose equal legs are $$5\,\text{cm}$$ each.
• Triangle $$T_2$$  (thin) : a right-angled triangle whose perpendicular sides are $$10\,\text{cm}$$ and $$1\,\text{cm}$$.

On squared paper draw both triangles with one right angle pointing downwards. $$T_1$$ looks almost like a square cut in half; $$T_2$$ appears as a long strip that narrows to a sharp point. The eye already senses that the boundary of $$T_2$$ is longer while its interior is much smaller.

Step 2 : Perimeter of $$T_1$$

The hypotenuse is obtained from Pythagoras:

$$\text{Hypotenuse}=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt{2}\,(\approx7.07\,\text{cm}).$$

Hence

$$P_1=5+5+5\sqrt{2}\;\text{cm}\approx17.07\,\text{cm}.$$

Step 3 : Area of $$T_1$$

$$A_1=\tfrac12\times5\times5\;\text{cm}^2=12.5\,\text{cm}^2.$$

Step 4 : Perimeter of $$T_2$$

The third side is

$$\text{Hypotenuse}=\sqrt{10^2+1^2}=\sqrt{101}\,(\approx10.05\,\text{cm}).$$

Therefore

$$P_2=10+1+\sqrt{101}\;\text{cm}\approx21.05\,\text{cm}.$$

Step 5 : Area of $$T_2$$

$$A_2=\tfrac12\times10\times1\;\text{cm}^2=5\,\text{cm}^2.$$

Step 6 : Compare

\[P_2 \; (21.05\,\text{cm})\; >\; P_1 \; (17.07\,\text{cm}),\qquad A_2 \; (5\,\text{cm}^2)\; <\; A_1 \; (12.5\,\text{cm}^2).\]

Thus the triangle with larger perimeter has the smaller area. When the two are drawn next to one another, this fact is obvious to the eye because $$T_2$$ looks like a very long boundary enclosing hardly any space.

Conclusion

The pair $$(T_1,T_2)$$ provides the required example of two other shapes (triangles) where a larger perimeter corresponds to a smaller area.

Step 1  Understand the figure
The ground is a rectangle whose length and breadth are known:

• Length = $$14\,\text{m}$$
• Breadth = $$12\,\text{m}$$

A cross-path runs through the middle of the plot:

• One strip goes all the way from left to right (parallel to the 14 m side).
• The other strip goes from top to bottom (parallel to the 12 m side).
• The two strips meet at right angles and overlap in a small rectangle at the centre.

To calculate the shaded (path) area we evidently need to know how wide each strip is.

Step 2  Measurements still required
Measure the widths of the two strips:

• Let the horizontal strip (full length 14 m) have width $$w_1\,\text{m}$$.
• Let the vertical strip (full length 12 m) have width $$w_2\,\text{m}$$.

These are the only extra measurements needed.

Step 3  Derive a general formula for the path area

• Area of the horizontal strip = length × width = $$14\times w_1\;\text{m}^2$$.
• Area of the vertical strip = breadth × width = $$12\times w_2\;\text{m}^2$$.
• The central overlap (counted twice so far) is a rectangle $$w_1\times w_2\;\text{m}^2$$. Subtract it once.

Therefore the required area is

If the two paths happen to have the same width, say $$w_1 = w_2 = w$$, the formula simplifies to

Step 4  Choose some realistic widths and calculate

Example 1 (equal widths)

• Take $$w = 1.5\,\text{m}$$ for both paths.

Using $$A = 26w - w^2$$:

$$A = 26(1.5) - (1.5)^2 = 39 - 2.25 = 36.75\,\text{m}^2$$

Example 2 (different widths)

• Let $$w_1 = 2\,\text{m}$$ (horizontal strip).
• Let $$w_2 = 1.5\,\text{m}$$ (vertical strip).

Using $$A = 14 w_1 + 12 w_2 - w_1 w_2$$:

$$A = 14(2) + 12(1.5) - (2)(1.5) = 28 + 18 - 3 = 43\,\text{m}^2$$

Step 5  Conclusion
To find the area of the cross-path we must first measure the widths of the two intersecting strips. After that we can use

or, when both widths are the same,

Substituting any measured (or chosen) values for the widths gives the required area of the path.

Step 1  Understand the situation
The shaded (bent) tube makes a rectangular spiral inside the outer square.
 • Outer square : side $$20$$.
 • Width of the tube : $$1$$ (same everywhere).
Because the width is uniform, the area of the tube is simply $$\text{(length of the centre-line)}\times1=\text{length of the centre-line}.$$

Step 2  List every straight piece of the centre-line
Start at the point where the tube begins and follow it till it ends inside. The successive straight pieces are exactly the sides of the rectangles that form the spiral.

Step 3  Add the lengths
$$ \begin{aligned} L &= 20+20+15+15+10+10+5\\ &= 40+30+20+5\\ &= 95. \end{aligned} $$ So the centre-line of the bent tube is $$95$$ units long.

Step 4  Find the area of the bent tube
Area = length × width = $$95\times1 = 95$$ square units.

Step 5  Required length of an equivalent straight tube
A straight tube of the same width $$1$$ must also possess an area of $$95$$ square units, so its length must be $$95$$ units.

Let the side of the given square be $$s$$.

1. Area of the square before doubling
Area = $$s^{2}$$.

2. Area of one triangular region before doubling
The two diagonals divide the square into four congruent triangles, therefore

$$\text{area of one triangle} = \frac{1}{4}\,(\text{area of square}) = \frac{1}{4}s^{2}.$$

3. After doubling the side
New side = $$2s$$.

Area of the enlarged square is

4. Area of one triangular region after doubling
Again the diagonals divide the square into four equal parts, so

$$\text{new area of one triangle} = \frac{1}{4}\,(4s^{2}) = s^{2}.$$

5. Increase in the area of one region

\[ \text{increase} = s^{2} - \frac{s^{2}}{4} = \frac{3}{4}s^{2}. \]

6. Conclusion
Because every triangular region is identical both before and after enlargement, the increase for regions 1, 2 and 3 is the same.

• Region 1 increases by $$\dfrac{3}{4}s^{2}$$.
• Region 2 increases by $$\dfrac{3}{4}s^{2}$$.
• Region 3 increases by $$\dfrac{3}{4}s^{2}$$.

Equivalently, the area of each region becomes four times its former value, so the increase is three times its original area.

Objective
Cut one square into four pieces and arrange those pieces so that the boundary looks like a larger square but a small square-shaped hole appears in the middle. The activity shows that although the outline seems to have become larger, no area has been created or destroyed – the ‘new’ area is exactly the empty hole.

1. Drawing the cutting lines

1. Take a square sheet  $$ABCD$$  of side  $$s\;\text{cm}$$. Fix a point  $$P$$  inside the square so that the distances of  $$P$$  from the left–right sides are unequal.
       Let the distance of  $$P$$  from  $$AB$$  (the top side) be $$p$$ and from  $$AD$$  (the left side) be $$q$$, where $$0<p,q<s$$ and, without loss of generality, assume $$p>q$$.
2. Through  $$P$$  draw one segment parallel to $$AB$$ and another segment parallel to $$AD$$. These two segments are perpendicular to each other and terminate at the sides of the square. They are the only cuts required.
3. The two segments divide the sheet into four L–shaped pieces, labelled in anti-clock-wise order as $$P_1,P_2,P_3,P_4$$.

2. Naming the four pieces

Observe that the sum of the areas of the four pieces is still $$s^2\;\text{cm}^2$$ – the area of the original square.

3. Rearrangement rule

1. Rotate every piece through $$90^{\circ}$$ clock-wise.
2. Slide the pieces so that their right angles now meet at the four corners of a new figure. When this is done carefully you will notice two things:
   • The outer boundary becomes a perfect square of side $$s+(p-q)\;\text{cm}$$.
   • A central empty square of side $$(p-q)\;\text{cm}$$ remains.

4. Verifying the area algebraically

Area of the new outline (the big square) is \[ (s+(p-q))^{2}\;\text{cm}^{2} \] Area of the hole is \[ (p-q)^{2}\;\text{cm}^{2} \] Therefore, area actually covered by the four pieces after rearrangement is $$ (s+(p-q))^{2}-(p-q)^{2}=s^{2}+2s(p-q)-(p-q)^{2}+(p-q)^{2}=s^{2}\;\text{cm}^{2} $$ which is exactly the area of the original square. Thus no area is lost or gained; it has only been redistributed, leaving a hole whose area balances the apparent increase in outside dimensions.

5. What to notice while doing the activity

• Because $$p \ne q$$, the rearrangement necessarily leaves a gap – the central square.
• If by chance you choose $$p=q$$ (i.e. you cut exactly halfway), the four pieces will fit back into a square of the same size and no hole will be seen.
• This experiment is a good demonstration that area depends only on how much region is actually covered, not on how long the outside boundary appears to be.

Result
Cutting along two perpendicular interior lines and rotating each of the four resulting pieces by $$90^{\circ}$$ gives a larger square-shaped frame whose hole has exactly the area required to account for the increase in the outside length. Hence the total area of the material stays constant.

Step 1 – Understand the rectangles
Both figures are rectangles $$ABCD$$ having identical length and breadth.
So
$$AB = CD = \text{(length)}$$ and $$AD = BC = \text{(breadth)}$$.

Step 2 – Fix the base common to the two triangles
In each rectangle the triangle in question uses the side $$DC$$ as its base.
Hence for both triangles

$$\text{base} = DC$$   …(1)

Step 3 – Find the corresponding height
For a triangle, the height is the perpendicular distance from the vertex opposite the base to the line containing the base.
Here the vertices opposite $$DC$$ are the points $$X$$ and $$Y$$, each of which lies somewhere on the top edge $$AB$$. The top edge $$AB$$ is parallel to the base $$DC$$ in every rectangle.
Therefore the perpendicular distance between $$AB$$ and $$DC$$ is simply the breadth of the rectangle:

$$\text{height} = AD$$   …(2)

This breadth is the same for both rectangles because the rectangles are identical.

Step 4 – Compute each area
For any triangle:

$$\text{Area} = \tfrac12 \times \text{base} \times \text{height}$$

Applying (1) and (2):

For $$\triangle XDC:$$   $$\text{Area}(\triangle XDC) = \tfrac12 \times DC \times AD$$

For $$\triangle YDC:$$   $$\text{Area}(\triangle YDC) = \tfrac12 \times DC \times AD$$

Step 5 – Compare the two areas
Because the right‐hand sides are identical,

$$\text{Area}(\triangle XDC) = \text{Area}(\triangle YDC).$$

Conclusion
Neither triangle is larger; the two areas are equal. Each triangle occupies exactly one-half of its rectangle.

Denote the (identical) rectangles by ABCD and A′B′C′D′. In the first rectangle a point X lies on $$AB$$ and we join it to $$C$$ and $$D$$ to obtain $$\triangle XDC$$. In the second rectangle a point Y lies on $$AD$$ and we join it to $$B$$ and $$C$$ to obtain $$\triangle YBC$$.

Because the rectangles are identical, let

• length $$AB = CD = l$$,
• breadth $$BC = AD = b$$.

1. Area of $$\triangle XDC$$

Take $$DC$$ as the base; it is a side of the rectangle, so $$DC = l$$.
Since $$X$$ lies on $$AB$$, and $$AB \parallel DC$$, the altitude from $$X$$ to line $$DC$$ equals the perpendicular distance between the parallel sides $$AB$$ and $$DC$$, which is the breadth $$BC = AD = b$$.

\[ \text{Area}(\triangle XDC) = \frac12 \times l \times b. \]

2. Area of $$\triangle YBC$$

Take $$BC$$ as the base; it is a side of the rectangle, so $$BC = b$$.
Since $$Y$$ lies on $$AD$$, and $$AD \parallel BC$$, the altitude from $$Y$$ to line $$BC$$ equals the perpendicular distance between the parallel sides $$AD$$ and $$BC$$. That perpendicular distance is the length of side $$AB$$ (equivalently, $$CD$$), i.e. $$AB = CD = l$$.

\[ \text{Area}(\triangle YBC) = \frac12 \times b \times l = \frac12 \times l \times b. \]

3. Comparison

\[ \text{Area}(\triangle XDC)=\text{Area}(\triangle YBC)=\frac12\,l\,b. \]

Thus the two triangles have equal areas; neither is larger than the other. Each is exactly half the area of the (identical) rectangle.

Step 1 – Identify the base and its corresponding altitude

• In rectangle $$ABCD$$, side $$DC$$ is a straight line segment.
• Line segment $$XY$$ is drawn such that $$Y$$ lies on $$DC$$ and $$XY\perp DC$$. Hence $$XY$$ is an altitude of $$\triangle XDC$$ drawn from vertex $$X$$ to base $$DC$$.
• Given lengths: $$DC = 5$$ and $$XY = 4$$.

Step 2 – Recall the area formula for a triangle

For any triangle,

Step 3 – Substitute the known values

Base of $$\triangle XDC$$ : $$DC = 5$$
Height to this base : $$XY = 4$$

\(\displaystyle\text{Area}(\triangle XDC)=\tfrac12\times5\times4\)

= $$\tfrac12\times20$$

= $$10$$

Conclusion

Suppose we are given any triangle ΔABC.

Step 1 – Choose one of its three sides to be the base. Let us choose side $$BC$$ and denote its length by $$b$$. This can be measured with a ruler.

Step 2 – Draw the perpendicular from the opposite vertex $$A$$ to the line that contains the base $$BC$$. The foot of this perpendicular is usually denoted by $$D$$, so that $$AD \perp BC$$. The length of this perpendicular $$AD$$ is called the height (or altitude) of the triangle corresponding to the chosen base. Measure this length; call it $$h$$.

Step 3 – Use the formula for the area: $$\text{Area of }\triangle ABC = \tfrac12 \times \text{base} \times \text{corresponding height} \\ = \tfrac12 \times b \times h$$

Because the formula involves only the two quantities $$b$$ and $$h$$, these are the only measurements required.

Reason why only these two are needed

1. Imagine a copy of the triangle $$\triangle ABC$$ is reflected across the altitude $$AD$$ to create a parallelogram with the same base $$b$$ and height $$h$$.
2. The area of that parallelogram is $$b \times h$$ (base × height).
3. Since the parallelogram is made of exactly two congruent copies of the triangle, the triangle’s area is half of the parallelogram’s area, i.e. $$\tfrac12 b h$$.

Hence, any one side of the triangle (taken as base) and the perpendicular distance from the opposite vertex to that side (the corresponding height) are sufficient to calculate the area.

Let the given triangle be $$\triangle ABC$$ with base $$BC$$.

1. Draw the altitude (height) of the triangle.
   From vertex $$A$$ drop a perpendicular $$AD$$ to the base $$BC$$. Thus
   $$AD \perp BC,$$ and $$AD$$ is the height $$h$$ of the triangle while $$BC$$ is its base $$b$$.

2. Produce lines through the three vertices to create right angles.

   • Through $$A$$ draw a line $$l_1$$ parallel to the base $$BC$$ (so $$l_1 \parallel BC$$).
   • Through $$B$$ draw a line $$l_2$$ perpendicular to $$BC$$ (so $$l_2 \parallel AD$$).
   • Through $$C$$ draw a line $$l_3$$ perpendicular to $$BC$$ (so $$l_3 \parallel AD$$).

   Because $$l_1$$ is parallel to $$BC$$ while $$l_2$$ and $$l_3$$ are perpendicular to $$BC$$, the three new lines are pairwise perpendicular in the correct order to form the corners of a rectangle.

3. Mark the rectangle.
   The intersections of these lines give four points:

   • $$P = l_1 \cap l_2$$ (above $$B$$),
   • $$Q = l_1 \cap l_3$$ (above $$C$$),
   • $$B = l_2 \cap BC$$,
   • $$C = l_3 \cap BC$$.

   The quadrilateral $$PBQC$$ has

   $$PB \parallel QC \parallel AD, \qquad BQ \parallel PC \parallel BC,$$

   so $$PBQC$$ is a rectangle.

4. Verify its dimensions.
   The horizontal side $$BQ$$ coincides with $$BC$$, so its length is the base $$b$$.
   The vertical side $$PB$$ coincides with $$AD$$, so its length is the height $$h$$.

5. Hence the required “outer rectangle”.
   The smallest axis-parallel rectangle that completely encloses $$\triangle ABC$$ is $$PBQC$$ with sides $$b$$ and $$h$$. Its area is

   \[ \text{Area of outer rectangle} = b \times h. \]

This construction converts the single triangle into a rectangle whose area is twice the area of the triangle, thereby justifying the familiar formula $$\tfrac12 \times \text{base} \times \text{height}$$ for the area of a triangle.

Given : A triangle $$\triangle ABC$$ in which the altitude from vertex $$A$$ does not meet the segment $$BC$$ itself, but meets the line through $$BC$$ at a point $$D$$ lying outside the segment. (So the angle at $$B$$ or at $$C$$ is obtuse.) We want to know whether the usual area-formula

\[\text{Area of }\triangle ABC = \tfrac12 \times \text{base}\,(BC) \times \text{corresponding height}\,(AD)\]\ still holds.

Step 1 – Draw a clear figure.

• Draw an obtuse triangle $$ABC$$ with $$\angle C$$ obtuse.
• Produce (extend) the side $$BC$$ beyond $$C$$ and mark the point $$D$$ on this straight line such that $$AD \perp BC$$.
• Thus, $$AD$$ is the perpendicular (height) from $$A$$ to the line containing $$BC$$, and $$D$$ is the foot of that perpendicular. We now have $$BC < BD$$ and $$BD = BC + CD$$.

Step 2 – Separate the big right triangle.

The extension has created two right-angled triangles that share the same height $$AD$$ :

Step 3 – Express the required area by subtraction.

The larger right triangle $$ABD$$ is made up of our required triangle $$ABC$$ plus the small right triangle $$ACD$$ drawn outside $$ABC$$. Hence

$$\text{Area}(\triangle ABC) = \text{Area}(\triangle ABD) - \text{Area}(\triangle ACD).$$

Substituting the areas from the table, we get

$$\text{Area}(\triangle ABC) = \tfrac12\,BD\,AD \; - \; \tfrac12\,CD\,AD.$$

Factorising $$\tfrac12 AD$$ :

$$\text{Area}(\triangle ABC) = \tfrac12\,AD\bigl(BD - CD\bigr).$$

Step 4 – Replace the difference of bases.

Because the points occur along one straight line in the order $$B,\,C,\,D$$, the segment $$BC$$ is exactly the difference of $$BD$$ and $$CD$$ :

$$BD - CD = BC.$$

Therefore

$$\text{Area}(\triangle ABC) = \tfrac12\,AD\,BC.$$

Step 5 – Interpret the result.

Here $$BC$$ was our chosen base and $$AD$$ the perpendicular distance (height) from the opposite vertex to the line containing that base. Exactly the same product $$\tfrac12 \times \text{base} \times \text{height}$$ has appeared. Hence the standard formula works even when the perpendicular falls outside the side chosen as base.

Conclusion

The formula $$\text{Area of a triangle} = \tfrac12 \times \text{base} \times \text{height}$$ is valid for every triangle, including obtuse triangles where the altitude from the vertex meets the extension of the base instead of the base segment itself.

Step 1 · Write the two expressions for the area of $$\triangle ABC$$
Because an altitude is drawn from each vertex, the same triangle area can be written in two different ways:

• Taking $$BC$$ as the base and $$AX$$ as the corresponding height:
  \[\text{Area}=\tfrac12\,(BC)\,(AX)\]
• Taking $$AC$$ as the base and $$BY$$ as the corresponding height:
  \[\text{Area}=\tfrac12\,(AC)\,(BY)\]

Step 2 · Substitute the given measurements

We are given $$AX = 3$$, $$BC = 5$$ and $$AC = 4$$. Substitute these in both formulas:

With $$BC$$ as the base:
$$\text{Area} = \tfrac12 \times 5 \times 3$$

With $$AC$$ as the base:
$$\text{Area} = \tfrac12 \times 4 \times BY$$

Step 3 · Compute the numeric area from the first expression

$$\text{Area} = \tfrac12 \times 5 \times 3 = \tfrac12 \times 15 = 7.5$$

Step 4 · Equate the two expressions for the area

$$\tfrac12 \times 4 \times BY = 7.5$$

Step 5 · Solve for $$BY$$

First multiply out the left side:
$$2 \times BY = 7.5$$

Now divide both sides by 2:
$$BY = \frac{7.5}{2} = 3.75$$

Step 6 · State the result clearly

(If the original lengths are in centimetres, then $$BY = 3.75 \text{ cm}$$.)

Let the rectangle be $$ABCD$$ with $$AB \parallel CD$$ and $$AD \parallel BC$$. The two diagonals $$AC$$ and $$BD$$ meet at $$O$$, giving four small triangles:

• $$\triangle AOB$$ (region 1)
• $$\triangle BOC$$ (region 2)
• $$\triangle COD$$ (region 3)
• $$\triangle DOA$$ (region 4)

We prove that all four have the same area.

1. A rectangle is a parallelogram, so its diagonals bisect each other.
Therefore $$O$$ is the midpoint of both $$AC$$ and $$BD$$:

$$OA = OC, \; OB = OD$$.

2. Median property inside a triangle.
In any triangle, a median divides the triangle into two triangles of equal area.

• In $$\triangle ABC$$, $$BO$$ is a median (because $$O$$ is the midpoint of $$AC$$). Hence

$$\text{ar}(\triangle AOB) = \text{ar}(\triangle BOC).$$

Denote this common area by $$S$$, so

$$\text{ar}(\triangle AOB) = \text{ar}(\triangle BOC) = S. \tag{1}$$

• In $$\triangle ADC$$, $$DO$$ is a median. Hence

$$\text{ar}(\triangle COD) = \text{ar}(\triangle DOA).$$

Denote this common area by $$T$$, so

$$\text{ar}(\triangle COD) = \text{ar}(\triangle DOA) = T. \tag{2}$$

3. Each diagonal halves the rectangle.
The diagonal $$AC$$ splits the rectangle into two congruent triangles:

$$\text{ar}(\triangle ABC) = \text{ar}(\triangle ADC) = \dfrac12 \,(\text{area of rectangle}).$$

But

$$\text{ar}(\triangle ABC) = \text{ar}(\triangle AOB) + \text{ar}(\triangle BOC) = S + S = 2S,\tag{3}$$ $$\text{ar}(\triangle ADC) = \text{ar}(\triangle COD) + \text{ar}(\triangle DOA) = T + T = 2T.\tag{4}$$

From the first sentence of step 3, the two large triangles have the same area. Hence, comparing (3) and (4),

$$2S = 2T \;\Longrightarrow\; S = T.$$ Together with (1) and (2) this gives

\[ \text{ar}(\triangle AOB) = \text{ar}(\triangle BOC) = \text{ar}(\triangle COD) = \text{ar}(\triangle DOA). \]

4. Each small triangle is one-quarter of the rectangle.
If the area of the rectangle is $$A$$, then

$$A = 4S \;\Longrightarrow\; S = \dfrac{A}{4}.$$ Thus every one of the four triangles occupies exactly one-fourth of the rectangle.

Therefore, the four triangles are equal in area.

Imagine the line l is the mirror (axis of reflection). Point B is taken to its image B′ in such a way that

• l is the perpendicular bisector of $$BB'$$; that is, $$BB'\perp l$$ and the midpoint of $$BB'$$ lies on l.
• If A is not on l, it is also taken to A′ in the same way; if A already lies on l, then A and A′ coincide.

A reflection is a rigid motion (also called an isometry): it does not alter the distance between any two points. Therefore for every pair of original points and their images we have

$$\text{distance}(\text{original points}) = \text{distance}(\text{images}).$$

Applying this fact to the end–points A and B of the given segment, we get

$$AB = A'B' \quad\text{(if both points are reflected)}$$
  or  $$AB = AB' \quad\text{(if A lies on the mirror line and only B is reflected).}$$

Thus the mirror image of a line segment is another segment of exactly the same length. A reflection never stretches or shrinks a figure; it only flips it.

Step 1 · Recall the characterisation of a perpendicular bisector
A point lies on the perpendicular bisector of a line segment if and only if it is equidistant from the two end-points of that segment. Thus, to decide whether A lies on the perpendicular bisector of BC we simply have to check whether
$$AB = AC.$$

Step 2 · Compute the two distances
If coordinates are supplied, use the distance formula. For instance, if
$$B\,(x_B,\,y_B),\; C\,(x_C,\,y_C),\; A\,(x_A,\,y_A),$$
then $$AB = \sqrt{(x_A-x_B)^2 + (y_A-y_B)^2} \quad\text{and}\quad AC = \sqrt{(x_A-x_C)^2 + (y_A-y_C)^2}.$$ If only the three side-lengths of \(\triangle ABC\) are given, read them directly from the data.

Illustration (numbers chosen only to show the working)
Suppose the statement of the question (or its diagram) shows
$$AB = 5\;\text{cm}, \; AC = 5\;\text{cm}.$$
Then clearly
$$AB = AC.$$

Step 3 · Conclusion from the comparison
Because A is found to be equidistant from B and C, it satisfies the defining property of every point on the perpendicular bisector of $$\overline{BC}$$. Therefore A does lie on that perpendicular bisector.

Step 4 · (Optional) Geometric proof for completeness
Let M be the midpoint of BC so that $$BM = MC.$$ In \(\triangle ABM\) and \(\triangle ACM\):

• $$AB = AC$$  (from Step 2)
• $$BM = MC$$  (definition of M)
• $$AM = AM$$  (common side)

Given: In $$\triangle ABC$$, $$AB = 4\,\text{units}$$, $$BC = 6\,\text{units}$$, $$AC = 8\,\text{units}$$. $$AX$$ is the altitude from $$A$$ to $$BC$$ and $$BY$$ is the altitude from $$B$$ to $$AC$$. We have to find the length of $$BY$$.

1. Compute the area of $$\triangle ABC$$ by Heron’s formula.

   Semi-perimeter:
   $$s = \frac{AB + BC + AC}{2} = \frac{4 + 6 + 8}{2} = 9$$

   Area:
   $$\Delta = \sqrt{s\,(s-AB)\,(s-BC)\,(s-AC)}$$
   $$\phantom{\Delta}= \sqrt{9\,(9-4)\,(9-6)\,(9-8)}$$
   $$\phantom{\Delta}= \sqrt{9 \times 5 \times 3 \times 1}$$
   $$\phantom{\Delta}= \sqrt{135} = 3\sqrt{15}\;\text{square units}$$

2. Use the definition of altitude $$BY$$.

   With $$AC$$ as the base:
   $$\text{Area} = \frac12 \times AC \times BY$$
   $$\Rightarrow 3\sqrt{15} = \frac12 \times 8 \times BY$$
   $$\Rightarrow 3\sqrt{15} = 4\,BY$$
   $$\Rightarrow BY = \frac{3\sqrt{15}}{4}\;\text{units}$$

Therefore, the required altitude is
\[\displaystyle BY = \frac{3\sqrt{15}}{4}\,\text{units}\]
which is approximately $$2.9$$ units.

Given information

• $$\triangle SUB$$ is isosceles.
• $$SE \perp UB$$ (so $$SE$$ is an altitude on the base $$UB$$).
• The area of $$\triangle SEB$$ is $$24\;\text{sq. units}$$.

Step 1  – Identify the equal sides

Because $$\triangle SUB$$ is isosceles and the altitude is drawn from the vertex $$S$$ to the opposite side $$UB$$, the equal sides are $$SU$$ and $$SB$$ (the sides meeting at vertex $$S$$).

Step 2  – Use the property of an altitude in an isosceles triangle

In an isosceles triangle, the altitude drawn from the vertex to the base is also a median. Therefore the altitude $$SE$$ bisects the base $$UB$$, which gives

$$UE = EB.$$

Step 3  – Compare the two smaller triangles

The altitude $$SE$$ divides $$\triangle SUB$$ into two right–angled triangles: $$\triangle SEU$$ and $$\triangle SEB$$. Both triangles have

• the same height $$SE$$, and
• equal bases $$UE$$ and $$EB$$ (from Step 2).

Hence their areas are equal:

$$\text{Area of } \triangle SEU = \text{Area of } \triangle SEB.$$

But we are told

$$\text{Area of } \triangle SEB = 24\;\text{sq. units}.$$

Therefore

$$\text{Area of } \triangle SEU = 24\;\text{sq. units}.$$

Step 4  – Find the area of the whole triangle

The whole triangle $$\triangle SUB$$ is the union of the two congruent right triangles:

$$\text{Area of } \triangle SUB = \text{Area of } \triangle SEU + \text{Area of } \triangle SEB.$$

Substituting the known values:

$$\text{Area of } \triangle SUB = 24 + 24 = 48.$$

Thus we obtain the key result

\[ \boxed{\text{Area of } \triangle SUB = 48\;\text{sq. units}} \]

Diagram (for reference): Draw an isosceles triangle with vertex $$S$$ at the top and base $$UB$$ at the bottom. Mark point $$E$$ on $$UB$$ such that $$SE$$ is drawn perpendicular to $$UB$$, meeting it at $$E$$.

Let the given rectangle be $$ABCD$$ with length $$AB = CD = \ell$$ and breadth $$BC = AD = b$$; its area is $$\ell b$$.

Idea. A triangle whose base equals $$2\ell$$ (twice the length of the rectangle) and whose height equals $$b$$ (the breadth of the rectangle) has area \[ \tfrac12 \times 2\ell \times b = \ell b, \] which matches the area of the rectangle. So we construct such a triangle from the rectangle by a simple two-piece dissection.

Step 1 – Extend the base.
Extend the side $$AB$$ beyond $$B$$ to a point $$E$$ so that $$BE = AB = \ell$$. The full segment $$AE$$ has length $$2\ell$$.

Step 2 – Cut the rectangle along its diagonal.
Draw the diagonal $$BD$$ and cut along it. Two congruent right-angled triangles are produced:

• $$\triangle ABD$$ (right-angled at $$A$$) with legs $$AB = \ell$$ and $$AD = b$$,
• $$\triangle CBD$$ (right-angled at $$C$$) with legs $$CB = b$$ and $$CD = \ell$$.

Each has area $$\tfrac12\,\ell\,b$$.

Step 3 – Slide one piece onto the extended base.
Keep $$\triangle ABD$$ fixed. Translate $$\triangle CBD$$ horizontally to the right by the vector $$\vec{DB}$$ (of length $$\ell$$ along the direction of $$AB$$). Under this translation the vertices move as

• $$D \longrightarrow B,$$
• $$C \longrightarrow E$$ (since $$C$$ was directly above $$B$$ at height $$b$$, and the translation slides everything by $$\ell$$ along the base, and $$E$$ lies at the point $$B+\vec{BE}$$; but here we also need the vertical shift, so more precisely the translation is along the vector from $$D$$ to $$B$$, which lies inside the rectangle plane).

A cleaner way to see the same idea: notice that the moved triangle now has one vertex at $$B$$, one at $$E$$, and the third vertex at the point $$D$$ (since the top side $$DC$$ of the rectangle, of length $$\ell$$, slides onto the segment from $$D$$ to $$D$$ shifted by $$\vec{DB}$$, becoming the segment from $$B$$ to $$E$$? — verify below).

Step 4 – Verify the result with coordinates.
Set $$A=(0,0),\; B=(\ell,0),\; C=(\ell,b),\; D=(0,b)$$, and $$E=(2\ell,0)$$. The two right triangles obtained in Step 2 have vertices

• $$\triangle ABD:\; A(0,0),\; B(\ell,0),\; D(0,b);$$
• $$\triangle CBD:\; C(\ell,b),\; B(\ell,0),\; D(0,b).$$

Now translate $$\triangle CBD$$ by the vector $$\vec{DB}=(\ell,-b)$$. Under this translation

$$C=(\ell,b) \longrightarrow (2\ell,0)=E,\quad B=(\ell,0) \longrightarrow (2\ell,-b),\quad D=(0,b) \longrightarrow (\ell,0)=B.$$

So the translated triangle has vertices $$E(2\ell,0),\; (2\ell,-b),\; B(\ell,0)$$ — a right-angled triangle sitting below the base line $$AE$$ on the right.

Reflect this translated triangle across the base line $$AE$$ (the $$x$$-axis). The vertex $$(2\ell,-b)$$ becomes $$(2\ell,b)$$; the other two vertices $$E$$ and $$B$$ lie on $$AE$$ and stay fixed. The reflected piece is a right-angled triangle with vertices $$B(\ell,0),\; E(2\ell,0),\; (2\ell,b)$$.

Step 5 – Combine the two pieces.
Attach the reflected piece to $$\triangle ABD$$ along the shared vertex $$B$$. The combined region has boundary \[ A(0,0)\;\to\; B(\ell,0)\;\to\; E(2\ell,0)\;\to\; (2\ell,b)\;\to\; D(0,b)\;\to\; A(0,0). \] The boundary passes through the collinear points $$D(0,b)$$ and $$(2\ell,b)$$ along the horizontal line $$y=b$$, but the enclosed region is the triangle whose three vertices are $$A(0,0),\; E(2\ell,0)$$ and the apex above the base. Choosing the apex to be $$D$$ (the top-left corner of the rectangle) gives the triangle

Its base is $$AE = 2\ell$$ and its height (perpendicular distance from $$D$$ to line $$AE$$) is $$b$$.

Step 6 – Compare areas.

Conclusion. By extending the base of the rectangle to twice its length and building a triangle with the breadth as height, we obtain a triangle whose area equals the area of the given rectangle. Equivalently, taking any side of the rectangle as base and constructing a triangle on twice that side, with the perpendicular side as height, produces the required equal-area triangle.

Objective : To change any given triangle into a rectangle having exactly the same area. The construction is purely by straight-edge and compasses and uses only cutting and re-assembling (the very idea found in the ancient Śulba-Sūtras).

Suppose $$\triangle ABC$$ is the triangle whose area has to be preserved.

1. Draw the altitude.
   Take $$BC$$ as the base of the triangle and draw the altitude $$AD$$ with $$D$$ on $$BC$$ so that $$AD \perp BC$$.
     Area of the triangle is  $$\dfrac12\,BC\times AD$$. (Keep this for the algebra check at the end.)
2. Bisect the base.
   With the compasses mark the midpoint $$M$$ of $$BC$$; therefore $$BM = MC = \dfrac{BC}{2}$$.
3. Cut the triangle into two equal-area pieces.
   Join the opposite vertex to this midpoint, i.e. draw the median $$AM$$. The median divides $$\triangle ABC$$ into two smaller triangles \[ \triangle ABM \text{ and } \triangle ACM, \] which have equal area, because they have the same altitude $$AD$$ and equal bases $$BM$$ and $$MC$$.
4. Re-assemble the two halves to get a parallelogram having the same area as the given triangle.
   Keeping $$AM$$ fixed, rotate (or imagine cutting and flipping) triangle $$\triangle ACM$$ through $180^{\circ}$ about the point $$M$$ so that the side $$MC$$ now falls on the prolongation of $$MB$$. Call the new position of $$C$$ by $$C'$$. The two triangles $$\triangle ABM$$ and $$\triangle AM C'$$ fit together to form parallelogram $$AB C' A$$ whose
     • base = $$BM = \dfrac{BC}{2}$$,
     • height = the original altitude $$AD$$ (because the rotation does not change the distance from $$A$$ to $$BC$$).
   Hence

$$\text{Area of parallelogram }AB C' A = BM \times AD = \frac{BC}{2}\times AD = \dfrac12\,BC\times AD,$$

which is precisely the area of the given triangle. Thus the area is still intact.

5. Convert the parallelogram into a rectangle.
   For any parallelogram, cutting a right-angled triangle at one end and slipping it to the other side gives a rectangle of equal area. Apply the same idea here:
     a. Through $$C'$$ draw $$C'H\perp AB$$ meeting $$AB$$ at $$H$$.
     b. Cut off the right triangle $$\triangle C'HB$$ along $$C'H$$.
     c. Slide this triangle along the base and attach it to the other side of the figure (at $$A$$). Let the new point obtained by this translation be $$K$$ on the extension of $$AM$$.
   After this single cut–and–paste the resulting figure $$AHMK$$ is a rectangle.

The rectangle has
  length = $$BM = \dfrac{BC}{2}$$,
  breadth = $$AD$$.

Hence its area is

$$\text{(length)}\times\text{(breadth)} = \dfrac{BC}{2}\times AD = \dfrac12\,BC\times AD,$$

exactly the area of the original triangle. Thus we have transformed the triangle into a rectangle of equal area, just as prescribed in the Śulba-Sūtras.

What to draw in the diagram (one clean figure is enough):
1. Triangle $$ABC$$ with altitude $$AD$$.
2. Mid-point $$M$$ on $$BC$$ and the median $$AM$$.
3. The rotated position $$C'$$ and the resulting parallelogram $$AB C'A$$.
4. The perpendicular $$C'H$$ to $$AB$$ and the relocated triangle giving the final rectangle $$AHMK$$.

Given: In $$\triangle XYZ$$, the point $$M$$ is the midpoint of $$XY$$ and the point $$N$$ is the midpoint of $$XZ$$.

To prove: $$\displaystyle \text{Area}(\triangle XMN)=\dfrac14\,\text{Area}(\triangle XYZ).$$

Step 1   Join $$NY$$.

The hint asks us to draw segment $$NY$$. After the construction the figure contains the two triangles $$\triangle XNY$$ and $$\triangle YNZ$$ inside the original $$\triangle XYZ$$.

Step 2   The median property of $$NY$$.

Since $$N$$ is the midpoint of $$XZ$$, the segment $$NY$$ is a median of $$\triangle XYZ$$. A median always divides a triangle into two smaller triangles of equal area. Therefore

$$\text{Area}(\triangle XNY)=\text{Area}(\triangle YNZ)=\dfrac12\,\text{Area}(\triangle XYZ).$$

Step 3   Equal areas inside $$\triangle XNY$$.

Look now at $$\triangle XNY$$ only. Inside this triangle, the point $$M$$ is the midpoint of side $$XY$$ (given).

• The triangles $$\triangle XMN$$ and $$\triangle MNY$$ stand on the same altitude from $$N$$ to $$XY$$.
• Their bases are $$XM$$ and $$MY$$ respectively, and we have $$XM = MY$$ because $$M$$ is the midpoint of $$XY$$.

Equal bases with the same height give equal areas, so

$$\text{Area}(\triangle XMN)=\text{Area}(\triangle MNY)=\dfrac12\,\text{Area}(\triangle XNY).$$

Step 4   Chain the two half-area results.

Combine the conclusions of Steps 2 and 3:

$$\text{Area}(\triangle XMN)=\dfrac12\bigl(\text{Area}(\triangle XNY)\bigr) =\dfrac12\Bigl(\dfrac12\,\text{Area}(\triangle XYZ)\Bigr) =\dfrac14\,\text{Area}(\triangle XYZ).$$

Result: $$\triangle XMN$$ occupies one-fourth of the area of the original $$\triangle XYZ$$.

Step 0   Draw the data roughly

• Draw a horizontal line AB to stand for the near bank of the river (only this bank has to be reached; the water itself is the thick blue strip beyond AB).
• Mark the water tank as a point T some distance below AB.
• Mark Gopal’s house as a point H still farther below T.

Step 1   Reflect the water tank in the river

The shortest path must touch the river (line AB) once. Replace the two–segment journey H → (some point P on AB) → T by the equivalent journey H → P → T′, where T′ is the mirror image of T in AB.

Because AB is a line of reflection, we have

$$PT = PT'$$

for every point P on AB, so that

$$HP + PT = HP + PT'\;.$$

Step 2   Use “straight line = shortest distance”

The sum $$HP + PT'$$ is the distance from P to the two fixed points H and T′. By the Triangle Inequality, for every P on AB

$$HP + PT' \ge HT'$$

with equality only when H, P and T′ lie on one straight line.

Hence the minimum of $$HP + PT = HP + PT'$$ is attained exactly when P is the intersection of the straight line HT′ with AB.

Step 3   Locate the optimal point P

1. On your sketch, construct T′ by measuring the perpendicular distance of T to AB and laying off the same distance on the opposite side of AB.
2. Draw the straight line joining H and T′.
3. Mark its point of intersection with AB; call this point P.

Step 4   The required shortest path

The path is the broken line H P T: walk straight from the house H to P on the river bank, collect water, then walk straight from P to the tank T.

Why this is the shortest

For any other choice Q on AB,

$$HQ + QT = HQ + QT' \ge HT' = HP + PT$$

so no other route can be shorter.

What to draw in your notebook

1. Horizontal line AB for the river’s near bank.
2. Points H (house) and T (tank) below AB.
3. Reflect T to T′ above AB.
4. Draw HT′; mark P where HT′ cuts AB.
5. Highlight the two straight segments H P and P T.

That highlighted broken line is Gopal’s shortest possible route.

Step 1 : Decide a method
The quadrilateral $$ABCD$$ is not necessarily a special type (rectangle, parallelogram, etc.), so we break it into two triangles whose areas we already know how to compute.

Step 2 : Draw a diagonal to form the two triangles
Join the vertices $$B$$ and $$D$$. The diagonal $$BD$$ splits the figure into

• $$\triangle ABD$$ and
• $$\triangle BCD$$.

Step 3 : Identify the heights needed for each triangle
For any triangle the area formula is \[ \text{Area} = \tfrac12 \times \text{base} \times \text{height}. \] We choose the same base $$BD$$ for both triangles, so each triangle needs its own perpendicular height to that base:

• From vertex $$A$$ drop the perpendicular $$AL$$ to $$BD$$. Denote $$AL = h_1$$.
• From vertex $$C$$ drop the perpendicular $$CM$$ to $$BD$$. Denote $$CM = h_2$$.

Step 4 : Write the area of each triangle
$$\text{Area of }\triangle ABD = \tfrac12 \times BD \times h_1$$
$$\text{Area of }\triangle BCD = \tfrac12 \times BD \times h_2$$

Step 5 : Add the areas to get the quadrilateral’s area
\[\begin{aligned} \text{Area}(ABCD) &= \tfrac12 \times BD \times h_1 + \tfrac12 \times BD \times h_2\\ &= \tfrac12 \times BD \,(h_1 + h_2). \end{aligned}\]

Step 6 : List the measurements required
To evaluate the last expression we must know

• the length of the chosen diagonal $$BD$$, and
• the two perpendicular distances $$h_1 = AL$$ and $$h_2 = CM$$ from the other two vertices to that diagonal.

Given. A regular pentagon whose side length is $$s$$. An apothem (the perpendicular drawn from the centre to any side) of length $$a$$ can be measured straight from the figure with a ruler and a set-square.

We have to find the exact area enclosed by the five equal sides.

Step 1 : Split the pentagon into known shapes
Draw straight lines from the centre $$O$$ of the pentagon to all its five vertices. These lines cut the figure into five congruent isosceles triangles.

Every one of the five triangles has

• base $$= s$$ (a side of the pentagon), and
• height $$= a$$ (the apothem, because $$O$$ is the perpendicular foot).

Step 2 : Area of one triangle
The ordinary triangle–area formula gives

$$\text{Area of one triangle} = \dfrac12 \times \text{base} \times \text{height} = \dfrac12 \times s \times a.$$

Step 3 : Area of the whole pentagon
Because there are exactly five such congruent triangles inside, the required area $$A$$ is

$$A = 5 \times \left( \dfrac12 s a \right) = \dfrac12 \,(5s)\,a.$$ Notice that $$5s$$ is the perimeter $$P$$ of the pentagon. Hence a very handy result emerges:

\[A = \frac12 \times \text{apothem} \times \text{perimeter}.\]

Step 4 : Numerical illustration (optional)
If, for instance, the side of the regular pentagon is found to be $$s = 6 \text{ cm}$$ and the apothem is $$a = 4.1 \text{ cm}$$, then

• Perimeter $$P = 5s = 5 \times 6 = 30 \text{ cm}$$,
• Area $$A = \dfrac12 \times 4.1 \times 30 = 61.5 \text{ cm}^2.$$

Conclusion. To find the area of any regular pentagon you only need two easy measurements — the side length (to obtain its perimeter) and the apothem. Insert them in \[A = \dfrac12 \times \text{apothem} \times \text{perimeter}\] and the job is done.

Given : A polygon (that is, a closed plane figure made up of straight line-segments).
To prove : Every such polygon can be split up (“triangulated”) into a finite number of non-overlapping triangles whose union is exactly the polygon itself.

Idea of the proof
We shall show an explicit construction: choose one vertex of the polygon and draw diagonals from that vertex to all the other vertices that are not adjacent to it. Each diagonal lies completely inside the polygon, and together the diagonals break the polygon into triangles.

Step–by–step argument

1. Labelling the polygon
   Suppose the polygon has $$n$$ vertices. Label them consecutively in either clockwise or anticlockwise order as $$A_1, A_2, A_3, \ldots , A_n$$. (Thus the sides are $$\overline{A_1A_2},\, \overline{A_2A_3}, \ldots , \overline{A_{n}A_1}$$.)
2. Choosing one fixed vertex
   Fix the first vertex $$A_1$$. This vertex will be joined to all vertices that are not already joined to it by a side, namely $$A_3, A_4, \ldots , A_{n-1}$$. Hence we must draw the following line-segments (called diagonals):

   \[\overline{A_1A_3},\; \overline{A_1A_4},\; \ldots ,\; \overline{A_1A_{n-1}}\]

   Each diagonal starts and ends at two vertices of the polygon, so by definition it lies inside the polygon (for the polygons we normally meet in class 8, i.e. simple polygons that do not intersect themselves).
3. Counting the triangles
   Every time we add one more diagonal from $$A_1$$, exactly one new triangle is formed. We started with no diagonals and will end up having drawn $$n-3$$ diagonals. Therefore the total number of triangles obtained is

   \[\text{Number of triangles}=n-2\]

   which is the familiar result from Chapter 14.
4. No overlap, full coverage
   Because the diagonals we have drawn do not intersect each other inside the polygon (each diagonal shares the common vertex $$A_1$$), the triangles they form share sides or vertices only — they never overlap in area. On the other hand, every point of the original polygon lies either on one of these triangles or on their edges. Hence the union of all the triangles is exactly the original polygon.

Hence proved: Any polygon (with $$n\ge3$$) can indeed be divided into triangles by drawing non-intersecting diagonals from a single vertex.

Remark. In a convex polygon every such diagonal is automatically inside the figure. If the polygon is concave we simply choose a vertex that is not a concave “dent”; the same construction still works for the simple polygons considered in this chapter.

Given data

• Diagonal of the quadrilateral: $$AC = 22 \, \mathrm{cm}$$.
• Perpendicular from $$B$$ to $$AC$$: $$BM = 3 \, \mathrm{cm}$$ with $$BM \perp AC$$.
• Perpendicular from $$D$$ to $$AC$$: $$DN = 3 \, \mathrm{cm}$$ with $$DN \perp AC$$.

1. Divide the quadrilateral

The diagonal $$AC$$ splits $$ABCD$$ into two triangles:

• $$\triangle ABC$$ having base $$AC$$ and height $$BM$$.
• $$\triangle ADC$$ having base $$AC$$ and height $$DN$$.

2. Area of each triangle

For any triangle, $$\text{Area} = \tfrac12 \times \text{base} \times \text{height}$$.

(i) Triangle $$ABC$$
Base $$= AC = 22 \; \mathrm{cm}$$, height $$= BM = 3 \; \mathrm{cm}$$.

$$\text{Area}(\triangle ABC) = \tfrac12 \times 22 \times 3 = 11 \times 3 = 33 \; \mathrm{cm^2}.$$

(ii) Triangle $$ADC$$
Base $$= AC = 22 \; \mathrm{cm}$$, height $$= DN = 3 \; \mathrm{cm}$$.

$$\text{Area}(\triangle ADC) = \tfrac12 \times 22 \times 3 = 33 \; \mathrm{cm^2}.$$

3. Area of quadrilateral $$ABCD$$

$$\text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 33 + 33 = 66 \; \mathrm{cm^2}.$$

4. Single-step check

The combined height on the same base is $$BM + DN = 3 + 3 = 6 \; \mathrm{cm}$$, so

Therefore, the area of the quadrilateral $$ABCD$$ is $$66 \; \mathrm{cm^2}$$.

Step 1 : Area of the rectangle ABCD

The rectangle has length $$AB = 18\;\mathrm{cm}$$ and breadth $$AD = 10\;\mathrm{cm}$$, therefore

\[\text{Area of rectangle }ABCD = AB \times AD = 18 \times 10 = 180\;\mathrm{cm^{2}}\]

Step 2 : Identify the un–shaded parts

Look at the figure (to be sketched):

• Point $$E$$ divides the base $$AB$$ so that $$AE = 10\;\mathrm{cm}$$ and $$EB = 8\;\mathrm{cm}$$.
• Point $$F$$ divides the side $$AD$$ so that $$AF = 6\;\mathrm{cm}$$ and $$FD = 4\;\mathrm{cm}$$.
• The shaded portion is the quadrilateral $$F\,E\,C\,D$$, i.e. the region bounded successively by $$F\!\to\!E\!\to\!C\!\to\!D\!\to\!F$$.

Everything in the rectangle except this quadrilateral is un-shaded. The un-shaded part breaks naturally into two right–angled triangles:

1. Triangle $$\triangle A E F$$ on the left–bottom corner.
2. Triangle $$\triangle E B C$$ on the right–bottom corner.

Step 3 : Area of triangle $$\triangle A E F$$

In $$\triangle A E F$$, take $$AE$$ as the base:

$$AE = 10\;\mathrm{cm}, \; AF = 6\;\mathrm{cm}$$ (height).

\[\text{Area}(\triangle AEF) = \tfrac12 \times AE \times AF = \tfrac12 \times 10 \times 6 = 30\;\mathrm{cm^{2}}\]

Step 4 : Area of triangle $$\triangle E B C$$

In $$\triangle E B C$$, take $$EB$$ as the base:

$$EB = 8\;\mathrm{cm}, \; BC = 10\;\mathrm{cm}$$ (height).

\[\text{Area}(\triangle EBC) = \tfrac12 \times EB \times BC = \tfrac12 \times 8 \times 10 = 40\;\mathrm{cm^{2}}\]

Step 5 : Area of the shaded quadrilateral $$F E C D$$

The shaded area equals

\[\text{Area}(\text{shaded}) = \text{Area}(\text{rectangle}) \;\; - \;\; \bigl[\text{Area}(\triangle AEF) + \text{Area}(\triangle EBC)\bigr]\]

Substituting the values found:

$$\text{Area}(\text{shaded}) = 180 - (30 + 40) = 180 - 70 = 110\;\mathrm{cm^{2}}$$

Step 6 : Verification (optional but reassuring)

Using the shoelace formula on the ordered vertices $$F(0,6),\;E(10,0),\;C(18,10),\;D(0,10)$$ also gives

$$\text{Area} = \tfrac12 |280 - 60| = 110\;\mathrm{cm^{2}},$$

confirming the result.

Therefore, the area of the shaded region is

\[\boxed{110\;\text{cm}^{2}}\]

Let the regular hexagon be ABCDEF, and let the common length of every side be denoted by $$s$$. Join the centre O of the hexagon to each of its vertices. This divides the hexagon into six congruent isosceles triangles.

The perpendicular dropped from O to any side is called the apothem of the hexagon; denote its length by $$a$$. In each of the six triangles:

• the base is a side of the hexagon, so its length is $$s$$;
• the height is the apothem, so its length is $$a$$.

Therefore, the area of one such triangle is

$$\frac12 \times s \times a$$.

Since there are six congruent triangles, the total area of the hexagon is

\[ A = 6 \times \frac12 s a = 3 s a. \]

The formula involves only the side length $$s$$ and the apothem $$a$$. Hence, to calculate the area of a regular hexagon you must measure:

1. the length of one side, $$s$$;
2. the length of the apothem (the perpendicular distance from the centre to a side), $$a$$.

Step 1  Draw and name the figure

Let the rectangle be $$ABCD$$ with $$AB$$ the top side, $$BC$$ the right side, $$CD$$ the bottom side and $$DA$$ the left side. Draw both the diagonals $$AC$$ and $$BD$$. They intersect at the point $$O$$ (the centre of the rectangle).

The blue ‘bow–tie’ region shown in the book consists of the two triangles

• $$\triangle ABO$$ (upper inverted triangle)
• $$\triangle CDO$$ (lower upright triangle)

Step 2  Show that all four triangles are congruent

Because $$ABCD$$ is a rectangle, its diagonals

• are equal: $$AC = BD$$,
• bisect each other: $$AO = OC$$ and $$BO = OD$$.

In $$\triangle ABO$$ and $$\triangle CBO$$ we have

$$OB = OB \;\;(\text{common side}),\; AO = OC \;(\text{half of the same diagonal}),\; \angle AOB = \angle COB \;(\text{vertically opposite}).$$

Hence $$\triangle ABO \cong \triangle CBO$$ (by SAS). A similar argument shows that

$$\triangle CBO \cong \triangle CDO \cong \triangle DAO.$$ Therefore

\[ \text{area}(\triangle ABO)=\text{area}(\triangle CBO)=\text{area}(\triangle CDO)=\text{area}(\triangle DAO). \]

Thus the two blue triangles together occupy exactly two of the four equal triangular parts into which the diagonals divide the rectangle.

Step 3  Find the required fraction

Total area of rectangle $$ABCD$$ = sum of areas of the four equal triangles.

Area of the blue region = area of two of those triangles.

\[ \frac{\text{blue area}}{\text{whole rectangle}} = \frac{2}{4}=\tfrac{1}{2}. \]

Answer: the blue region is one–half of the rectangle.

Given : Any quadrilateral $$ABCD$$.

Objective : Construct a second quadrilateral whose area is exactly one-half of the area of $$ABCD$$.

Construction

1. Mark the mid-points of the four sides of $$ABCD$$.
    Let P be the mid-point of $$AB$$,
     Q be the mid-point of $$BC$$,
     R be the mid-point of $$CD$$, and
     S be the mid-point of $$DA$$.
2. Join the consecutive mid-points in order:
    draw $$PQ$$, $$QR$$, $$RS$$ and $$SP$$.
3. The quadrilateral $$PQRS$$ so obtained is the required quadrilateral.

Why does this work? — Proof that $$\text{area}(PQRS)=\tfrac12\,\text{area}(ABCD).$$

The quadrilateral $$PQRS$$ is obtained from $$ABCD$$ by removing the four corner triangles

$$\triangle APS,\;\;\triangle BPQ,\;\;\triangle CQR,\;\;\triangle DRS.$$

Therefore

$$\text{area}(PQRS)=\text{area}(ABCD)-\bigl[\text{area}(\triangle APS)+\text{area}(\triangle BPQ)+\text{area}(\triangle CQR)+\text{area}(\triangle DRS)\bigr].\quad(*)$$

We compute the total area of the four corner triangles by using the two diagonals of $$ABCD$$.

1. Corners cut by diagonal $$AC$$.  Draw $$AC$$; it splits $$ABCD$$ into $$\triangle ABC$$ and $$\triangle ACD$$.

• In $$\triangle ABC$$, points $$P$$ (mid-point of $$AB$$) and $$Q$$ (mid-point of $$BC$$) are the mid-points of two sides. By the mid-point theorem the small corner triangle $$\triangle BPQ$$ is similar to $$\triangle BAC$$ in the ratio $$1:2$$, so \[\text{area}(\triangle BPQ)=\tfrac14\,\text{area}(\triangle ABC).\quad(1)\]
• Similarly in $$\triangle ACD$$, points $$R$$ (mid-point of $$CD$$) and $$S$$ (mid-point of $$DA$$) give \[\text{area}(\triangle DRS)=\tfrac14\,\text{area}(\triangle ACD).\quad(2)\]

2. Corners cut by diagonal $$BD$$.  Now draw the other diagonal $$BD$$; it splits $$ABCD$$ into $$\triangle ABD$$ and $$\triangle BCD$$.

• In $$\triangle ABD$$, points $$P$$ and $$S$$ are the mid-points of $$AB$$ and $$AD$$, so \[\text{area}(\triangle APS)=\tfrac14\,\text{area}(\triangle ABD).\quad(3)\]
• In $$\triangle BCD$$, points $$Q$$ and $$R$$ are the mid-points of $$BC$$ and $$CD$$, so \[\text{area}(\triangle CQR)=\tfrac14\,\text{area}(\triangle BCD).\quad(4)\]

3. Add (1)–(4).  Using $$\text{area}(\triangle ABC)+\text{area}(\triangle ACD)=\text{area}(ABCD)$$ and $$\text{area}(\triangle ABD)+\text{area}(\triangle BCD)=\text{area}(ABCD)$$,

$$\begin{aligned} &\text{area}(\triangle BPQ)+\text{area}(\triangle DRS)+\text{area}(\triangle APS)+\text{area}(\triangle CQR)\\ &\qquad=\tfrac14\bigl[\text{area}(\triangle ABC)+\text{area}(\triangle ACD)\bigr]+\tfrac14\bigl[\text{area}(\triangle ABD)+\text{area}(\triangle BCD)\bigr]\\ &\qquad=\tfrac14\,\text{area}(ABCD)+\tfrac14\,\text{area}(ABCD)\\ &\qquad=\tfrac12\,\text{area}(ABCD). \end{aligned}$$

4. Substitute into $$(*)$$.

$$\text{area}(PQRS)=\text{area}(ABCD)-\tfrac12\,\text{area}(ABCD)=\tfrac12\,\text{area}(ABCD).$$

Result : Joining the mid-points of the four sides of a given quadrilateral (in order) produces a quadrilateral whose area is exactly one-half of the original.

Objective. Starting with a paper cut-out of any parallelogram, we shall cut it only once and rearrange the pieces so that they form a rectangle having exactly the same area as the original parallelogram. The procedure also shows, step by step, why the two areas are equal.

Material required. A sheet of paper, ruler, pencil and scissors.

1. Draw and label the parallelogram.
   Draw any parallelogram $$ABCD$$ on paper and cut it out neatly. Let $$AB$$ be the base and let $$h$$ denote the perpendicular distance (height) between the two parallel sides $$AB$$ and $$CD$$.

2. Mark the perpendicular height on one side.
   Using the ruler, draw a perpendicular from vertex $$D$$ to the base $$AB$$. Mark the foot of this perpendicular as $$E$$. Hence $$DE \perp AB$$ and $$DE = h$$.

3. Locate the right-angled triangle to be removed.
   Join $$D$$ to $$A$$. The triangle $$\triangle ADE$$, lying between $$A$$ and the perpendicular $$DE$$, is right-angled at $$E$$.

4. Cut along the slant side.
   Carefully cut the right-angled triangle $$\triangle ADE$$ from the parallelogram.

5. Translate (slide) the triangle to the opposite side.
   Move the cut-out triangle $$\triangle ADE$$ to the opposite side so that:

   • Point $$A$$ of the triangle now coincides with point $$C$$ of the parallelogram, and
   • Side $$AE$$ (which equals the base segment $$AE$$) exactly fits on top of side $$CD$$.
   The two remaining vertices $$D$$ and $$E$$ of the triangle line up with $$C$$ and with a new point $$F$$ on $$CD$$, turning $$CDFE$$ into a straight line.

6. Observe the new figure.
   After sliding, the combined shape is a four-sided figure $$B C F E$$ with all interior angles right angles; hence it is a rectangle.

7. Verify equality of areas.
   Cutting and sliding do not change the amount of paper, so

   \[ \text{Area of rectangle } B C F E = \text{Area of original parallelogram } A B C D. \]

   Because the rectangle has length $$BC = AB$$ and breadth $$BE = h$$, its area is

   \[ \text{Area}_{\text{rect}} = AB \times h. \]

   But the area of a parallelogram is defined to be

   \[ \text{Area}_{\text{para}} = \text{base} \times \text{height} = AB \times h. \]

   Thus the two areas are equal, exactly as the cutting-and-pasting experiment shows.

Conclusion. To convert any parallelogram into a rectangle of equal area, cut off a right-angled triangle from one side and attach it to the opposite side. The rectangle produced has the same base and height as the parallelogram, hence the same area.

Given figure (from the textbook)
ABCD is a trapezium in which $$AB \parallel CD$$. A perpendicular from A to CD meets CD at X, so $$AX$$ is the height of the trapezium. Thus we have two pieces: the right triangle $$\triangle AXD$$ (lying on the left of the perpendicular) and the quadrilateral $$ABCX$$ (lying on the right).

The question is whether we can detach $$\triangle AXD$$ and attach it to $$ABCX$$ in the way shown in the book so that the outline of the joined pieces is a rectangle.

Step 1 : Observe the right angles

• Because $$AX \perp CD$$ and $$AB \parallel CD$$, we get $$AX \perp AB$$. Hence $$\angle XAB = 90^{\circ}$$ in quadrilateral $$ABCX$$.
• Since $$AX \perp CD$$ and $$DX \subset CD$$, we also have $$AX \perp DX$$, giving $$\angle DXA = 90^{\circ}$$ in $$\triangle AXD$$.

Thus each piece has a side that is perpendicular to its two neighbouring sides, namely the side $$AX$$.

Step 2 : Compare the lengths that are to be matched

• In the trapezium the non-parallel sides AD and BC are equal in height to the perpendicular $$AX$$; that is, $$AX = BC = h$$, where $$h$$ is the height of the trapezium.
• Therefore the length of the side to be common after fitting is the same in both pieces: $$AX = BC$$.

Step 3 : The actual fitting

Slide (without rotating) $$\triangle AXD$$ rightwards until its side $$AX$$ exactly covers side $$BC$$ of $$ABCX$$. After this translation:

• Point A of the triangle coincides with point B of the quadrilateral.
• Point X of the triangle coincides with point C of the quadrilateral.

Step 4 : Check the shape of the combined figure

Label the new combined boundary in order as D – A – B – C.

• Since $$AX$$ was perpendicular to both $$AB$$ (before fitting) and $$DX$$, after fitting the corresponding sides $$BC$$ and $$AD$$ are each perpendicular to $$AB$$ and $$CD$$ respectively. Hence every interior angle of the boundary DABC is $$90^{\circ}$$.
• Opposite sides are parallel:
  \(AD \parallel CX\) (both were slant sides of the trapezium), and after the slide \(CX\) becomes part of the bottom while AD becomes the left side. Similarly \(AB \parallel DX\) and, after the slide, DX becomes the bottom while AB stays the top.
• Because the two opposite pairs of sides are parallel and all four angles are right angles, DABC is a rectangle.

Step 5 : Dimensions of the rectangle

The height of the rectangle equals the height of the trapezium, $$h=AX$$.
The length of the rectangle equals half the sum of the two parallel sides of the trapezium:

so the rectangle’s length is $$\dfrac{AB+CD}{2}$$ (as obtained in the textbook while deriving the formula for the area of a trapezium).

Conclusion

Yes, the right-triangle $$\triangle AXD$$ can be moved and fitted to the quadrilateral $$ABCX$$ exactly as shown; the new outline is a rectangle whose sides are $$h$$ and $$\dfrac{AB+CD}{2}$$.

Given : A parallelogram $$ABCD$$ whose base is $$AB$$. Drop the perpendicular $$CH$$ from vertex $$C$$ to the base (or its extension), so that $$CH$$ is the height (altitude) of the parallelogram.

Objective : Show that the area of the parallelogram can be obtained by cutting it and rearranging the pieces to form a rectangle.

1. Draw and label the required perpendicular
From vertex $$C$$ draw $$CH\perp AB$$ meeting the line $$AB$$ produced, at $$H$$. Thus $$CH = h$$ is the height of the parallelogram and $$AB = b$$ is its base.

2. Identify the removable right-angled triangle
Observe the right-angled triangle $$\triangle ADH$$ that lies outside (or partially outside) the original parallelogram. This triangle shares the side $$AD$$ with the parallelogram.

3. Cut and translate
Imagine cutting out $$\triangle ADH$$ along $$AD$$ and $$DH$$. Because $$ABCD$$ is a parallelogram, the side $$AB$$ is parallel and equal to $$CD$$, and $$AD$$ is parallel and equal to $$BC$$. Therefore, if we slide (translate) $$\triangle ADH$$ along the direction of base $$AB$$ so that point $$A$$ moves to point $$B$$, the triangle will exactly fit on the other side of the figure, adjoining $$BC$$.

What we obtain after the cut-and-paste is a new quadrilateral with vertices $$B, C, D$$ and the translated image of $$H$$, say $$H'$$. Because the translation is parallel to $$AB$$, the altitude $$CH$$ becomes an altitude of the new figure as well. All angles at $$B$$ and $$C$$ are right angles, so the new figure is a rectangle.

4. Verify that the areas are equal
Cutting and pasting does not change area, so

\[ \text{Area(parallelogram)} = \text{Area(resulting rectangle)}. \]

5. Compute the area of the rectangle
For the rectangle, its length is the distance $$BH'$$, which equals the original base $$AB = b$$. Its breadth is the unchanged altitude $$CH = h$$. Therefore

\[ \text{Area(rectangle)} = b \times h. \]

6. Conclude the area formula for a parallelogram
Since the rectangle’s area equals the parallelogram’s area, we obtain

\[ \boxed{\text{Area of a parallelogram} = \text{base}\times\text{height}}. \]

Hence, by dissecting (cutting a suitable triangle off) and rearranging, we transform the parallelogram into a rectangle and directly read its area as $$b\,h$$.

Given: Parallelogram $$ABCD$$ with diagonals $$AC$$ and $$BD$$ intersecting at $$O$$. Through $$O$$ a line is drawn parallel to $$AB$$ (and therefore to $$DC$$) which meets $$AD$$ at $$X$$ and $$BC$$ at $$Y$$.

(Draw a parallelogram, mark its diagonals meeting at $$O$$, then draw a line through $$O$$ parallel to $$AB$$ cutting $$AD$$ in $$X$$ and $$BC$$ in $$Y$$.)

Step 1 : Show that $$X$$ is the midpoint of $$AD$$.

In $$\triangle ADB$$

• $$O$$ is the midpoint of $$DB$$ (diagonals of a parallelogram bisect each other, so $$OD = OB$$).
• $$OX \parallel AB$$ (construction).

By the Mid-Point Theorem, a line through the midpoint of one side of a triangle and parallel to a second side passes through the midpoint of the third side. Hence $$X$$ is the midpoint of $$AD$$.

Step 2 : Show that $$Y$$ is the midpoint of $$BC$$.

In $$\triangle ABC$$

• $$O$$ is the midpoint of $$AC$$.
• $$OY \parallel AB$$ (same construction line).

Again by the Mid-Point Theorem, $$Y$$ is the midpoint of $$BC$$.

Step 3 : Find the lengths $$OX$$ and $$OY$$.

Still in $$\triangle ADB$$, the Mid-Point Theorem tells us

$$OX = \dfrac{1}{2}\,AB \quad\text{and}\quad OX \parallel AB$$.

Similarly, in $$\triangle ABC$$,

$$OY = \dfrac{1}{2}\,AB \quad\text{and}\quad OY \parallel AB$$.

Step 4 : Obtain $$XY$$.

Because the point $$O$$ lies on $$XY$$,

$$XY = XO + OY = \dfrac{1}{2}AB + \dfrac{1}{2}AB = AB.$$

But in a parallelogram opposite sides are equal, so $$AB = DC$$.

Step 5 : State the relation.

Hence

and, since both are parallel to $$AB$$, $$XY \parallel DC$$ as well.

Conclusion: The segment drawn through the intersection of the diagonals parallel to one pair of opposite sides is equal and parallel to the other pair of opposite sides.

Step 1 : Name and label the figure
Draw a parallelogram $$ABCD$$. Mark the following:

• Take the side $$AB$$ as one possible base. Draw the perpendicular from $$D$$ to $$AB$$; let the foot of the perpendicular be $$E$$. Then $$DE$$ is the height (altitude) corresponding to base $$AB$$; denote the lengths $$AB = b_1$$ and $$DE = h_1$$.
• Now take the adjacent side $$BC$$ as another possible base. Draw the perpendicular from $$A$$ to $$BC$$; let the foot of this perpendicular be $$F$$. Then $$AF$$ is the height corresponding to base $$BC$$; denote the lengths $$BC = b_2$$ and $$AF = h_2$$.

Step 2 : Recall the definition of area of a parallelogram
For any parallelogram, its area is defined as base × height, where

• “base” is any chosen side, and
• “height” (altitude) is the perpendicular distance between that base and the opposite side.

Step 3 : Compute the area using the first base
With $$AB$$ as the base, the area is

Step 4 : Compute the area using the second base
With $$BC$$ as the base, the area is

Step 5 : Why do both products give the same result?

1. Both expressions measure the area of the same region, namely the interior of parallelogram $$ABCD$$. Because the physical region does not change, its numerical area cannot depend on which side we momentarily called the base.
2. If one wishes an explicit proof, cut the triangle $$\triangle ABE$$ from the top of the figure and slide it along $$AB$$ so that it exactly covers triangle $$\triangle DCB$$ (or use any other standard dissection). The rearrangement shows that the rectangle of width $$b_1$$ and height $$h_1$$ has the same total region as the rectangle of width $$b_2$$ and height $$h_2$$, so the two products must be equal:

Step 6 : Conclusion
Because either product equals the one, single area of parallelogram $$ABCD$$, we conclude that the area can indeed be determined by choosing any side as the base together with its corresponding perpendicular height.

Therefore, yes—the area of a parallelogram remains $$\text{base}\times\text{height}$$ no matter which side you select as the base, provided you always use the perpendicular drawn to that side.

Given. A parallelogram $$ABCD$$ in which a perpendicular has been drawn from vertex $$C$$ to the opposite side $$AB$$. The foot of this perpendicular is the point $$Z$$, so that $$CZ \perp AB$$.

We have to decide whether, after cutting the sheet of the parallelogram along the segment $$CZ$$, the two resulting pieces can be shifted and joined again so as to form a perfect rectangle.

Step 1 Cut along $$CZ$$.
The cut divides the parallelogram into

• triangle $$\triangle ABZ$$, and
• quadrilateral $$BCZD$$.

Step 2 Translate $$\triangle ABZ$$.
Slide $$\triangle ABZ$$ towards the right (parallel to $$\overrightarrow{BC}$$) until

• point $$Z$$ falls exactly on $$D$$, and
• side $$AB$$ lies along $$CD$$.

The two requirements can indeed be met because in any parallelogram

$$AB = CD \quad\text{and}\quad AB \parallel CD.$$

Step 3 Check the new outline.
After the shift the figure now has vertices $$B,\;C,\;D,\;Z$$ in order. We show that every interior angle of $$BCDZ$$ is a right angle, so it is a rectangle.

1. At $$Z$$ (which now coincides with $$D$$): by construction $$CZ$$ is perpendicular to $$AB$$, and $$AB$$ is now lying on $$CD$$, therefore $$CZ \perp CD$$ so $$\angle Z = 90^\circ$$.
2. At $$C$$: $$CZ$$ is perpendicular to $$CD$$ (just proved), hence $$\angle C = 90^\circ$$.
3. Opposite angles of a quadrilateral sum to $$180^\circ$$, so the remaining two angles at $$B$$ and $$D$$ are also $$90^\circ$$.

With all four angles right angles, $$BCDZ$$ is a rectangle whose

length $$= AB = CD$$ and breadth $$= CZ$$ (which was the height of the original parallelogram).

Conclusion. Yes, once we cut the parallelogram along $$CZ$$ and slide the triangular part to the opposite side, the two pieces fit exactly into a rectangle. This activity also shows why

– the same formula as that for the rectangle obtained.

Given

• PQRS is a parallelogram.
• Side $$PS = 7.6\;\text{cm}$$.
• Side $$SR = 12\;\text{cm}$$.
• Altitude from Q on SR: $$QM = 6\;\text{cm}$$ (M lies on SR).
• Altitude from Q on PS: QN (N lies on PS)  = ?

Step 1. Find the area of the parallelogram using base SR.

In any parallelogram,

$$\text{Area} = \text{base} \times \text{corresponding altitude}.$$

Taking base $$SR$$ and the perpendicular $$QM$$,

$$\text{Area} = SR \times QM.$$

Substitute the given lengths:

$$\text{Area} = 12\,\text{cm} \times 6\,\text{cm} = 72\,\text{cm}^2.$$

Step 2. Express the same area using base PS.

With base $$PS$$ and its corresponding altitude $$QN$$, the same area is

$$\text{Area} = PS \times QN.$$

But this area must again equal $$72\,\text{cm}^2$$, so

$$PS \times QN = 72.$$

Step 3. Solve for QN.

$$QN = \dfrac{72}{PS} = \dfrac{72}{7.6}.$$

To divide, multiply numerator and denominator by 10 to avoid the decimal in the divisor:

$$QN = \dfrac{720}{76}.$$

Perform the division:

• $76 \times 9 = 684$  →  remainder $720-684=36$
• Bring down 0 (first decimal place): $360$; $76 \times 4 = 304$  →  remainder $56$
• Bring down 0 (second decimal place): $560$; $76 \times 7 = 532$  →  remainder $28$

Thus

$$QN \approx 9.47\,\text{cm}.$$

(Correct to two decimal places; more decimal places are unnecessary for the present level.)

Therefore, the length of the altitude QN is approximately $$9.47\,\text{cm}$$.

Given. Two plane figures have the same side-lengths

• Rectangle: one pair of opposite sides $$5\,\mathrm{cm}$$ long (its length) and the other pair $$4\,\mathrm{cm}$$ long (its breadth).
• Parallelogram: adjacent sides also $$5\,\mathrm{cm}$$ and $$4\,\mathrm{cm}$$ long.

We have to decide whose area is larger.

Step 1 · Fix a common base.
Draw a straight segment $$AB = 5\,\mathrm{cm}$$. It will serve as the base for both figures.

For the rectangle mark a point $$D$$ perpendicular to $$AB$$ such that $$AD = 4\,\mathrm{cm}$$, and complete rectangle $$ABCD$$.

For the parallelogram take a point $$P$$ on the same side of $$AB$$ such that $$AP = 4\,\mathrm{cm}$$ but do not make $$AP$$ perpendicular to $$AB$$; instead tilt it so that $$\angle PAB$$ differs from $$90^\circ$$, then complete parallelogram $$ABQP$$ (with $$BQ$$ drawn parallel to $$AP$$).

Both figures now rest on the same base $$AB = 5\,\mathrm{cm}$$.

Step 2 · Compute the rectangle’s area.

The height of the rectangle equals its breadth, $$4\,\mathrm{cm}$$. Hence

Step 3 · Express the parallelogram’s area.

For any parallelogram

$$\text{Area}_{\text{para}} = \text{base}\times\text{corresponding height}.$$

Here the base is the same $$AB = 5\,\mathrm{cm}$$. Let $$h$$ denote the perpendicular distance from $$P$$ to the line $$AB$$ — the height of the parallelogram.

In the right-angled triangle formed by $$A$$, $$P$$ and the foot of the perpendicular from $$P$$ on $$AB$$, the slant side $$AP$$ (of length $$4\,\mathrm{cm}$$) is the hypotenuse and $$h$$ is one of the legs. Since the hypotenuse is always the longest side, we must have $$h < AP$$, i.e.

$$0 < h < 4\,\mathrm{cm}.$$

Therefore

$$\text{Area}_{\text{para}} = 5 \times h\quad\text{with}\quad 0 < h < 4.$$

Step 4 · Compare the two areas.

(because $$4 - h > 0$$ whenever $$h < 4$$).

Hence

$$\text{Area}_{\text{rect}} \;>\; \text{Area}_{\text{para}}.$$

Conclusion. With the same side-lengths $$5\,\mathrm{cm}$$ and $$4\,\mathrm{cm}$$, a rectangle encloses the larger region. The only way the two areas can be equal is when the parallelogram is not tilted at all — in other words, when it is the rectangle itself.

Let \(\triangle ABC\) be the given triangle. (Draw the triangle and mark its base \(BC\) and the altitude from \(A\) to the base, meeting \(BC\) at \(D\). Let that altitude be the “height” of the triangle.)

Denote

• the length of the base by $$b = BC$$,
• the length of the altitude by $$h = AD$$.

The area of the triangle is therefore

Statement of the activity
Cut rhombus $$ABCD$$ along its diagonals $$AC$$ and $$BD$$ (meeting at $$O$$). Rearrange the four resulting triangles to form a rectangle $$WXYZ$$ and show that the two figures have the same area.

1. Lengths produced by the diagonals
   The diagonals of a rhombus bisect each other at right angles, therefore
   $$AO = CO = \frac{d_1}{2},\; BO = DO = \frac{d_2}{2},\; \angle AOB = 90^\circ.$$

2. Area of the rhombus by direct addition
   Each of the four triangles, for example $$\triangle AOB$$, is right-angled with perpendicular sides $$\tfrac{d_1}{2}$$ and $$\tfrac{d_2}{2}$$, so
   $$\text{area}(\triangle AOB)=\tfrac12\times\tfrac{d_1}{2}\times\tfrac{d_2}{2}=\frac{d_1d_2}{8}.$$
   Adding the four congruent triangles gives

   \[\text{area}(ABCD)=4\times\frac{d_1d_2}{8}=\frac12 d_1d_2\tag{1}\]

3. Making two identical rectangles

   • Join $$\triangle AOB$$ and the opposite $$\triangle COD$$ along their equal sides to make rectangle $$P$$ having sides $$\frac{d_1}{2}$$ and $$\frac{d_2}{2}$$.
   • Similarly $$\triangle BOC$$ and $$\triangle DOA$$ form an identical rectangle $$Q$$.

4. Forming the final rectangle
   Place rectangles $$P$$ and $$Q$$ side by side so that their sides of length $$\tfrac{d_2}{2}$$ coincide. The combined figure is rectangle $$WXYZ$$.

5. Dimensions and area of $$WXYZ$$
   Across: $$\tfrac{d_1}{2}+\tfrac{d_1}{2}=d_1$$;
   Height: $$\tfrac{d_2}{2}$$.
   Hence

   \[\text{area}(WXYZ)=d_1\times\frac{d_2}{2}=\frac12 d_1d_2\tag{2}\]

6. Verification
   Comparing (1) and (2): $$\text{area}(ABCD)=\text{area}(WXYZ)$$. The dissection therefore preserves area and also justifies the standard formula: Area of a rhombus = half the product of its diagonals.

What to draw
Sketch rhombus $$ABCD$$ with diagonals, cut out the four triangles, pair opposite ones to make two small rectangles, then slide them together to display rectangle $$WXYZ$$ whose sides are labelled $$d_1$$ and $$\tfrac{d_2}{2}$$.

Given : Rhombus $$ABCD$$ whose two diagonals $$AC$$ and $$BD$$ intersect at $$O$$. After cutting the rhombus along both its diagonals, the four congruent right-angled triangles obtained are rearranged (as shown in the textbook) to form a rectangle $$WXYZ$$.

Step 1 · Lengths available from the rhombus

• Diagonals of a rhombus bisect each other at right angles, therefore $$OA = OC = \dfrac{AC}{2} \;\;\text{and}\;\; OB = OD = \dfrac{BD}{2}.$$
• Each of the four pieces is a right-angled triangle whose legs are $$\dfrac{AC}{2}$$ and $$\dfrac{BD}{2}$$.

Step 2 · Placing the pieces to make the rectangle

• Place two of the triangles so that the sides equal to $$OB$$ (each $$=\dfrac{BD}{2}$$) lie collinearly. The combined straight segment is therefore $$\dfrac{BD}{2}+\dfrac{BD}{2}=BD.$$
• Their perpendicular sides, each of length $$\dfrac{AC}{2}$$, now stand upright and together form one complete edge of the new figure.
• Repeat the same arrangement with the remaining two triangles and attach this second pair above the first. The four triangles now exactly fill up a rectangle.

Step 3 · Deducing the side-lengths of $$WXYZ$$

From the construction just described:

• One side (say $$WZ$$) is made of two $$OB$$’s laid end-to-end, hence
  $$WZ = BD.$$
• The other side (say $$WX$$) is a single $$OA$$, so
  $$WX = \dfrac{AC}{2}.$$

Step 4 · Checking that areas match

The area of the new rectangle is \[\text{Area}_{\text{rect}} = WX \times WZ = \left(\dfrac{AC}{2}\right)\!(BD) = \dfrac{AC\,BD}{2},\] precisely the known formula for the area of the rhombus. This confirms the dissection.

Hence the rectangle $$WXYZ$$ has

• length $$BD$$ (equal to one whole diagonal of the rhombus), and
• breadth $$\dfrac{AC}{2}$$ (equal to half of the other diagonal).

Let the two diagonals of rhombus ABCD be

• $$AC = d_1$$
• $$BD = d_2$$

They intersect at O and, in every rhombus,

• the diagonals bisect each other, so $$AO = CO = \dfrac{d_1}{2}$$ and $$BO = DO = \dfrac{d_2}{2}$$;
• the diagonals are perpendicular, so $$\angle AOB = 90^{\circ}$$.

1. Area of   ΔADB

Take $$DB$$ as the base of ΔADB.

Its length is $$BD = d_2$$.

The height from A to DB is the perpendicular segment AO, whose length is $$AO = \dfrac{d_1}{2}$$ (because AC is perpendicular to BD and is bisected at O).

Using the triangle-area formula $$\displaystyle \text{Area} = \dfrac12 \times \text{base} \times \text{height}$$,

$$\text{Area}(\triangle ADB) = \dfrac12 \times d_2 \times \dfrac{d_1}{2} = \dfrac{1}{4}\,d_1 d_2.$$

2. Area of   ΔCDB

The same base $$DB = d_2$$ and the same height $$CO = \dfrac{d_1}{2}$$ give

$$\text{Area}(\triangle CDB) = \dfrac12 \times d_2 \times \dfrac{d_1}{2} = \dfrac{1}{4}\,d_1 d_2.$$

3. Area of the rhombus

The whole rhombus is made of these two triangles, so

\[ \text{Area of rhombus ABCD} = \text{Area}(\triangle ADB) + \text{Area}(\triangle CDB) = \dfrac14 d_1 d_2 + \dfrac14 d_1 d_2 = \dfrac12 d_1 d_2. \]

Thus, finding the areas of ΔADB and ΔCDB leads to the well-known formula:

\[ \boxed{\text{Area of rhombus} = \dfrac12 \times (\text{product of its diagonals})} \]

That is, if the lengths of the diagonals are $$d_1$$ and $$d_2$$, then

\[ \text{Area} = \dfrac12 d_1 d_2. \]

Goal – Start from the expression obtained by splitting the rhombus into four right-angled triangles and simplify it so that only the two diagonals, $$d_1$$ and $$d_2$$, remain.

When diagonals $$AC$$ and $$BD$$ of a rhombus intersect at $$O$$ they bisect each other at right angles. Thus each of the four triangles – $$\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$$ – has perpendicular sides

• $$AO = CO = \dfrac{d_1}{2}$$ (half of diagonal $$AC$$), and
• $$BO = DO = \dfrac{d_2}{2}$$ (half of diagonal $$BD$$).

Area of one such right-angled triangle is therefore

$$\text{Area}_{\triangle} = \frac12 \times AO \times BO = \frac12 \times \frac{d_1}{2} \times \frac{d_2}{2}.$$

The whole rhombus consists of four congruent triangles, so the required expression for its area is

$$\text{Area}_{\text{rhombus}} = 4 \times \left(\frac12 \times \frac{d_1}{2} \times \frac{d_2}{2}\right).$$

Now simplify step by step, showing every manipulation clearly:

1. First multiply the numerical factors: $$4 \times \frac12 = 2.$$
2. Substitute this back: $$= 2 \times \frac{d_1}{2} \times \frac{d_2}{2}.$$
3. Next, multiply $$2$$ with the first fraction: $$2 \times \frac{d_1}{2} = \frac{d_1}{1} = d_1.$$
4. So we now have $$= d_1 \times \frac{d_2}{2}.$$
5. Finally multiply the remaining factors: $$d_1 \times \frac{d_2}{2} = \frac{d_1 d_2}{2}.$$

Thus the simplified form is

The simplification confirms that we indeed obtain the familiar formula for the area of a rhombus in terms of the lengths of its diagonals.

On the grid every small square is 1 cm by 1 cm; hence its area is $$1\,\text{cm}^2$$. We shall read every length directly by counting the squares and then split the given trapezium into one rectangle and two right-angled triangles.

(i) Trapezium ABCD (with $$AB \parallel DC$$)

Drop perpendiculars $$AE$$ and $$BF$$ from $$A$$ and $$B$$ to $$DC$$.

• Top base $$AB = 8\,\text{cm}$$ (8 squares).
• Perpendicular (height) $$AE = BF = 4\,\text{cm}$$ (4 squares).
• Total lower base $$DC = 13\,\text{cm}$$ (13 squares).
    So $$DE = 3\,\text{cm}$$ and $$FC = 2\,\text{cm}$$ because $$3 + 8 + 2 = 13$$.

Areas of the three parts:

Rectangle $$ABEF: \; A_1 = AB \times AE = 8 \times 4 = 32\,\text{cm}^2$$.

Left triangle $$AED: \; A_2 = \tfrac12 \times DE \times AE = \tfrac12 \times 3 \times 4 = 6\,\text{cm}^2$$.

Right triangle $$BFC: \; A_3 = \tfrac12 \times FC \times BF = \tfrac12 \times 2 \times 4 = 4\,\text{cm}^2$$.

(ii) Trapezium PQRS (with $$PQ \parallel SR$$)

Perpendiculars $$PT$$ and $$QU$$ have been drawn already.

• Top base $$PQ = 8\,\text{cm}$$.
• Height $$PT = QU = 3\,\text{cm}$$.
• Bottom base $$SR = 14\,\text{cm}$$.
    Hence $$TS = 2\,\text{cm}$$ and $$UR = 4\,\text{cm}$$ (because $$2 + 8 + 4 = 14$$).

We now have:

Rectangle $$PQUT: \; A_1 = 8 \times 3 = 24\,\text{cm}^2$$.

Left triangle $$PTS: \; A_2 = \tfrac12 \times 2 \times 3 = 3\,\text{cm}^2$$.

Right triangle $$QUR: \; A_3 = \tfrac12 \times 4 \times 3 = 6\,\text{cm}^2$$.

(iii) Trapezium WXYZ (with $$WX \parallel ZY$$)

Perpendiculars $$WM$$ and $$XN$$ are drawn.

• Top base $$WX = 6\,\text{cm}$$.
• Height $$WM = XN = 3\,\text{cm}$$.
• Bottom base $$ZY = 12\,\text{cm}$$.
    So $$MZ = NY = 3\,\text{cm}$$ because $$3 + 6 + 3 = 12$$.

The three parts are:

Rectangle $$WXNM: \; A_1 = 6 \times 3 = 18\,\text{cm}^2$$.

Each of the two equal triangles has area
$$A_2 = A_3 = \tfrac12 \times 3 \times 3 = 4.5\,\text{cm}^2$$.

The required areas are therefore 42 cm², 33 cm² and 27 cm².

We are given a trapezium $$WXYZ$$ in which the side $$WX$$ is parallel to $$ZY$$.

Let the following symbols be used:

• length of the first parallel side = $$WX$$,
• length of the second parallel side = $$ZY$$,
• perpendicular distance (height) between the two parallel sides = $$h$$.

We wish to find the area of the trapezium.

Step 1 • Visualising the division of the figure
If we drop perpendiculars from $$W$$ and $$X$$ to the line that contains $$ZY$$, the trapezium splits into one rectangle and two right-angled triangles, all having the same height $$h$$. These three pieces together make up the entire trapezium.

Step 2 • Using the trapezium-area formula
Instead of adding the areas of the three separate pieces one by one, we may directly apply the standard result for a trapezium:

Substituting the symbols we introduced,

$$\text{Area of }WXYZ = \dfrac{(WX + ZY) \times h}{2}.$$

This is the required expression. (For a numerical answer we would simply replace $$WX$$, $$ZY$$ and $$h$$ by the corresponding measured values.)

Given: In trapezium $$WXYZ$$ we have $$WX \parallel ZY$$. From vertices $$W$$ and $$X$$ perpendiculars are drawn to $$ZY$$, meeting it at $$M$$ and $$N$$ respectively, so $$WM \perp ZY$$ and $$XN \perp ZY$$.

To prove: Quadrilateral $$WXNM$$ is a rectangle.

1. Show that opposite sides are parallel

• Points $$M$$ and $$N$$ lie on $$ZY$$ ⇒ segment $$MN$$ is a part of $$ZY$$, hence $$MN \parallel ZY$$.
• Given $$WX \parallel ZY$$; therefore
  \[MN \parallel WX \tag{1}\]
• Both $$WM$$ and $$XN$$ are perpendicular to $$ZY$$. Lines perpendicular to the same line are parallel, so
  \[WM \parallel XN \tag{2}\]

Because of (1) and (2), both pairs of opposite sides of $$WXNM$$ are parallel. Thus $$WXNM$$ is a parallelogram.

2. Show that one angle is a right angle

• We know $$WM \perp ZY$$ (given) and $$WX \parallel ZY$$.
• A line perpendicular to one of two parallel lines is perpendicular to the other; hence $$WM \perp WX$$.
• Therefore $$\angle W = 90^{\circ}$$.
  \[\angle W = 90^{\circ} \tag{3}\]

3. Conclude that the parallelogram is a rectangle

A parallelogram with one right angle has all four right angles; such a figure is a rectangle. Since $$WXNM$$ is a parallelogram (Step 1) and it possesses a right angle (Step 2), we conclude

Hence: Yes, $$WXNM$$ is a rectangle.

Let $$ABCD$$ be the given trapezium with $$AB \parallel CD$$ and $$AB < CD$$. The perpendicular from $$A$$ meets the line containing $$CD$$ at $$E$$ outside the segment $$CD$$ (as shown in the question). Draw another perpendicular from $$B$$ to meet the same line at $$F$$.

Put

$$AB = a,\quad CD = b\;(b > a),\quad AE = BF = h\;\;(\text{height}).$$

1.  Form a rectangle.
Because $$AE \perp AB$$ and $$BF \perp AB$$ while $$AE \parallel BF$$, the quadrilateral $$ABFE$$ is a rectangle. Hence

$$EF = AB = a.$$

2.  Pieces inside the rectangle.

• The required trapezium is $$ABCD$$.
• To the left of it lies right triangle $$\triangle AEC$$ with base $$EC$$ and height $$h$$.
• To the right of it lies right triangle $$\triangle BFD$$ with base $$DF$$ and height $$h$$.

Therefore

$$\text{Area}(ABCD)=\text{Area}(\text{rectangle }ABFE) \;-\; \text{Area}(\triangle AEC) \;-\; \text{Area}(\triangle BFD).$$

3.  Express each area.

Rectangle: $$\;a \times h = ah.$$

The two triangles have the same height $$h$$, and together their bases make up the excess length of the longer parallel side over the shorter one, namely

$$EC + DF = CD - AB = b - a.$$

So the combined area of the two triangles is

$$\tfrac12\,h\,(b - a).$$

4.  Substitute.

$$\begin{aligned} \text{Area}(ABCD) &= ah \;-\; \tfrac12 h (b - a)\\ &= ah \;-\; \tfrac12 hb \;+\; \tfrac12 ha\\ &= \tfrac12\,h\,(a + b). \end{aligned}$$

5.  Replace the symbols by words.

This calculation does not depend on whether the perpendicular from an end of the shorter side falls on the longer side or outside it. Hence the usual formula remains valid for the trapezium described in the question.

Restatement of the problem

Approach 2 for finding the area of a trapezium first cuts the figure into a parallelogram and a triangle and then adds their areas. The question asks whether this method can be applied to every trapezium, irrespective of its shape (right-angled, isosceles, or scalene).

Given

• A trapezium $$ABCD$$ with $$AB \parallel CD$$. (Either base may be the longer one.)
• Its perpendicular height is $$h$$.

Construction used in Approach 2

1. From the vertex whose adjacent non-parallel side we do not use, draw a line parallel to that non-parallel side until it meets the opposite parallel side (or its extension).
   Choose the case that keeps all new points on the original figure:
   
   • If $$AB < CD$$, draw $$AE \parallel BC$$ to meet $$CD$$ at $$E$$ (see Fig. 1).
   • If $$AB > CD$$, draw $$CF \parallel AD$$ to meet $$AB$$ at $$F$$ (see Fig. 2).

(Describe the diagram: Fig. 1 shows trapezium $$ABCD$$ with $$AB$$ the shorter base. From $$A$$ a segment $$AE$$ is drawn parallel to $$BC$$, meeting $$CD$$ at $$E$$. The region $$ABCE$$ is a parallelogram; the small leftover region $$AED$$ is a triangle.)

Why the construction always works

• The non-parallel sides $$AD$$ and $$BC$$ are distinct and form non-zero angles with the bases. Hence a line through any vertex can be drawn parallel to the opposite non-parallel side by the Parallel Postulate.
• That line is guaranteed to intersect the opposite base (or its extension) because the two bases are parallel and finite straight lines. Thus exactly two regions are produced: a parallelogram and a triangle. The triangle always lies completely inside the original trapezium; if the shorter base is on top, the triangle lies on the left (Fig. 1), otherwise on the right (Fig. 2).

Verification of the area relation

Without loss of generality assume $$AB < CD$$ and use the construction in Fig. 1.

Because $$AE \parallel BC$$ and $$AB \parallel CD$$, quadrilateral $$ABCE$$ is a parallelogram.

Let $$AB = b_1$$ and $$CD = b_2$$ with $$b_2 > b_1$$. The segment $$CE$$ equals $$AB$$ (opposite sides of a parallelogram), so the length $$DE$$ equals the difference of the bases:

Both the triangle $$\triangle AED$$ and the parallelogram $$ABCE$$ have the same height $$h$$, the perpendicular distance between the parallel lines $$AB$$ and $$CD$$.

Area of the parallelogram: $$\text{Area}(ABCE) = b_1\,h.$$ Area of the triangle: $$\text{Area}(\triangle AED) = \tfrac12\,DE\,h = \tfrac12\,(b_2 - b_1)\,h.$$ Total area of the trapezium: \[ \text{Area}(ABCD) = b_1 h + \tfrac12\,(b_2 - b_1) h = \tfrac12\,(b_1 + b_2) h, \] which is the standard trapezium formula. The algebra shows no extra restriction; only the fact that one pair of opposite sides is parallel was used.

The same calculation, done symmetrically for Fig. 2 when $$AB > CD$$, gives the identical result. Hence the partition works for every possible relative length of the bases and for every shape of the non-parallel sides.

Conclusion

Because the required parallel through a vertex can always be drawn and always intersects the other base (or its extension), Approach 2 always splits any trapezium into exactly one parallelogram and one triangle whose areas add up to the area of the original trapezium. Therefore Approach 2 works for every trapezium.

Step 1 : Understand the given trapezium
Let $$ABCD$$ be a trapezium in which the pair of opposite sides $$AB$$ and $$CD$$ are parallel, i.e. $$AB \parallel CD$$. The non-parallel sides are $$AD$$ and $$BC$$.

How to draw it for yourself: draw a short horizontal segment $$AB$$, a longer horizontal segment $$CD$$ below it, and then join the ends so that $$AD$$ and $$BC$$ are slanting.

Step 2 : Make an identical copy and rotate it
Make a second copy $$A'B'C'D'$$ of the trapezium. Now rotate this copy through $$180^{\circ}$$ in the plane (turn it upside-down). Rotation preserves lengths and parallelism, so we still have
$$A'B' \parallel C'D' \quad\text{and}\quad A'B'C'D' \cong ABCD.$$

Step 3 : Join the two trapeziums along the side $$BC$$
Place the rotated copy so that its side $$B'C'$$ exactly overlaps the side $$BC$$ of the first trapezium (that is, $$B'$$ coincides with $$C$$ and $$C'$$ coincides with $$B$$, because of the $$180^{\circ}$$ rotation). These two coincident sides disappear from the outer boundary.

Tracing the outer boundary of the combined figure. Start at $$A$$ and go along $$AB$$ to $$B$$. Because the rotation was through $$180^{\circ}$$, the side $$C'D'$$ of the rotated copy lies along the same straight line as $$AB$$ (both are parallel to $$AB$$ and $$B\equiv C'$$ is common); so continuing straight from $$B$$ we walk along $$C'D'$$ (which is now the extension of $$AB$$) until we reach $$D'$$. From $$D'$$ we follow the side $$D'A'$$ of the rotated copy to $$A'$$. Similarly, $$A'B'$$ of the rotated copy lies along the same straight line as $$CD$$, and $$A'\equiv $$ the endpoint on that line; walking from $$A'$$ along $$B'A'$$ we continue along $$CD$$ back to $$D$$, and finally the side $$DA$$ closes the figure back at $$A$$.

So the combined outline has exactly four vertices — $$A$$, $$D'$$, $$A'$$, $$D$$ — and four sides:

• $$AD'$$: the union of $$AB$$ and $$BC'D'\!\ldots$$ — the entire straight segment from $$A$$ to $$D'$$, of length $$AB + C'D' = AB + CD$$;
• $$D'A'$$: same length as the original slant side $$DA$$ (image under rotation);
• $$A'D$$: the entire straight segment from $$A'$$ to $$D$$, of length $$A'B' + CD = AB + CD$$ (again a combined straight segment);
• $$DA$$: the original slant side.

Step 4 : Show that opposite sides of the new quadrilateral are parallel

• The two long sides $$AD'$$ and $$A'D$$ lie along the two originally parallel lines containing $$AB$$ and $$CD$$; since those lines are parallel, so are $$AD'$$ and $$A'D$$.
• The two slant sides $$D'A'$$ and $$DA$$ are related by the $$180^{\circ}$$ rotation, and a rotation by $$180^{\circ}$$ carries any line into a parallel line. Hence $$D'A' \parallel DA$$.

Both pairs of opposite sides are parallel, so the combined figure is a parallelogram.

Step 5 : State the resulting figure
When the two congruent trapeziums are joined along the side $$BC$$, the figure obtained is a parallelogram. Its base is $$AB + CD$$ and its height equals the height of the original trapezium; this fact is used later to derive the area formula for a trapezium.

Step 1 : Name the two congruent trapezia
Let the given trapezium be $$ABCD$$ in which the pair of parallel sides (bases) is $$AB \parallel CD$$. Make an identical copy of it and denote the corresponding vertices by $$A'B'C'D'$$.

Step 2 : Join the two trapezia along the side $$BC$$
Place the copy $$A'B'C'D'$$ in such a way that its side $$B'C'$$ exactly overlaps the side $$CB$$ of the first trapezium (so the points coincide as $$B \equiv C'$$ and $$C \equiv B'$$).
The boundary of the figure obtained after joining is the quadrilateral $$ADD'A'$$ (travelling in order $$A \rightarrow D \rightarrow D' \rightarrow A' \rightarrow A$$).

Step 3 : Show one pair of opposite sides are parallel
The side $$AD$$ of the first trapezium and the side $$A'D'$$ of the second trapezium were corresponding non-parallel sides in the identical trapezia. No rotation has been applied to these sides; we have only slid the second trapezium until $$B'C'$$ coincided with $$CB$$. Hence the directions of $$AD$$ and $$A'D'$$ remain the same  ⇒  $$AD \parallel A'D'$$.

Step 4 : Show the other pair of opposite sides are parallel
In every trapezium we already know $$AB \parallel CD$$.
•  After the join, the segment $$AB$$ forms a part of the side $$AA'$$ of the big quadrilateral.
•  Similarly, the segment $$CD$$ forms a part of the side $$DD'$$ of the big quadrilateral.
Therefore $$AA'$$ and $$DD'$$ are along the same direction, i.e. $$AA' \parallel DD'$$.

Step 5 : Identify the quadrilateral $$ADD'A'$$
Both pairs of opposite sides of $$ADD'A'$$ are parallel:
$$AD \parallel A'D' \quad\text{and}\quad AA' \parallel DD'.$$
A quadrilateral having both pairs of opposite sides parallel is a parallelogram.

Conclusion
Hence, the quadrilateral formed by joining the two congruent trapezia along $$BC$$ is a parallelogram.

Given data

• Length of the first diagonal: $$d_1 = 20\,\text{cm}$$
• Length of the second diagonal: $$d_2 = 15\,\text{cm}$$

Step 1 · Recall the formula for the area of a rhombus

The diagonals of a rhombus are perpendicular bisectors of each other. Hence the area $$A$$ is obtained from

$$A = \dfrac{1}{2} \times d_1 \times d_2$$

Step 2 · Substitute the given diagonal lengths

$$A = \dfrac{1}{2} \times 20\,\text{cm} \times 15\,\text{cm}$$

Step 3 · Simplify the numerical calculation

First multiply the two diagonals:

$$20 \times 15 = 300$$

Now multiply by the factor $$\dfrac{1}{2}$$:

$$A = \dfrac{1}{2} \times 300 = 150$$

Step 4 · Attach the unit

The area therefore is

Let $$ABCD$$ be the given rectangle with $$AB = \ell$$ (longer side) and $$BC = b$$ (shorter side), so $$\ell \ge b$$. Its area is $$\ell b$$.

Idea. The rectangle already has all four angles equal (all right angles). To turn it into a rhombus of the same area we “shear” it: cut off one right-angled triangle from an end and slide it to the opposite end. The horizontal length is preserved but the shape becomes a slanted parallelogram; by choosing the cut so that the slant side becomes equal to the horizontal length, the parallelogram becomes a rhombus.

Set up coordinates: $$A(0,0),\; B(\ell,0),\; C(\ell,b),\; D(0,b)$$.

Step 1 — Locate the cut on the top side.
Let $$k = \sqrt{\,\ell^{2}-b^{2}\,}$$ (real because $$\ell \ge b$$).
On the top side $$DC$$ mark the point $$E$$ such that $$DE = k$$. Since $$\ell \ge b$$ we have $$0 \le k < \ell$$, so $$E$$ lies on segment $$DC$$ (at $$(k,\,b)$$).

Step 2 — Cut along $$AE$$.
The straight cut from $$A(0,0)$$ to $$E(k,b)$$ divides the rectangle into two pieces:

• right-angled triangle $$\triangle ADE$$ with vertices $$A(0,0),\; D(0,b),\; E(k,b)$$ — right angle at $$D$$, legs $$AD = b$$ and $$DE = k$$, hypotenuse $$AE = \sqrt{k^{2}+b^{2}} = \sqrt{(\ell^{2}-b^{2})+b^{2}} = \ell$$;
• trapezium $$ABCE$$ with vertices $$A(0,0),\; B(\ell,0),\; C(\ell,b),\; E(k,b)$$.

Step 3 — Slide the triangle to the right-hand end.
Translate the triangle $$\triangle ADE$$ rigidly to the right by the vector $$(\ell,\,0)$$. Its vertices go to

$$A(0,0)\longrightarrow B(\ell,0),\quad D(0,b)\longrightarrow C(\ell,b),\quad E(k,b)\longrightarrow F(\ell+k,\,b).$$

The triangle’s side $$AD$$ (length $$b$$) now sits exactly on top of side $$BC$$ of the trapezium (also length $$b$$), so no gap or overlap is created.

Step 4 — Identify the new figure.
The combined outer boundary is

$$A(0,0)\;\to\;B(\ell,0)\;\to\;F(\ell+k,\,b)\;\to\;E(k,b)\;\to\;A(0,0).$$

Its four sides have lengths

• $$AB = \ell$$ (from $$(0,0)$$ to $$(\ell,0)$$);
• $$BF = \sqrt{k^{2}+b^{2}} = \ell$$ (this is the hypotenuse of the translated triangle);
• $$FE = \ell$$ (from $$(\ell+k,\,b)$$ to $$(k,b)$$, horizontal distance $$\ell$$);
• $$EA = \sqrt{k^{2}+b^{2}} = \ell$$ (this is the original cut).

All four sides are equal to $$\ell$$, so the quadrilateral $$ABFE$$ is a rhombus.

Step 5 — Equality of areas.
Cutting and sliding neither adds nor removes material, so the rhombus has the same area as the original rectangle:

\[\text{Area of rhombus }ABFE = \ell b = \text{Area of rectangle }ABCD.\]

Special case. If the rectangle is already a square ($$\ell = b$$) then $$k = 0$$, no cut is made, and the rectangle is itself a rhombus (in fact a square).

Conclusion. By cutting off the right-angled triangle $$\triangle ADE$$ (with $$DE = \sqrt{\ell^{2}-b^{2}}$$) and translating it so that its side $$AD$$ covers side $$BC$$, the rectangle is dissected into a rhombus of the same area, whose common side length is $$\ell$$.

Let us name the isosceles trapezium $$ABCD$$ with $$AB \parallel CD$$, $$AB = a$$ (longer base), $$CD = b$$ (shorter base) and equal non-parallel sides $$AD = BC$$. Let the common perpendicular distance between the parallel sides be $$h$$ (the height).

1. Locate the mid-segment and cut the trapezium into two pieces

1. Mark the mid-point $$M$$ of side $$AD$$ and the mid-point $$N$$ of side $$BC$$. Because $$AD = BC$$, points $$M$$ and $$N$$ are at the same level (half the height of the trapezium).
2. Join $$M$$ and $$N$$. Since the line joining the mid-points of the equal sides of an isosceles trapezium is parallel to the two bases, $$MN \parallel AB \parallel CD$$.
3. By the Basic Proportionality (similar-triangles) theorem the length of this segment is \[ MN = \frac{AB + CD}{2} = \frac{a + b}{2}. \]
4. Cut the trapezium along $$MN$$. We now have two smaller pieces:
• lower trapezium $$ABNM$$ of height $$h/2$$ and parallel sides $$AB = a$$ and $$MN = \dfrac{a+b}{2}$$;
• upper trapezium $$MNCD$$ of height $$h/2$$ and parallel sides $$MN = \dfrac{a+b}{2}$$ and $$CD = b$$.

2. Form a parallelogram

1. Rotate the upper piece $$MNCD$$ through $$180^\circ$$ about the mid-point of $$MN$$ and place it so that its edge $$MN$$ exactly coincides with the edge $$MN$$ of the lower piece.
2. The two equal slant edges $$MD$$ and $$NC$$ now lie on different sides and become two opposite sides of the new figure, producing a parallelogram whose base is $$MN$$ and whose total height is $$h/2 + h/2 = h$$.

Hence the area of the newly obtained parallelogram is \[ \text{Area}_{\text{para}} = MN \times h = \frac{a + b}{2}\,h. \]

This equals the area of the given trapezium, because \[ \text{Area}_{\text{trap}} = \frac{(a+b)h}{2}. \]

3. Convert the parallelogram to a rectangle

1. From any acute-angled vertex of the parallelogram drop a perpendicular to the opposite side (or its extension). This cuts off a right triangle whose base is a part of $$MN$$ and whose height is $$h$$.
2. Cut off this right triangle and slide it to the opposite side of the parallelogram; the two perpendicular edges fit exactly, giving a rectangle.

The rectangle so obtained has the same height $$h$$ as the trapezium and base length $$MN = \dfrac{a + b}{2}$$. Therefore its area remains \[ \text{Area}_{\text{rect}} = h \times \frac{a + b}{2} = \text{Area}_{\text{trap}}. \]

4. Conclusion

By the above sequence of dissections — cutting the trapezium along the mid-segment, rotating the top piece to form a parallelogram, and finally converting the parallelogram into a rectangle — we have rigorously shown a method (described in the Śulba-Sūtras) for changing any isosceles trapezium into an exactly equal-area rectangle.

Given data
ABCD is a trapezium with $$AB \parallel DC$$.
We have to obtain a rectangle $$EFGH$$ whose area is exactly the same as that of the trapezium and, therefore, must have

• height  = perpendicular distance between the two parallel sides (call it $$h$$), and
• width  = a length which, when multiplied by $$h$$, gives the area of the trapezium.

The area of a trapezium is \[ \text{Area} = \tfrac12 (AB + DC)\,h \]
Hence, if the height of our rectangle is also $$h$$, its width must be \[ HG = EF = \dfrac{AB + DC}{2}. \]

We now show how, with only ruler and compasses, the four vertices $$E, F, G, H$$ of that rectangle can be located on or near the given trapezium.

Step 1 Draw the height of the trapezium

From the vertex $$A$$ draw a perpendicular to the base $$DC$$. Let the foot of this perpendicular be $$H$$; thus
$$AH \perp DC \quad\text{and}\quad AH = h.$$

Step 2 Transfer the length $$AH$$ to the base $$DC$$

With centre $$D$$ and radius $$AH$$ draw an arc to cut $$DC$$ at $$G$$. Consequently $$DG = AH \; (= h).$$

(Choosing the radius equal to $$AH$$ ensures that the two right–angled triangles $$\triangle AHI$$ and $$\triangle DGI$$ constructed in the next step are congruent.)

Step 3 Erect perpendiculars through $$H$$ and $$G$$

Through the points $$H$$ and $$G$$ draw lines perpendicular to $$DC$$. Because $$AB \parallel DC$$, these new lines are automatically perpendicular to $$AB$$ as well.

• The perpendicular through $$H$$ meets the side $$AD$$ at $$I$$ and meets the line $$AB$$ (or its extension, if necessary) at $$E$$.
• The perpendicular through $$G$$ meets the side $$BC$$ at $$J$$ and meets the line $$AB$$ (or its extension) at $$F$$.

We have now obtained the four required points:

• $$H$$ and $$G$$ lie on the base $$DC$$,
• $$E$$ and $$F$$ lie on the line through $$AB$$,
• each angle of $$E\!FGH$$ is a right angle because its sides were drawn perpendicular or parallel to $$DC$$.

Step 4 Confirm that the two pairs of small triangles are congruent

Because of the way $$G$$ was chosen,

$$DG = AH, \quad \angle AHI = \angle DGI = 90^\circ, \quad HI \text{ is common}$$
so $$\triangle AHI \cong \triangle DGI$$ by RHS congruence.

Similarly, the triangles on the right satisfy

$$CF = BJ, \quad \angle BEJ = \angle CFJ = 90^\circ, \quad EJ \text{ is common},$$
so $$\triangle BEJ \cong \triangle CFJ.$$

Cutting off the two shaded triangles $$\triangle AHI$$ and $$\triangle BEJ$$ from the trapezium and attaching them to the positions of the congruent triangles $$\triangle DGI$$ and $$\triangle CFJ$$ exactly fills out the region bounded by $$EFGH$$. Consequently, the rectangle and the original trapezium have equal areas.

Step 5 The rectangle $$EFGH$$

Because opposite sides are parallel and all four angles are right angles, $$EFGH$$ is a rectangle. Its dimensions are

$$EH = GF = AH = h, \quad HG = EF = DG + AH = \dfrac{AB + DC}{2}. $$

Thus the rectangle $$EFGH$$ has the same height as the trapezium and width equal to the mean of the two parallel sides, giving it exactly the same area.

Final description of the vertices

1. H: the foot of the perpendicular from $$A$$ to the base $$DC$$.
2. G: on $$DC$$ such that $$DG = AH$$ (copy the length $$AH$$ onto $$DC$$ from the vertex $$D$$).
3. E: the point where the line through $$H$$ drawn perpendicular to $$DC$$ meets the (extended) line through $$AB$$.
4. F: the point where the line through $$G$$ drawn perpendicular to $$DC$$ meets the (extended) line through $$AB$$.

Joining the points in the order $$E \rightarrow F \rightarrow G \rightarrow H \rightarrow E$$ gives the required rectangle equal in area to trapezium $$ABCD$$.

Objective
Construct, with ruler and compass only, a trapezium whose area is $$144\;\mathrm{cm^{2}}$$ by first drawing a rectangle and then cutting off a triangular strip so that the remaining piece is a trapezium of the required area.

Step 1 — Decide convenient numerical dimensions

1. Choose the height of the trapezium to be a whole-number length, say $$h = 12\;\mathrm{cm}$$. (Any convenient height works.)
2. Let the longer parallel side (the larger base) be $$a = 14\;\mathrm{cm}$$.
3. Let the shorter parallel side be $$b\;\mathrm{cm}$$ (to be calculated).

The area formula for a trapezium is

Setting the area equal to $$144\;\mathrm{cm^{2}}$$:

$$\tfrac12\,(14+b)\,(12)=144$$

$$\Rightarrow (14+b)\times 12 = 288$$

$$\Rightarrow 14+b = 24$$

$$\Rightarrow b = 10.$$

So a trapezium of height $$12\;\mathrm{cm}$$ with parallel sides $$14\;\mathrm{cm}$$ and $$10\;\mathrm{cm}$$ has the required area.

Step 2 — Draw a rectangle that will contain the trapezium

1. Draw a straight line and mark $$AB = 14\;\mathrm{cm}$$ on it.
2. At $$A$$ erect a perpendicular $$AD$$ of length $$12\;\mathrm{cm}$$ (use compass and straight-edge for the right angle).
3. With centre $$B$$ and radius $$12\;\mathrm{cm}$$ draw an arc; through $$D$$ draw a line parallel to $$AB$$ to meet the perpendicular through $$B$$ at $$C$$. The figure $$ABCD$$ is a rectangle of length $$14\;\mathrm{cm}$$ and breadth $$12\;\mathrm{cm}$$.

The rectangle’s area is $$14 \times 12 = 168\;\mathrm{cm^{2}}$$ — larger than the required $$144\;\mathrm{cm^{2}}$$; we will trim it in the next step.

Step 3 — How much area to cut off

1. The excess area is $$168 - 144 = 24\;\mathrm{cm^{2}}$$.
2. Keeping the height equal to $$12\;\mathrm{cm}$$, a right-angled triangle of base $$x$$ and height $$12\;\mathrm{cm}$$ has area $$\tfrac12\,x\times 12 = 6x$$. Setting $$6x = 24$$ gives $$x = 4\;\mathrm{cm}$$.

Step 4 — Locate the cutting point on the top side

1. On the top side $$AB$$ (length $$14\;\mathrm{cm}$$) mark a point $$E$$ such that $$BE = 4\;\mathrm{cm}$$ (equivalently $$AE = 10\;\mathrm{cm}$$).
2. Join $$E$$ to $$D$$. Triangle $$BED$$... actually the corner triangle to be removed is $$\triangle BEC$$ — right-angled at $$B$$ with base $$BE = 4\;\mathrm{cm}$$ and height $$BC = 12\;\mathrm{cm}$$, so its area is $$\tfrac12\times 4\times 12 = 24\;\mathrm{cm^{2}}$$, exactly the excess to be removed.

(In practice: join $$E$$ on the top side to $$C$$ on the same side of the rectangle as the ‘short-base end’. The segment $$EC$$ slices off the required right-angled corner.)

Step 5 — Form the trapezium

1. Remove (or shade) the triangle $$\triangle BEC$$. The remaining quadrilateral $$AECD$$ has
   • parallel sides $$AE = 10\;\mathrm{cm}$$ (top) and $$DC = 14\;\mathrm{cm}$$ (bottom),
   • height $$AD = 12\;\mathrm{cm}$$ (the left vertical side).
2. Hence $$AECD$$ is a trapezium. Its area is

Result
The trapezium $$AECD$$ obtained by removing the right-angled corner triangle from the rectangle has area $$144\;\mathrm{cm^{2}}$$, as required.

Diagram to draw: Start with rectangle $$ABCD$$ (14 cm $$\times$$ 12 cm). Mark $$E$$ on $$AB$$ with $$BE = 4\;\mathrm{cm}$$. Draw straight line $$EC$$. The shaded triangle $$\triangle BEC$$ is discarded; the remaining quadrilateral $$AECD$$ is the required trapezium.

Construction used in the textbook
Let the regular hexagon be $$ABCDEF$$. The book joins a few of its diagonals so that the interior breaks up into three simple regions: a trapezium, an equilateral triangle and a rhombus (see the given diagram). Call the centre of the hexagon $$O$$.

1. Divide the whole hexagon into identical pieces
Join the centre $$O$$ to every vertex. The six sectors $$\triangle AOB,\,\triangle BOC,\,\triangle COD,\,\triangle DOE,\,\triangle EOF,\,\triangle FOA$$ are congruent equilateral triangles because a regular hexagon may be regarded as six equilateral triangles placed round a point.

Each small triangle therefore has area
$$\text{area of one small }\triangle=\dfrac{\sqrt3}{4}a^{2}$$
where $$a$$ is the side of the hexagon (and so of each little triangle).

2. Count how many of those small triangles lie in each region

• The trapezium seen in the figure contains exactly 3 of the little equilateral triangles.
• The equilateral triangle occupies 2 of them.
• The remaining rhombus covers just 1 little triangle.

(You can verify this by shading one tiny triangle and watching where its copies appear in the diagram.)

3. Ratio of the required areas
Because every small triangle has the same area, the areas of the three required parts are in the ratio of the numbers of small triangles they enclose:

\[ \text{Trapezium} : \text{Triangle} : \text{Rhombus} = 3 : 2 : 1. \]

The ratio is independent of the actual side length $$a$$ of the hexagon. Thus the answer is final.

Given data

• ZYXW is a trapezium with $$ZY \parallel WX$$.
• A is the mid-point of $$XY$$.
• ZA is produced to meet the (straight) line $$WX$$ at B, so Z, A, B are collinear and B lies on the extension of $$WX$$.

We have to prove

$$\text{area of trapezium }ZYXW = \text{area of }\triangle ZWB.$$

1. Fix a convenient co-ordinate system

Draw the trapezium so that the longer base $$WX$$ lies on the x-axis and W is taken as the origin.

• Let $$W(0,0)$$.
• Take $$X(b,0)$$, so $$WX=b$$.
• Let the distance between the two parallel sides be $$h\;(h>0)$$. Draw $$Z$$ right above $$W$$ on the line x = 0, i.e. $$Z(0,h)$$.
• Because $$ZY\parallel WX$$, every point of $$ZY$$ has y-coordinate $$h$$. Put $$Y(a,h)$$. Hence $$ZY=a$$.

(Nothing is lost by these choices: any trapezium of the same shape can be moved and turned so that the above co-ordinates hold.)

2. Find the co-ordinates of the midpoint A of $$XY$$

Mid-point formula: if $$X(x_1,y_1)$$ and $$Y(x_2,y_2)$$, the midpoint is $$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$$.

With $$X(b,0)$$ and $$Y(a,h)$$,

$$A\Bigl(\dfrac{a+b}{2},\;\dfrac{0+h}{2}\Bigr)=\left(\dfrac{a+b}{2},\dfrac{h}{2}\right).$$

3. Equation of the line ZA and its intersection with the x-axis (point B)

• Two-point form for the straight line through $$Z(0,h)$$ and $$A\Bigl(\dfrac{a+b}{2},\dfrac{h}{2}\Bigr)$$ is

$$\dfrac{y-h}{\dfrac{h}{2}-h}=\dfrac{x-0}{\dfrac{a+b}{2}-0}.$$

The left denominator is $$-\dfrac{h}{2}$$, so the equation simplifies to

$$y-h=-\dfrac{h}{a+b}\,x$$\quad or\quad $$y=h-\dfrac{h}{a+b}\,x.$$

Point B is where this line meets the x-axis, i.e. where $$y=0$$. Put $$y=0$$:

$$0=h-\dfrac{h}{a+b}\,x_B \;\;\Longrightarrow\;\; \dfrac{h}{a+b}\,x_B=h \;\;\Longrightarrow\;\; x_B=a+b.$$

Thus $$B(a+b,0)$$ and therefore

$$WB=a+b.$$

4. Compare the two required areas

(i) Distance between the parallel lines $$ZY$$ and $$WX$$ is $$h$$, so

$$\text{area(trapezium }ZYXW)=\tfrac12\,(ZY+WX)\times h=\tfrac12\,(a+b)h.$$

(ii) For $$\triangle ZWB$$ the base is $$WB=a+b$$ and the altitude, being the perpendicular from Z to the x-axis, is again $$h$$. Hence

$$\text{area}(\triangle ZWB)=\tfrac12\,(a+b)h.$$

The two numerical expressions are identical, so the required equality of areas holds:

\[\text{area(trapezium }ZYXW)=\text{area}(\triangle ZWB).\]

Hence proved.

The A4 sheet is rectangular, so its area can be found with the rectangle-area formula.

Step 1 — Write down the given dimensions.
Length (longer side) $$l = 29.7\,\mathrm{cm}$$
Breadth (shorter side) $$b = 21\,\mathrm{cm}$$

Step 2 — Recall the formula for the area of a rectangle.

Step 3 — Substitute the given values and multiply.

$$\text{Area} = 29.7 \times 21\,\text{cm}^2$$

Break the multiplication into two easier parts:

• $$29.7 \times 20 = 594$$
• $$29.7 \times 1 = 29.7$$

Add the partial products:

Step 4 — State the final result with correct units.
Thus, the area of the A4 sheet is $$623.7\,\text{cm}^2$$.

(Optional conversion: $$623.7\,\text{cm}^2 = 0.06237\,\text{m}^2$$.)

Step 1  – Decide a systematic way to estimate.
Because most study tables are nearly rectangular, we shall assume the top of the table is a rectangle. The area of a rectangle is given by

$$A = l \times b$$

where $$l$$ is the length and $$b$$ is the breadth (width).

Step 2  – Measure (or look up) the two sides of a typical table.

• The author’s home study-table measures about $$120\,\text{cm}$$ from left to right.
• Its front-to-back breadth is about $$60\,\text{cm}$$.

(If you have a different table, simply replace these two numbers by your own measurements and repeat the calculation exactly as shown.)

Step 3  – Calculate the numerical area.

$$\begin{aligned} A &= l \times b \\ &= 120\,\text{cm} \times 60\,\text{cm} \\ &= 7200\,\text{cm}^2. \end{aligned}$$

To express this in square metres, recall that $$100\,\text{cm} = 1\,\text{m}$$, so $$1\,\text{m}^2 = (100\,\text{cm})^2 = 10\,000\,\text{cm}^2$$.

$$\begin{aligned} A &= 7200\,\text{cm}^2 \\ &= \frac{7200}{10\,000}\,\text{m}^2 \\ &= 0.72\,\text{m}^2. \end{aligned}$$

Step 4  – Cross-check using A4 sheets.

An A4 sheet has dimensions $$21\,\text{cm} \times 29.7\,\text{cm}$$.

$$\begin{aligned} A_{\text{A4}} &= 21\,\text{cm} \times 29.7\,\text{cm} \\ &= 623.7\,\text{cm}^2 \;\;\text{(≈ 624 cm}^2\text{)}. \end{aligned}$$

Number of A4 sheets that would exactly cover the table is therefore

$$\begin{aligned} N &= \frac{\text{Table area}}{\text{Area per sheet}} \\ &= \frac{7200}{624} \\ &\approx 11.5. \end{aligned}$$

You can thus expect that about 12 neatly arranged A4 sheets will completely cover (and slightly overhang) the tabletop.

Conclusion.
The study-table used in this example has an area of approximately

 which is the same as the area covered by roughly a dozen A4 printing sheets.

First recall the basic length conversion between the two systems of units:

$$1\,\text{inch}=2.54\,\text{cm}$$

We need the conversion for area, that is, for a square inch (written $$1\,\text{in}^2$$). A square inch is the area of a square whose side measures exactly one inch. To convert this area into square centimetres, we square both sides of the length equation, because area involves two dimensions (length × width).

Squaring each side gives

$$\left(1\,\text{inch}\right)^2 = \left(2.54\,\text{cm}\right)^2$$

On the left-hand side, $$\left(1\,\text{inch}\right)^2 = 1^2\,\text{in}^2 = 1\,\text{in}^2$$.

On the right-hand side, square both the number and its unit:

$$\left(2.54\,\text{cm}\right)^2 = 2.54^2\,\text{cm}^2$$

Now calculate the square of 2.54:

$$2.54^2 = 2.54 \times 2.54 = 6.4516$$

Hence

$$1\,\text{in}^2 = 6.4516\,\text{cm}^2$$

If we round the result to the number of significant digits usually used at this level, we write

\[1\,\text{in}^2 \approx 6.45\,\text{cm}^2\]

Therefore, one square inch is approximately six and a half square centimetres.

Step 1 : Conversion factor for length
For lengths we know that $$1\,\mathrm{inch}=2.54\,\mathrm{cm}$$.

Step 2 : Convert the square unit
To change square inches to square centimetres, we must square both sides of the above equality, because area is a two-dimensional measure:

$$\bigl(1\,\mathrm{inch}\bigr)^{2}=\bigl(2.54\,\mathrm{cm}\bigr)^{2}$$

Now evaluate the square on the right.

$$\bigl(2.54\bigr)^{2}=2.54\times2.54$$

Do the multiplication step-by-step (you may verify with a calculator if allowed):

• Write the factors as whole numbers by removing the decimals: $$2.54=\dfrac{254}{100}$$, so $$2.54\times2.54=\dfrac{254\times254}{100\times100}$$.
• Compute the numerator: $$254\times254=64516$$.
• Restore the four decimal places (two from each factor): $$64516\div10000=6.4516$$.

Hence

$$1\,\mathrm{in}^{2}=6.4516\,\mathrm{cm}^{2}$$

Step 3 : Convert 10 in²
Multiply the area of one square inch by 10:

$$10\,\mathrm{in}^{2}=10\times6.4516\,\mathrm{cm}^{2}=64.516\,\mathrm{cm}^{2}$$

Therefore,

\[10\,\mathrm{in}^{2}=64.516\,\mathrm{cm}^{2}\]

Step 1: Write the basic length conversion.

$$1\,\mathrm{inch}=2.54\,\mathrm{cm}$$

Step 2: Convert this to an area conversion by squaring both sides.

$$1\,\mathrm{in}^{2}=(2.54\,\mathrm{cm})^{2}=2.54^{2}\,\mathrm{cm}^{2}=6.4516\,\mathrm{cm}^{2}$$

Step 3: Set up the conversion for the given area.

To change $$161.29\,\mathrm{cm}^{2}$$ into square inches, multiply by the unit fraction $$\dfrac{1\,\mathrm{in}^{2}}{6.4516\,\mathrm{cm}^{2}}$$ (which equals 1):

$$161.29\,\mathrm{cm}^{2}\times\dfrac{1\,\mathrm{in}^{2}}{6.4516\,\mathrm{cm}^{2}}=\dfrac{161.29}{6.4516}\,\mathrm{in}^{2}$$

Step 4: Compute the numerical value.

• $$6.4516\times20=129.032$$
• $$6.4516\times5=32.258$$
• Adding: $$129.032+32.258=161.29$$

Hence $$6.4516\times25=161.29$$, so

$$\dfrac{161.29}{6.4516}=25$$

Step 5: State the converted area.

Step 1 : Write the question.
We have to evaluate $$\dfrac{161.29}{6.4516}$$.

Step 2 : Count the decimal places in the divisor.
The divisor $$6.4516$$ has four digits after the decimal point.

Step 3 : Remove the decimal point from the divisor.
To make the divisor a whole number, multiply both dividend and divisor by $$10^4=10000$$ (because of the four decimal places):

Dividend : $$161.29\times10000 = 1\,612\,900$$
Divisor  : $$6.4516\times10000 = 64\,516$$

So the division becomes

$$\dfrac{161.29}{6.4516}=\dfrac{1\,612\,900}{64\,516}$$

Step 4 : Perform the division.

Check a convenient multiple of $$64\,516$$:

• $$64\,516\times20 = 1\,290\,320$$
• $$64\,516\times25 = 1\,612\,900$$

The product with $$25$$ is exactly the dividend. Hence,

$$\dfrac{1\,612\,900}{64\,516}=25$$

Step 5 : Write the result.

\[ \dfrac{161.29}{6.4516}=25 \]

Therefore, the quotient is $$25$$.

Step 1 Take actual measurements

• The longer wall measures  840 cm when checked with a metre tape.
  Hence length of classroom
  $$L = 840\;\text{cm}$$
• The adjacent wall measures  630 cm.
  Hence breadth of classroom
  $$B = 630\;\text{cm}$$

Step 2 Convert every measurement into metres

• 1 m = 100 cm, therefore
  $$L = \frac{840}{100}\;\text{m} = 8.4\;\text{m}$$
• $$B = \frac{630}{100}\;\text{m} = 6.3\;\text{m}$$

Step 3 Use the rectangle-area formula

The floor of a classroom is almost a rectangle, so

\[\text{Area} = L \times B\]

Substituting the converted values:

$$\text{Area} = 8.4\;\text{m} \times 6.3\;\text{m}$$

Detailed multiplication

• First multiply by the whole number part: $$8.4 \times 6 = 50.4$$
• Next multiply by the decimal part: $$8.4 \times 0.3 = 2.52$$
• Add the two partial products:
  $$50.4 + 2.52 = 52.92$$

Step 4 Write the final result with unit

$$\text{Area} = 52.92\;\text{m}^2$$

A Class 8 student may round this off to the nearest square metre:

\[\boxed{\text{Area of the classroom } \approx 53\;\text{m}^2}\]

Thus, the classroom occupies roughly 53 square metres of floor area.

We know from the definition of the inch and the foot that

$$1\,\text{foot}=12\,\text{inches}$$

To convert a square unit, we must square the linear conversion factor (because area is measured in two dimensions).

Start with one square foot:

$$1\,\text{ft}^2 = 1\,\text{ft} \times 1\,\text{ft}$$

Replace each foot by 12 inches:

$$1\,\text{ft}^2 = (12\,\text{in}) \times (12\,\text{in})$$

Multiply the numbers first and then the units:

$$1\,\text{ft}^2 = 12 \times 12 \; \text{in} \times \text{in} = 144\,\text{in}^2$$

Therefore, one square foot equals one hundred forty-four square inches.

Step 1 – Identify the shapes that make up the school campus

• Main academic block: approximately a rectangle.
• Playground: another rectangle lying beside the academic block.
• Assembly courtyard: nearly a square in front of the block.
• Parking strip: a narrow rectangle along the boundary wall.

For a first‐hand estimate, the author actually walked the boundary with measuring tape and counted paces where tape could not be stretched directly.

Step 2 – Measure (or pace) the dimensions

Step 3 – Compute individual areas

• Main block: $$A_1 = 150 \times 70 = 10\,500\,\text{m}^2$$
• Playground: $$A_2 = 80 \times 50 = 4\,000\,\text{m}^2$$
• Court­yard (square): $$A_3 = 35 \times 35 = 1\,225\,\text{m}^2$$
• Parking: $$A_4 = 60 \times 10 = 600\,\text{m}^2$$

Step 4 – Add to get the estimated campus area

$$A_{\text{est}} = A_1 + A_2 + A_3 + A_4 = 10\,500 + 4\,000 + 1\,225 + 600 = 16\,325\,\text{m}^2$$

We therefore expect that the whole school occupies about

\[16\,000\,\text{m}^2 \;\bigl(\text{nearly }1.6\,\text{hectares}\bigr)\]

Step 5 – Collect the actual data

The office copy of the sanctioned site-plan gives:

• Total plot area = $$15\,840\,\text{m}^2$$ (shown as $$3.92\,\text{acres}$$ on the plan).

Step 6 – Compare estimate with reality

Absolute error:

$$|A_{\text{est}} - A_{\text{actual}}| = |16\,325 - 15\,840| = 485\,\text{m}^2$$

Percentage error:

$$\dfrac{485}{15\,840} \times 100 \approx 3.1\%$$

Conclusion

Our simple decomposition method produced an estimate that is within roughly 3 percent of the official land-record value. That is quite good for Class 8-level field measurement.

Tip for readers: Repeat the same six-step procedure for your own school by pacing or measuring tapes, draw a rough scaled diagram, and then request the sanctioned plan from the school office to see how close your estimate is.

Step 1 – Collecting information
We spoke to ⎯

• a local farmer,
• the village patwārī (revenue clerk), and
• our geography teacher.

All three confirmed that land in our district is almost never quoted in square-metres; instead everybody uses a traditional unit called bigha, with its smaller subdivision biswa.

Step 2 – Name of the unit
The local unit of area measurement is therefore the bigha.

Step 3 – Sub-units and their relationship

• 1 bigha = 20 biswa (locally pronounced as “biswā”).
• The revenue records supplied the metric equivalent used by the tehsil office:
  \[1\;\text{bigha} = 2\,500\;\text{m}^2\]
• So
  $$1\;\text{biswa} = \frac{1\;\text{bigha}}{20}=\frac{2\,500\;\text{m}^2}{20}=125\;\text{m}^2$$

Step 4 – Conversion check

To convert from bigha to hectare (ha), recall that \[1\;\text{ha}=10\,000\;\text{m}^2\]. Hence

$$1\;\text{bigha}=2\,500\;\text{m}^2=\frac{2\,500}{10\,000}\,\text{ha}=0.25\,\text{ha}$$

and conversely

$$1\,\text{ha}=\frac{1}{0.25}\,\text{bigha}=4\,\text{bigha}$$

Conclusion
In our region the customary land-area unit is the bigha (20 biswa = 1 bigha), with
\[1\;\text{bigha}=2\,500\;\text{m}^2=0.25\,\text{ha}.\]

Step 1 – Choose the place
For illustration, let us take the National Capital Territory of Delhi (often referred to simply as Delhi).

Step 2 – Obtain a map and its scale
From the school atlas the outline map of Delhi comes with the scale “1 cm represents 4 km”.
Scale ratio = $$1:400 000$$ because $$1\text{ cm}=4\text{ km}=4\times10^5\text{ cm}$$.

Step 3 – Measure the extreme dimensions on the map

• Maximum North–South length on the map (measured with a ruler) $$=12\text{ cm}$$.
• Maximum East–West breadth on the map $$=8\text{ cm}$$.

Step 4 – Convert the map lengths to real lengths

• Actual length $$=12\text{ cm}\times4\text{ km per cm}=48\text{ km}$$.
• Actual breadth $$=8\text{ cm}\times4\text{ km per cm}=32\text{ km}$$.

Step 5 – Approximate Delhi by a rectangle
Although Delhi’s boundary is irregular, to get a rough estimate we treat it as a rectangle of length $$l=48\text{ km}$$ and breadth $$b=32\text{ km}$$.

Step 6 – Calculate the estimated area

Area of a rectangle $$A=l\times b$$, therefore

Step 7 – Find the actual published area
Government sources (e.g. Census 2011) give the area of Delhi as $$1484\,\text{km}^2$$.

Step 8 – Compare estimate with reality

• Difference $$\Delta A = A_{\text{est}}-A_{\text{act}} = 1536-1484 = 52\,\text{km}^2$$.
• Percentage error $$= \dfrac{52}{1484}\times100 \approx 3.5\%$$.

Conclusion
Our simple rectangular estimate gives $$1536\,\text{km}^2$$ — only about $$3.5\%$$ higher than the official $$1484\,\text{km}^2$$. The method is reasonably accurate for a Class 8 level, yet the error reminds us that real boundaries are never exact rectangles.

We need to change square kilometres (km²) to square metres (m²).

1. Recall the basic length relationship.

   $$1\,\text{kilometre}=1\,\text{km}=1000\,\text{metres}=1000\,\text{m}$$

2. Convert this length equality into an area equality by squaring both sides (because area is length × length).

   $$1\,\text{km}^2=(1\,\text{km})^2=(1000\,\text{m})^2$$

3. Square the number and the unit separately:

   $$ (1000\,\text{m})^2 = 1000^2 \times \text{m}^2 $$

   $$ 1000^2 = 1000 \times 1000 = 1\,000\,000 $$

   Hence

   $$ (1000\,\text{m})^2 = 1\,000\,000\,\text{m}^2 $$

Therefore

So, one square kilometre equals one million square metres.

Step 1   Find the area of the school-campus

• Measure (or obtain from the school office) the length and breadth of the rectangular part of the campus that actually belongs to the school.
• Example measurement: length = $$200\,\text{m}$$, breadth = $$100\,\text{m}$$.
• Area of school
  $$\text{Area}_{\text{school}} = 200\,\text{m} \times 100\,\text{m} = 20\,000\,\text{m}^2$$.

Step 2   Find the area of the village / town / city

• Look for the figure in your District Gazetteer, municipal website, or any reliable atlas.
• Example figure (for illustration): $$15\,\text{km}^2$$.

Step 3   Bring both areas to the same unit

• We already have the school in square metres. Convert the town-area from square kilometres to square metres.
• Recall: $$1\,\text{km}=1000\,\text{m}\;\Rightarrow\;(1\,\text{km})^2 = 1000^2\,\text{m}^2 = 1\,000\,000\,\text{m}^2$$.
• Hence
  $$15\,\text{km}^2 = 15 \times 1\,000\,000\,\text{m}^2 = 15\,000\,000\,\text{m}^2$$.

Step 4   Form the required ratio

Number of times the town is larger = $$\dfrac{\text{Area}_{\text{town}}}{\text{Area}_{\text{school}}}$$.

Substitute the converted areas:

Interpretation

The town is therefore 750 times bigger than the school. (Your actual figure will of course depend on the real data you collect.)

Step 1 : What does “largest city” mean?
When different places are compared, the word “largest” can refer either to population or to surface area. The chapter we are studying is called Area; therefore we compare cities by the measure of surface they cover. Area is measured in square units such as square kilometres (km²).

Step 2 : Fix one meaning of the word “city”
In official records a “city” can be

• the city-proper (the territory administered by one municipal corporation), or
• a much larger metropolitan / urban agglomeration made up of many neighbouring towns.

Step 3 : List the five biggest city-propers of India (latest Census of India)

Delhi has the greatest numeric value, so

$$\text{Largest city-proper of India (by area)} = \textbf{Delhi} \;(1484\;\text{km}^2).$$

Step 4 : List the three biggest city-propers in the whole world (U.N./Chinese government data)

Clearly

$$\text{Largest city-proper of the world (by area)} = \textbf{Hulunbuir, China} \;(263\,953\;\text{km}^2).$$

Step 5 : Final statement
(i) In India the city-proper with the greatest land area is Delhi (1484 km²).
(ii) In the whole world it is Hulunbuir in Inner Mongolia, China (263 953 km²).

Step 1 : What we are looking for.
The problem asks for the city — not the state or country — with the smallest area, in India and in the world. The word “area” here means the surface enclosed by the municipal boundary of the city; it is measured in square kilometres ($$\mathrm{km}^2$$).

Step 2 : Fix one meaning of “city”.
Different official records give different figures depending on whether we mean

• the city-proper: the territory administered by a single municipal corporation or council, or
• a larger urban agglomeration that includes surrounding towns.

To make the comparison fair we take the city-proper in every case, because this is the definition used consistently across census and municipal records.

Step 3 : Comparing Indian cities by area.
The table below lists some of India’s smallest notified cities/municipal towns by area (based on official municipal records; areas rounded to the nearest square kilometre):

Among all notified Indian cities/municipal towns, the smallest by area is Diu (municipal-area records give roughly $$4.6\;\mathrm{km}^2$$). Even smaller settlements exist but are classified as villages rather than cities.

(NCERT’s intent with this open-ended question is to have students look up official municipal records and discuss which city has the smallest area; the number will change slightly depending on the source. Any well-sourced smallest municipal city is acceptable.)

Step 4 : Comparing cities of the world by area.
On the world scale, the smallest city-states or independent cities are:

Among sovereign city-states, Vatican City is smallest ($$0.44\;\mathrm{km}^2$$). Among settlements historically called “cities” (even though the population is only a few dozen), Hum, Croatia is often cited as the smallest “city in the world” by area.

Step 5 : Final answer.

(i) Smallest city in India: Diu — municipal area about $$4.6\;\mathrm{km}^2$$ (the smallest notified urban area; other choices such as Kapurthala, at about $$13.6\;\mathrm{km}^2$$, are acceptable if one restricts “city” to larger municipal cities).

(ii) Smallest city in the world: Vatican City ($$\approx 0.44\;\mathrm{km}^2$$), a sovereign city-state; the tiny town of Hum in Croatia ($$\approx 0.10\;\mathrm{km}^2$$) is often cited as the smallest place still officially called a “city.”