Chapter 6: Perimeter and Area

First, write down the measurements of the rectangular table-cloth:

• Length (ℓ) = $$3\,\text{m}$$
• Breadth (b) = $$2\,\text{m}$$

To put lace all around the edge, Akshi needs the perimeter of the rectangle.

Formula for the perimeter of a rectangle:

$$\text{Perimeter} = 2(\text{length} + \text{breadth})$$

Substitute the given values:

$$\text{Perimeter} = 2(3\,\text{m} + 2\,\text{m})$$

Add the numbers inside the bracket first (according to the order of operations):

$$3\,\text{m} + 2\,\text{m} = 5\,\text{m}$$

Now multiply by 2:

$$2 \times 5\,\text{m} = 10\,\text{m}$$

Therefore, the length of lace required to go all around the table-cloth is

\[10\,\text{metres}\]

Step 1 — Identify the side of the square park

The length of each side is given as $$75\,\text{m}$$.

Step 2 — Find the perimeter of the square

For any square,

$$\text{Perimeter} = 4 \times \text{side}$$

Substituting the side,

$$\text{Perimeter} = 4 \times 75\,\text{m}$$

$$\text{Perimeter} = 300\,\text{m}$$

Step 3 — Find the distance for three rounds

One round = perimeter, so three rounds are

$$\text{Distance} = 3 \times 300\,\text{m}$$

$$\text{Distance} = 900\,\text{m}$$

Conclusion

Usha travels a total distance of

\[900\,\text{metres}\]

Step 1 — Find the length of the wire.

The wire first forms a rectangle whose length is $$5\,\text{cm}$$ and breadth is $$3\,\text{cm}$$.

Perimeter of a rectangle is

$$\text{Perimeter}=2(\text{length}+\text{breadth})$$

Substituting the numbers,

$$\text{Perimeter}=2(5+3)=2\times8=16\,\text{cm}$$

So the wire is exactly $$16\,\text{cm}$$ long.

Step 2 — Let the wire form a square.

Suppose the side of the new square is $$s\,\text{cm}$$. A square has all four sides equal, hence its perimeter is

$$\text{Perimeter}=4s$$

The same piece of wire is used, so its length (the perimeter) stays $$16\,\text{cm}$$:

$$4s = 16$$

Step 3 — Solve for the side $$s$$.

Divide both sides of the equation by $$4$$:

$$s = \frac{16}{4} = 4$$

Thus the side length of the square is

Therefore, when the wire is bent into a square, each side of the square measures 4 cm.

Step 1 – Recall the idea of perimeter
For any triangle, the perimeter is the sum of the lengths of its three sides.

Step 2 – Introduce a variable for the unknown side
Let the third side be $$x$$ cm.

Step 3 – Translate the word statement into an equation
Since the perimeter is 55 cm and two of the sides are 20 cm and 14 cm, we write

$$20 + 14 + x = 55$$

Step 4 – Simplify the known sum
Add the known lengths on the left:

$$34 + x = 55$$

Step 5 – Isolate the variable
Subtract 34 from both sides:

$$x = 55 - 34$$

Step 6 – Compute the difference

$$x = 21$$

Step 7 – State the result clearly

The length of the third side of the triangle is

\[x = 21 \text{ cm}\]

Given: Length $$l = 150 \text{ m}$$, breadth $$b = 120 \text{ m}$$; fencing cost = ₹40 per metre.

Step 1 – Find the perimeter of the rectangular park.

For a rectangle, the perimeter $$P$$ is the sum of all four sides:

$$P = 2 \times (l + b)$$

Substituting the given values:

$$P = 2 \times (150 \text{ m} + 120 \text{ m})$$

$$P = 2 \times 270 \text{ m}$$

So,

\[P = 540 \text{ m}\]

Step 2 – Calculate the total cost of fencing.

Cost per metre = ₹40.

Total cost $$C$$ is:

$$C = P \times (\text{cost per metre})$$

$$C = 540 \text{ m} \times \text{₹}40/\text{m}$$

First multiply the numbers, ignoring the units:

$$540 \times 40 = 540 \times 4 \times 10 = 2160 \times 10 = 21600$$

Therefore,

\[C = \text{₹}\,21\,600\]

Conclusion: It will cost ₹21,600 to fence the park once all around.

The shape of the farmer’s field is a rectangle.

Step 1: Write the given dimensions.
Length of the rectangle: $$l = 230 \text{ m}$$
Breadth of the rectangle: $$b = 160 \text{ m}$$

Step 2: Recall the perimeter formula for a rectangle.
For a rectangle, $$\text{Perimeter} = 2\,(l + b).$$

Step 3: Substitute the given values.
$$l + b = 230 \text{ m} + 160 \text{ m} = 390 \text{ m}$$
Therefore, $$\text{Perimeter} = 2 \times 390 \text{ m} = 780 \text{ m}.$$

Step 4: Account for three rounds of fencing.
To fence the field three times around, multiply the perimeter by 3:
$$\text{Total rope} = 3 \times 780 \text{ m} = 2340 \text{ m}.$$

Conclusion.
The farmer needs a total of 2340 metres of rope.

First, find the length of one complete round (the perimeter) of the outer rectangular track that Akshi is running on.

The outer track is a rectangle with
• length = $$70\,\text{m}$$
• breadth = $$40\,\text{m}$$

The perimeter $$P$$ of a rectangle is given by:
$$P = 2 \times (\text{length} + \text{breadth})$$

Substitute the given measurements:

$$P = 2 \times (70\,\text{m} + 40\,\text{m})$$

$$P = 2 \times 110\,\text{m}$$

$$P = 220\,\text{m}$$

So, Akshi covers $$220\,\text{m}$$ in one round.

Akshi completes 5 rounds. The total distance $$D$$ she runs is:

$$D = 5 \times 220\,\text{m}$$

$$D = 1100\,\text{m}$$

Therefore, Akshi covers a total distance of 1100 metres in 5 rounds.

Step 1 – Perimeter of Toshi’s track (a rectangle)
The rectangle is $$60\,\text{m}$$ long and $$45\,\text{m}$$ wide.
Perimeter of a rectangle = $$2\times(\text{length}+\text{breadth})$$.

$$\text{Perimeter}=2\times(60\;\text{m}+45\;\text{m})$$

$$\text{Perimeter}=2\times105\;\text{m}$$

$$\text{Perimeter}=210\;\text{m}$$

Step 2 – Distance covered by Toshi in 7 rounds

Distance in one round = $$210\;\text{m}$$.
Distance in 7 rounds = $$210\;\text{m}\times7$$.

$$210\times7=1470$$

So, Toshi covers $$1470\;\text{m}$$.

Step 3 – Distance covered by the other runner (Bansi)
Bansi runs around a square of side $$75\,\text{m}$$ and completes 4 rounds.

Perimeter of a square = $$4\times\text{side}$$.

$$\text{Perimeter}=4\times75\;\text{m}=300\;\text{m}$$

Distance in 4 rounds = $$300\;\text{m}\times4=1200\;\text{m}$$.

Step 4 – Comparing the two distances

Toshi’s distance = $$1470\;\text{m}$$
Bansi’s distance = $$1200\;\text{m}$$.

Since $$1470\;\text{m}>1200\;\text{m}$$, Toshi ran the longer distance.

Step 1 – Write the perimeters of the two square tracks
Inner track side = 100 m ⇒ perimeter
$$P_1 = 4 \times 100 = 400 \text{ m}$$
Outer track side = 150 m ⇒ perimeter
$$P_2 = 4 \times 150 = 600 \text{ m}$$

Step 2 – Decide how much of each lap must be covered
Required race length = 350 m.
Therefore distance still un-run in one full lap is

• Inner track : $$400 - 350 = 50 \text{ m}$$
• Outer track : $$600 - 350 = 250 \text{ m}$$

So each runner has to start exactly this much behind the common finishing line, measured along the direction in which they will run.

Step 3 – Locate starting point A on inner track
The finishing line is at the centre of one side (side length = 100 m).
Half that side (centre to corner) = 50 m.
Because the required setback is also 50 m, point A is exactly the corner that lies immediately before the finishing-line side (when moving in the running direction).

Step 4 – Locate starting point B on outer track
Required setback on outer track = 250 m.
Move back from the finishing line:

1. 50 m to reach the first corner (centre → corner).
2. 150 m along the next full side to the second corner (total 200 m so far).
3. Another 50 m along the third side to its centre (total 250 m).

Hence point B is the mid-point of the third side counted in the backward direction from the finishing-line side.

Step 5 – What to draw
Draw the two concentric squares. Mark the common finishing line at the centre of the bottom side (for example).

• On the inner square mark point A at the bottom-left corner.
• On the outer square mark point B at the centre of the left side (the third side reached when going anti-clockwise from the bottom-middle).

Objective  To get a feel for the idea of perimeter by first guessing (estimating) the boundary length of a shape and then finding its actual value with a measuring tool.

Material used  Old newspaper sheet, scissors, 15 cm ruler, measuring-tape, pencil, string (for curved sides).

How the activity was done

1. Cutting the shapes
   We cut three different pieces:
   • Shape A – a roughly rectangular piece, nearly 16 cm long and 9 cm wide.
   • Shape B – a right-angled triangular piece that looked like the well-known 5–12–13 triangle.
   • Shape C – an irregular 5-sided piece (no two sides equal; one side slightly curved).
2. Estimation
   For each shape we ran a finger all round the edge and compared each side with the 15 cm scale we had in hand. We wrote down the guessed lengths, then added them mentally.
   • Shape A (rectangle) – we felt the longer side was a bit longer than the scale, so we guessed $$16\,\text{cm}$$; the shorter side looked a little more than half the scale so we guessed $$9\,\text{cm}$$. Hence
     $$\text{Estimated perimeter}=2\times(16+9)=50\,\text{cm}$$.
   • Shape B (triangle) – one side fitted one-third of the tape (≈5 cm), the second side almost the full 12 cm of the ruler, the hypotenuse a shade longer than the ruler (≈13 cm). So
     $$\text{Estimated perimeter}=5+12+13=30\,\text{cm}.$$
   • Shape C (irregular) – moving point by point we guessed the sides as 6 cm, 4 cm, 5 cm, 7 cm and 6 cm (the tiny curve was felt equal to about 4 cm). Therefore
     $$\text{Estimated perimeter}=6+4+5+7+6=28\,\text{cm}.$$
3. Actual measurement
   We now placed the ruler (for straight parts) and a tight piece of string (for the curved corner in Shape C) exactly on every edge, marked the end point and read the scale. All readings were written in a table.

Verification

• Shape A : estimated $$50\,\text{cm}$$, measured $$50\,\text{cm}$$  →  exact match.
• Shape B : estimated $$30\,\text{cm}$$, measured $$30\,\text{cm}$$  →  exact match.
• Shape C : estimated $$28\,\text{cm}$$, measured $$28\,\text{cm}$$  →  exact match.

What we learnt The perimeter of a closed figure is simply the sum of the lengths of all its sides, straight or curved. With a little practice our “eye-estimates” can become quite close to the actual values.

Step 1 : Identify the three sides on the dot–grid.

Each small square of the grid has side $$1$$ unit.

• The horizontal side joins two dots that are $$4$$ squares apart, so its length is $$4$$ units.
• The vertical side joins two dots that are $$3$$ squares apart, so its length is $$3$$ units.
• The third (slanting) side joins the free ends of these two sides. Together, the three sides form a right-angled triangle, because the horizontal and vertical sides meet at a right angle on the grid.

Step 2 : Find the length of the slanting side.

For a right-angled triangle, the length of the slanting side (the hypotenuse) is given by

$$(\text{slanting side})^{2} = (\text{horizontal side})^{2} + (\text{vertical side})^{2}$$

$$(\text{slanting side})^{2} = 4^{2} + 3^{2} = 16 + 9 = 25$$

Therefore, the slanting side $$= \sqrt{25} = 5$$ units.

Step 3 : Calculate the perimeter.

$$\text{Perimeter} = 4 + 3 + 5 = 12\text{ units}$$

The key result is

\[P = 12\;\text{units}\]

Step 4 : Whose claim is correct?

Akshi's answer (9 units) is too small because she did not measure the slanting side correctly (she seems to have counted only the horizontal and vertical sides, missing the slanting one, or counted it as a short diagonal of one square).

Toshi is correct: the perimeter is more than 9 units. In fact, the perimeter is exactly 12 units, because the slanting side of a right-angled triangle on the grid is always longer than either of the two perpendicular sides.

Let one horizontal/vertical step on the dot grid be a “straight unit’’ and denote its length by the symbol $$s$$.
Let one 45° step on the grid be a “diagonal unit’’ and denote its length by the symbol $$d$$.
To obtain the perimeter of any drawing we only have to count how many straight and how many diagonal units have been used and then form the sum.

(i) Letter F
• A vertical backbone of 4 straight units.
• A top arm of 3 straight units.
• A middle arm of 2 straight units.
There are no slanting lines.
Number of straight units  = $$4+3+2=9$$,  number of diagonal units = 0.
Therefore $$P_F = 9s$$.

(ii) Letter M
• Two vertical sides, each 4 straight units  ⇒ $$2\times4s = 8s$$.
• Two slanting strokes, each 2 diagonal units  ⇒ $$2\times2d = 4d$$.
Hence $$P_M = 8s + 4d$$.

(iii) Letter U
• Two vertical sides, each 3 straight units  ⇒ $$2\times3s = 6s$$.
• Two bottom slants, each 1 diagonal unit  ⇒ $$2\times1d = 2d$$.
Thus $$P_U = 6s + 2d$$.

(iv) Letter N
• Two vertical sides, each 4 straight units  ⇒ $$2\times4s = 8s$$.
• One long diagonal made of 4 diagonal units  ⇒ $$4d$$.
Consequently $$P_N = 8s + 4d$$.

So the required perimeters, expressed in the short form “straight + diagonal”, are listed below.

Step 1 : Recall what “regular” means

A regular shape has all sides equal in length. When the sides are equal, the perimeter can be found just by counting how many sides there are and then multiplying by the common side length.

Step 2 : Collect some regular objects from the surroundings

1. Square paper coaster
   Measured side length  =  $$8\,\text{cm}$$ (all four sides equal).

   Perimeter  =  $$8\,\text{cm}+8\,\text{cm}+8\,\text{cm}+8\,\text{cm}$$
   =  $$4\times8\,\text{cm}=32\,\text{cm}$$.

2. Equilateral–triangle traffic sign
   Measured side length  =  $$30\,\text{cm}$$ (three equal sides).

   Perimeter  =  $$30\,\text{cm}+30\,\text{cm}+30\,\text{cm}$$
               =  $$3\times30\,\text{cm}=90\,\text{cm}$$.

3. Regular hexagonal floor tile
   Measured side length  =  $$15\,\text{cm}$$ (six equal sides).

   Perimeter  =  $$6\times15\,\text{cm}=90\,\text{cm}$$.

Step 3 : Look for a pattern

• Square  ($$4$$ sides)  →  $$P=4\times s$$.
• Equilateral triangle  ($$3$$ sides)  →  $$P=3\times s$$.
• Regular hexagon  ($$6$$ sides)  →  $$P=6\times s$$.

In every case we are multiplying the common side length by the number of sides.

Step 4 : General rule for any regular polygon

Let

• $$n$$ = number of equal sides (so a square has $$n=4$$, a pentagon $$n=5$$, a hexagon $$n=6$$, …)
• $$s$$ = length of each side.

Adding all $$n$$ equal sides gives

Thus,

Perimeter of any regular $$n$$-sided polygon = $$n \times s$$.

Step 5 : Check the formula on the objects we measured

Our measurements confirm the general rule. Whenever you see a regular polygon, just multiply the number of its equal sides by the common side length to get its perimeter.

Given : A rectangle of size 6 cm × 4 cm is cut into two equal parts in the middle of its 6 cm side.

1. Size of each piece
Width is halved : $$\frac{6\,\text{cm}}{2}=3\,\text{cm}$$
So every part is a rectangle 3 cm × 4 cm.

2. How can the two pieces be joined again?

• Join the 4 cm sides → we get the old 6 cm × 4 cm rectangle.
    Perimeter  $$=2(6+4)=20\,\text{cm}\;(\neq 22\,\text{cm}).$$
• Join the 3 cm sides → one piece is placed exactly above the other.
    New length $$=4\,\text{cm}+4\,\text{cm}=8\,\text{cm},$$ breadth $$=3\,\text{cm}.$$
    Perimeter   \[P = 2(8+3)=2\times11=22\,\text{cm}.\]

3. Required arrangement
Place the two 3 cm × 4 cm rectangles one above the other so that their 3 cm edges overlap. The resulting 3 cm × 8 cm rectangle has the demanded perimeter of 22 cm.

Diagram (to be drawn) : Draw a rectangle 3 cm wide and 8 cm high; mark the join halfway along the 8 cm side to show the two original pieces.

Given: Length of the floor $$l = 5\,\text{m}$$ and breadth $$b = 4\,\text{m}$$.
Side of the square carpet $$s = 3\,\text{m}$$.

1. Area of the rectangular floor

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

Substituting: $$\text{Area of floor} = 5 \times 4 = 20\,\text{m}^2$$

2. Area of the square carpet

Formula: $$\text{Area of square} = s \times s$$ (that is $$s^{2}$$)

Substituting: $$\text{Area of carpet} = 3 \times 3 = 9\,\text{m}^2$$

3. Area of the floor that is not carpeted

$$\text{Uncovered area} = \text{Area of floor} - \text{Area of carpet}$$

$$= 20\,\text{m}^2 - 9\,\text{m}^2 = 11\,\text{m}^2$$

Therefore, the area of the floor that is not carpeted is $$11\,\text{m}^2$$.

Step 1 : Find the area of the whole rectangular piece of land

Its length is 12 m and its breadth is 10 m.

Area of a rectangle  =  $$\text{length}\times\text{breadth}$$

So,

$$\text{Area of land}=12\,\text{m}\times10\,\text{m}=120\,\text{m}^2$$

Step 2 : Find the area of one square flower bed

Each side of a square bed is 4 m.

Area of a square  =  $$\text{side}\times\text{side}$$

Therefore,

$$\text{Area of one bed}=4\,\text{m}\times4\,\text{m}=16\,\text{m}^2$$

Step 3 : Find the total area covered by the four beds

There are 4 identical beds, so

$$\text{Total bed area}=4\times16\,\text{m}^2=64\,\text{m}^2$$

Step 4 : Find the area of the remaining land

Remaining area  =  $$\text{(area of land)}-\text{(total bed area)}$$

$$\text{Remaining area}=120\,\text{m}^2-64\,\text{m}^2=56\,\text{m}^2$$

Answer

The area of the remaining part of the land is

Step 1 – Write down the data.
Length of the garden: $$L = 25\,\text{m}$$
Area of the garden: $$A = 300\,\text{m}^2$$

Step 2 – Recall the area formula for a rectangle.
For any rectangle, the area is given by
$$A = L \times B$$
where $$B$$ is the width (also called breadth).

Step 3 – Solve the formula for the unknown width $$B$$.
Divide both sides by $$L$$:
$$B = \frac{A}{L}$$

Step 4 – Substitute the known values.
$$B = \frac{300}{25}\,\text{m}$$

Step 5 – Carry out the division.
300 ÷ 25 = 12 because 25 × 12 = 300.
Hence $$B = 12\,\text{m}$$.

Conclusion.

First find the area of the rectangular plot.

Length = $$500\,\text{m}$$,   Breadth = $$200\,\text{m}$$

Area = Length × Breadth

$$\text{Area}=500\,\text{m}\times200\,\text{m}=100\,000\,\text{m}^2$$

The cost of tiling is quoted at $$\text{₹}8$$ per $$100\,\text{m}^2$$.

First compute how many blocks of $$100\,\text{m}^2$$ are there in the total area:

Number of $$100\,\text{m}^2$$ blocks = $$\dfrac{100\,000}{100}=1\,000$$

Now multiply by the rate:

$$\text{Total cost}=1\,000\times\text{₹}8=\text{₹}8\,000$$

Hence the cost of tiling the plot is

\[\boxed{\text{₹}\,8\,000}\]

Step 1 – Write down the dimensions of the grove
Length = $$100\,\text{m}$$,  Breadth = $$50\,\text{m}$$.

Step 2 – Find the area of the rectangular grove
The area $$A$$ of a rectangle is given by
$$A = \text{length} \times \text{breadth}$$.
Substituting the given numbers,
$$A = 100\,\text{m} \times 50\,\text{m} = 5000\,\text{m}^2.$$

Step 3 – Write down the area needed for one coconut tree
Each tree requires $$25\,\text{m}^2$$ of space.

Step 4 – Calculate the number of trees that can fit
Number of trees $$= \displaystyle \frac{\text{Total area of grove}}{\text{Area required per tree}}$$
$$= \frac{5000\,\text{m}^2}{25\,\text{m}^2} = 200.$$

Step 5 – Interpret the result
Since we cannot plant a fraction of a tree, the whole-number result $$200$$ already represents the maximum possible.

Therefore, the greatest number of coconut trees that can be planted in the grove is

Step 1 : Pick a small piece as the unit of area.

Look at the tangram and notice that pieces F and G are the two smallest pieces — both are small right-angled triangles of exactly the same size. Let the area of one such small triangle be our unit of area, and call it 1 unit.

$$\text{Area of F} = \text{Area of G} = 1\text{ unit}.$$

Step 2 : Find how many small triangles fit into each of the other pieces.

Place the small triangles (F and G) one by one over each of the remaining pieces. You will see that:

• The small square C is exactly covered by 2 small triangles, so its area is 2 units.
• The medium triangle D is exactly covered by 2 small triangles, so its area is 2 units.
• The parallelogram E is exactly covered by 2 small triangles, so its area is 2 units.
• Each of the large triangles A and B is exactly covered by 4 small triangles, so each has area 4 units.

Step 3 : Put the counts in a table.

Adding all seven areas: $$4+4+2+2+2+1+1 = 16$$ small-triangle units, which is exactly the area of the whole tangram square. So the count is consistent.

Step 4 : Compare the areas.

• Pieces A and B each have area 4 units, so they have the same area.
• Pieces C, D and E each have area 2 units, so these three pieces have the same area.
• Pieces F and G each have area 1 unit, so they have the same area.

Conclusion. The seven tangram pieces fall into three groups of equal area:

\[ \text{(i)}\; A \text{ and } B, \qquad \text{(ii)}\; C,\, D,\, E, \qquad \text{(iii)}\; F \text{ and } G. \]

Step 1 – Find the area of Shape C

The figure is drawn on a square grid in which every small square represents
1 square unit. By counting the shaded squares in Shape C we get

$$\text{Area of Shape C}=6\;\text{square units}$$

Step 2 – Find the area of Shape D

Counting the shaded squares inside Shape D in the same way gives

$$\text{Area of Shape D}=12\;\text{square units}$$

Step 3 – Compare the two areas

To know how many times bigger Shape D is than Shape C, divide the two areas:

$$\dfrac{\text{Area of D}}{\text{Area of C}}=\dfrac{12}{6}=2$$

So Shape D is twice (2 times) as big as Shape C.

Step 4 – Find the area of Shape E

Again counting the grid squares for Shape E:

$$\text{Area of Shape E}=18\;\text{square units}$$

Step 5 – Write the relationship among C, D and E

• Shape C → 6 sq units
• Shape D → 12 sq units
• Shape E → 18 sq units

Notice that

$$12=2\times6\quad\text{and}\quad18=3\times6$$

Therefore

• Shape D is made of 2 copies of Shape C.
• Shape E is made of 3 copies of Shape C.
• Consequently, Shape E is $$\dfrac{18}{12}=\tfrac{3}{2}$$ times (1.5 times) Shape D.

In ratio form the three areas are

$$\text{C} : \text{D} : \text{E}=6:12:18=1:2:3$$

So the three shapes are simply successive enlargements of the same basic rectangle: each time we join one more copy of Shape C we pass from C to D to E.

Idea of method A square grid is drawn below every figure. One small square of the grid represents exactly one square unit. Hence, the area of a figure is obtained simply by adding up

• every grid square that lies completely inside the figure (count = 1 each), and
• all the partial grid squares cut by the outline. Two half–squares are taken to be equal to one full square.

We now apply this procedure separately to Shape D and Shape F.

1. Counting the completely filled squares

• Shape D encloses 10 whole grid squares.
• Shape F encloses 12 whole grid squares.

2. Counting the partly filled squares

When the boundary of a shape passes through a square, we estimate the fraction that is inside the figure; two halves make one whole.

• Shape D: 6 half-squares ⇒ $$\frac{6}{2}=3$$ full squares
• Shape F: 4 half-squares ⇒ $$\frac{4}{2}=2$$ full squares

3. Total area of each figure

4. Comparison

Because

Shape F occupies a larger region.

Reason: On the common grid Shape F covers one extra unit square compared with Shape D; therefore Shape F has the greater area.

Step 1 – Copy the two shapes on square paper
Both Shape F and Shape G in the textbook are drawn on the same square grid. Each tiny square of the grid represents one square unit of area.

Step 2 – Count the squares that are fully inside each shape

Step 3 – Count the half-filled (approximately) squares
A square that is roughly half–covered is taken as $$\tfrac12$$ of a square unit.

Step 4 – Convert half-squares into whole squares

• For Shape F: $$6 \times \tfrac12 = 3$$ full squares.
• For Shape G: $$4 \times \tfrac12 = 2$$ full squares.

Step 5 – Find the total area of each shape

• Shape F
  $$\text{Area of F} = 11 + 3 = 14\;\text{square units}$$
• Shape G
  $$\text{Area of G} = 13 + 2 = 15\;\text{square units}$$

Step 6 – Compare
Since $$15 \gt 14$$, Shape G covers the larger surface.

Reason
The comparison is made by counting how many full grid squares (and equivalent full squares formed from half-covered ones) lie inside each outline. The shape containing the greater total number of unit squares has the larger area.

Step 1 – Choosing a unit of area
Let the area of one of the smallest triangles (Shape C or Shape E) be taken as one unit.

So  $$\text{Area of Shape C}=\text{Area of Shape E}=1\;\text{unit}.$$

Step 2 – Areas we already know from the hint

• Shape D can be exactly covered by Shapes C and E, therefore
  $$\text{Area of Shape D}=1+1=2\;\text{units}.$$
• Shapes A and B are congruent, hence have equal areas (exact value to be found next).

Step 3 – Finding the area of Shape G
Place two small triangles (for example, C and E) on Shape G; they fit it perfectly without gaps or overlaps.

Hence  $$\text{Area of Shape G}=1+1=2\;\text{units}.$$

Step 4 – Finding the area of Shape A
Arrange four small triangles so that their hypotenuses form the boundary of one large right-angled isosceles triangle. That large triangle is exactly Shape A.

Therefore  $$\text{Area of Shape A}=1+1+1+1=4\;\text{units}.$$

Step 5 – Comparing the two areas

So Shape A covers twice as much area as Shape G (not four times).

Let the area of the little square (Shape C) be denoted by $$A_C\,(\text{square units}).$$

The seven tangram pieces and their areas compared with $$A_C$$ are:

Add the areas of all seven pieces:

$$\text{Area of big square}=\big(\underbrace{2A_C}_{\text{large}}+\underbrace{2A_C}_{\text{large}}\big)+\underbrace{A_C}_{\text{medium}}+\underbrace{A_C}_{\text{parallelogram}}+\underbrace{A_C}_{\text{square C}}+\big(\underbrace{\tfrac{1}{2}A_C}_{\text{small}}+\underbrace{\tfrac{1}{2}A_C}_{\text{small}}\big).$$

Simplifying step by step (each step written out for the learner):

$$2A_C+2A_C=4A_C,$$

$$\tfrac{1}{2}A_C+\tfrac{1}{2}A_C=A_C,$$

so

$$4A_C+A_C+A_C+A_C=8A_C.$$

This deserves a highlighted final line:

\[\boxed{\text{Area of the big square}=8\,A_C}\]

In words: the big square formed by all seven tangram pieces has an area eight times the area of Shape C.

Step 1 – Understand the pieces
In the tangram the 7 pieces are named A, B, C, D, E, F and G.
For comparison we will take the area of Shape C as one unit and call it $$1\text{ C-unit}$$.

Every tangram piece can be built up out of the smallest right-isosceles triangle that occurs in the figure. Let the area of that smallest triangle be $$1\text{ T-unit}$$. Counting those triangles inside each piece (you can actually cut out the pieces from paper and place the triangles on them) gives:

(Because Shape C itself contains exactly 2 of the smallest triangles, $$1\text{ C-unit}=2\text{ T-units}$$.)

Step 2 – Total area of the 7 pieces
Add the areas expressed in C-units:

$$\text{Total area}=2C+2C+1C+1C+\dfrac{1}{2}C+\dfrac{1}{2}C+1C$$
$$\phantom{\text{Total area}}=(2+2+1+1+0.5+0.5+1)C$$
$$\phantom{\text{Total area}}=8C$$

Step 3 – Forming the rectangle
When you physically place the seven pieces edge to edge (for example, keep the two large triangles along the long sides, put the square and parallelogram in the middle and fill the corners with the three remaining triangles) they exactly cover a perfect rectangle. Rearrangement does not change the amount of paper used, so the area of that rectangle is still the total area calculated in Step 2.

Step 4 – Conclusion

Step 1 – Recall what “area” means

The area of a closed figure is the measure of the region inside its boundary. If two figures have the same boundary length (perimeter), the one that spreads farther away from its centre in every direction will usually cover more region and hence have the larger area.

Step 2 – Observe the two outlines

• Figure (a) is a smooth, wavy curve that swells outward almost everywhere. There are no deep in-cuts; the boundary is always “pushing out”.
• Figure (b) is jagged and star-like. Between every two outward spikes there is a sharp inward dent. These dents pull the boundary back toward the centre and, therefore, leave some space that is not included in the interior.

Step 3 – Reason out which one encloses more region

Because of the inward dents in figure (b), the region it encloses is “eaten away” many times. Figure (a) keeps most of those portions because its boundary does not dip sharply inside. Hence the region inside figure (a) must be larger.

Step 4 – (Optional practical check)

If you trace both figures on squared paper and count the complete and half squares inside each outline (a standard Class 6 method of estimating area), you will find that the count for figure (a) is greater.

Conclusion

Therefore, without any exact calculation we can confidently guess that figure (a) has the larger area.

Given: The four letters F, M, U and N are drawn on a square-dot grid.
Each smallest square on the grid is a unit square, so its area is $$1\text{ square unit}$$.
To get the area of a letter we only have to count how many unit squares (or half–squares that make one full square) lie completely inside its boundary.

Below, for every letter we first state how the counting is done and then write the calculation very clearly.

(i) Letter F

• Look at the long vertical strip of F – it covers $$5$$ unit squares.
• The top horizontal strip is $$5$$ squares long. One of those squares (the one at the left end) has already been counted in the vertical strip, so we add only the remaining $$5-1=4$$ squares.
• The middle horizontal strip is $$3$$ squares long. Again the first of these has already been counted, so we add $$3-1=2$$ squares.

Total number of squares for F

Therefore,

Area of F = $$11\text{ square units}$$.

(ii) Letter M

• The two outer vertical strips contribute $$5+5=10$$ unit squares.
• The two slanting strokes that meet in the middle contribute $$3$$ squares each. (You may mark them lightly and count one by one.) None of these six squares has been counted before.

Total number of squares for M

Therefore,

Area of M = $$16\text{ square units}$$.

(iii) Letter U

• Each of the two side strips contains $$5$$ full squares, so together they give $$5+5=10$$ squares.
• The bottom horizontal strip is $$5$$ squares long. The two corner squares are already in the side strips, so we add only $$5-2=3$$ new squares.

Total number of squares for U

Therefore,

Area of U = $$13\text{ square units}$$.

(iv) Letter N

• The left and right vertical strips together give $$5+5=10$$ unit squares.
• The slanting strip joining the top of the left bar to the bottom of the right bar covers $$4$$ more squares that have not yet been counted.

Total number of squares for N

Therefore,

Area of N = $$14\text{ square units}$$.

Final result

Step 1 — Make the rectangle

1. Draw any rectangle ABCD on a sheet of paper (the longer sides are $$AB$$ and $$CD$$, the shorter sides are $$BC$$ and $$DA$$).
2. Measure its length $$l$$ and breadth $$b$$ with a ruler so that you know its exact size. For example, let us first take $$l = 8\,\text{cm}$$ and $$b = 5\,\text{cm}$$.

Step 2 — Draw a diagonal and cut

1. With a pencil join opposite corners, say join $$A$$ to $$C$$. This segment $$AC$$ is called a diagonal.
2. Cut the rectangle carefully along $$AC$$. You now have two separate pieces — two triangles: △ABC and △ADC.

Step 3 — Check whether the two triangles overlap exactly

• Place △ABC on top of △ADC. All three sides coincide one on another:
  • They share the side $$AC$$ (the diagonal).
  • Both have one side of length $$l$$ (either $$AB$$ or $$CD$$).
  • Both have one side of length $$b$$ (either $$BC$$ or $$DA$$).
  Since the corresponding three sides are equal, the two triangles are congruent by the SSS rule (or by the RHS rule because each triangle is right-angled at $$B$$ and $$D$$).
• Congruent figures fit exactly one over the other, so the two triangles do overlap each other completely.

Step 4 — Compare their areas

The area of the whole rectangle is

$$A_{\text{rect}} = l \times b = 8\,\text{cm} \times 5\,\text{cm} = 40\,\text{cm}^2.$$

The diagonal divides the rectangle into the two congruent triangles, so each triangle has half of this area:

$$A_{\text{one triangle}} = \tfrac12 \times 40\,\text{cm}^2 = 20\,\text{cm}^2.$$

Algebraically for any rectangle:

\[\text{Area of each triangle} = \tfrac12 \times l \times b\]

Therefore the two triangles always have the same area.

Step 5 — Try more rectangles

Every time, the two triangular pieces are identical in shape and equal in area.

Step 6 — Special case: a square

• Take a square of side $$s$$; its area is $$s^2$$.
• The diagonal again cuts it into two congruent right-angled isosceles triangles, each of area $$\tfrac12 s^2$$.
• When you place one triangle on the other they match exactly, confirming equal area.

Conclusion

• The diagonal of any rectangle (including a square) divides it into two congruent right-angled triangles.
• Because the triangles are congruent, they overlap perfectly and have equal area.
• Each triangle’s area is always exactly one-half of the area of the original rectangle.

Step 1 – Draw a rectangle ABCD
Let the longer side be $$AB = l$$ and the shorter side be $$AD = b$$.

Step 2 – Area of the rectangle
For any rectangle:
$$\text{Area}_{\text{rectangle}} = \text{length} \times \text{breadth}$$
Therefore,
$$\text{Area}_{\text{ABCD}} = l \times b$$.

Step 3 – Cut along a diagonal
Draw the diagonal $$AC$$. It divides the rectangle into two triangles: $$\triangle ABC$$ and $$\triangle ADC$$.

Step 4 – Congruence of the two triangles
Both triangles have the same three side lengths (rectangle sides and the common diagonal), so they are congruent. Hence they cover equal areas.

Step 5 – Area of each triangle
The diagonal splits the rectangle into two equal parts, so
$$\text{Area}_{\triangle ABC} = \text{Area}_{\triangle ADC} = \dfrac{1}{2} \times (l \times b)$$.

Step 6 – Connecting with the triangle formula
In $$\triangle ABC$$, take $$AB$$ as the base and $$AD$$ (or the perpendicular from $$C$$ to $$AB$$) as the height. Thus
\[ \text{Area of a triangle} = \dfrac{1}{2} \times \text{base} \times \text{height} \]

Inference
Cutting a rectangle along its diagonal always gives two congruent triangles, each occupying exactly half the area of the rectangle. Therefore, the area of any triangle equals half the product of its base and its corresponding height.

Step 1 : Observe the two shapes
In the book you see a blue rectangle and a yellow triangle. Look carefully–the yellow triangle has exactly the same base and the same height as the blue rectangle.

Step 2 : Recall the two area formulae

• For a rectangle:  $$\text{Area of rectangle}=\text{length}\times\text{breadth}$$
• For a triangle:  $$\text{Area of triangle}=\tfrac12\times\text{base}\times\text{height}$$

Step 3 : Put the actual symbols
Let the common base be $$b$$ and the common height (or breadth of the rectangle) be $$h$$.

Blue rectangle:

$$\text{Area}_{\text{rect}} = b\times h$$

Yellow triangle:

$$\text{Area}_{\triangle} = \tfrac12\times b\times h$$

Step 4 : Compare the two areas

$$\frac{\text{Area}_{\text{rect}}}{\text{Area}_{\triangle}} = \frac{b\,h}{\tfrac12\,b\,h} = \frac{1}{1/2} = 2$$

This means

or, the other way round,

Step 1 · Recall the idea of perimeter
Every edge of every unit square that is exposed to the outside air contributes 1 unit to the perimeter. Two neighbouring squares share the edge between them, so that inner common edge is not counted in the perimeter.

Step 2 · Notice what can change when we attach one more unit square
A fresh unit square by itself possesses 4 edges.
When we press it against the existing figure it will certainly share at least one full edge with that figure (otherwise it would touch only at a corner and would drop off!). Let $$s$$ be the number of whole edges it shares.

• Those $$s$$ edges were already counted in the old perimeter, so each shared edge removes 1 unit from the old perimeter.
• The new square still brings its own 4 edges, but the $$s$$ shared ones are now inside the figure, so only $$4-s$$ of its edges are exposed. They add to the perimeter.

Step 3 · Compute the change in perimeter

Change  =  (edges added) − (edges deleted) = $$\bigl(4-s\bigr) - s = 4 - 2s$$.

Step 4 · See every possible value of $$s$$

Values $$s=0$$ (no contact) and $$s=4$$ (completely surrounded) are impossible here, so these three cases are all we need.

Step 5 · Apply the result to the given figure
The present perimeter is 24 units:

• If we attach so that $$s=1$$, the new perimeter will be $$24+2=26$$ units — it increases.
• If we attach so that $$s=2$$, the new perimeter will be $$24+0=24$$ units — it stays the same.
• If we attach so that $$s=3$$, the new perimeter will be $$24-2=22$$ units — it decreases.

Thus by choosing a suitable place for the extra square we can make the perimeter (a) increase, (b) decrease, or (c) remain unchanged.

How to draw the three situations
Take the given outline of unit squares (perimeter 24). Then:
• For (a) press the new square flat against any straight edge on the outside.
• For (b) put it in the missing corner of an “L” shaped indentation.
• For (c) slide it into a U-shaped hollow that already has three sides.

Step 1 – Find the area of each given rectangle.

• For the rectangle that is $$5\,\text{m} \times 10\,\text{m}$$:
  Area = $$5\,\text{m}\times10\,\text{m}=50\,\text{m}^2$$
• For the rectangle that is $$2\,\text{m} \times 7\,\text{m}$$:
  Area = $$2\,\text{m}\times7\,\text{m}=14\,\text{m}^2$$

Step 2 – Add the two areas.

Total area required = $$50\,\text{m}^2+14\,\text{m}^2=64\,\text{m}^2$$

Step 3 – Find whole-number pairs whose product is $$64\,\text{m}^2$$.

Any pair of whole numbers whose product is $$64$$ can serve as the length and breadth of the required rectangle:

• $$1\,\text{m}\times64\,\text{m}$$  (because $$1\times64=64$$)
• $$2\,\text{m}\times32\,\text{m}$$  (because $$2\times32=64$$)
• $$4\,\text{m}\times16\,\text{m}$$  (because $$4\times16=64$$)
• $$8\,\text{m}\times8\,\text{m}$$   (because $$8\times8=64$$)

Conclusion. Any of the above pairs gives a rectangle whose area equals the sum of the areas of the two given rectangles.

We are given:

• Length of the rectangular garden  $$= 50 \text{ m}$$
• Area of the rectangular garden  $$= 1000 \text{ sq m}$$

For a rectangle, the relation between area, length, and width is

$$\text{Area} = \text{Length} \times \text{Width}$$

Let the width of the garden be $$w\,\text{m}$$.

Substituting the known values:

$$1000 = 50 \times w$$

To isolate $$w$$, divide both sides by $$50$$:

$$\frac{1000}{50} = \frac{50 \times w}{50}$$

Simplifying each side gives:

$$20 = w$$

Thus, the width of the garden is

A quick check: $$50 \times 20 = 1000$$, which matches the given area, so the calculation is correct.

Step 1 – Find the area of the floor
The floor is a rectangle.
Formula: $$\text{Area of rectangle}=\text{length}\times\text{breadth}$$
$$\text{Area of floor}=5\,\text{m}\times4\,\text{m}=20\,\text{m}^2$$

Step 2 – Find the area of the square carpet
Formula: $$\text{Area of square}=\text{side}\times\text{side}$$
$$\text{Area of carpet}=3\,\text{m}\times3\,\text{m}=9\,\text{m}^2$$

Step 3 – Find the uncovered area
$$\text{Uncovered area}=\text{Area of floor}-\text{Area of carpet}$$
$$\text{Uncovered area}=20\,\text{m}^2-9\,\text{m}^2=11\,\text{m}^2$$

Therefore,

\[\boxed{\text{Area not carpeted}=11\,\text{m}^2}\]

Step 1 – Area of the whole garden
The garden is a rectangle $$15\,\text{m}$$ long and $$12\,\text{m}$$ wide.
So,
$$\text{Area of garden}=15\,\text{m}\times12\,\text{m}=180\,\text{m}^2$$

Step 2 – Area of one flower bed
Each flower bed is also a rectangle, $$2\,\text{m}$$ long and $$1\,\text{m}$$ wide.
$$\text{Area of one bed}=2\,\text{m}\times1\,\text{m}=2\,\text{m}^2$$

Step 3 – Total area occupied by four beds
There are $$4$$ such beds, so
$$\text{Area of 4 beds}=4\times2\,\text{m}^2=8\,\text{m}^2$$

Step 4 – Area available for the lawn
Subtract the area of all flower beds from the area of the whole garden:
$$\text{Lawn area}=180\,\text{m}^2-8\,\text{m}^2=172\,\text{m}^2$$

Therefore, the area now available for laying the lawn is

\[172\,\text{m}^2\]

Step 1 : Decide convenient shapes.

Since the question only mentions area and perimeter, rectangles are the easiest shapes to handle. For a fixed area, a very long and thin rectangle has a big perimeter, whereas a nearly-square rectangle has a small perimeter. So we look for

• a rectangle whose area is $$18\,\text{sq units}$$ but whose perimeter is as large as we like (Shape A), and
• another rectangle whose area is $$20\,\text{sq units}$$ but whose perimeter is smaller than the one chosen for Shape A (Shape B).

Step 2 : Pick the side lengths.

Step 3 : Verify the two conditions.

Areas: $$18\,\text{sq units}$$ is less than $$20\,\text{sq units}$$, so Shape A is smaller in area than Shape B. (correct)

Perimeters: Shape A has perimeter $$38\,\text{units}$$ and Shape B has perimeter $$18\,\text{units}$$. Since $$38\,\text{units}$$ is greater than $$18\,\text{units}$$, Shape A indeed has the longer perimeter, as required. (correct)

Step 4 : What to draw.

1. Shape A: draw a long thin rectangle, 1 unit wide and 18 units long. Mark its length and breadth, and label:
   • Area = $$18\,\text{sq units}$$
   • Perimeter = $$38\,\text{units}$$
2. Shape B: beside it, draw another rectangle, 4 units by 5 units. Mark the sides and label:
   • Area = $$20\,\text{sq units}$$
   • Perimeter = $$18\,\text{units}$$

These two rectangles satisfy all the conditions of the question.

Step 1 – Measure the page
Use your ruler to find:

• the length of the page; call it $$L\text{ cm}$$, and
• the breadth (width) of the page; call it $$B\text{ cm}$$.

Step 2 – Find the length and breadth of the border rectangle

• The border is drawn $$1\text{ cm}$$ below the top edge and $$1\text{ cm}$$ above the bottom edge.
    So the border’s inside length is
    $$\text{inside length}=L-1-1=L-2\text{ cm}$$.
• The border is also $$1.5\text{ cm}$$ away from the left edge and from the right edge.
    So the border’s inside breadth is
    $$\text{inside breadth}=B-1.5-1.5=B-3\text{ cm}$$.

Step 3 – Perimeter of the border rectangle

The perimeter $$P$$ of any rectangle is given by $$P=2(\text{length}+\text{breadth})$$. Hence

Step 4 – Insert your own measurements
Measure your page, substitute those values for $$L$$ and $$B$$ in the formula above, and work out the numerical perimeter in centimetres.

Step 1 : Find the area of the outer rectangle

Given length $$L = 12\text{ units}$$ and breadth $$B = 8\text{ units}$$.

Step 2 : Area required for the inner rectangle

The inner rectangle must occupy exactly one-half of this area.

Step 3 : Choose suitable whole-number side lengths whose product is 48

Hence let the inner rectangle have length $$l = 8\text{ units}$$ and breadth $$b = 6\text{ units}$$.

Step 4 : Work out the margins

• Horizontal spare length: $$12 - 8 = 4$$ units ⇒ $$\tfrac{4}{2} = 2$$ units margin on both left and right.
• Vertical spare breadth: $$8 - 6 = 2$$ units ⇒ $$\tfrac{2}{2} = 1$$ unit margin at the top and bottom.

Step 5 : Construction instructions

1. Draw rectangle ABCD with AB = 12 units and AD = 8 units.
2. From AB and DC, mark points 2 units inwards; draw two vertical guide lines.
3. From AD and BC, mark points 1 unit inwards; draw two horizontal guide lines.
4. The intersection of these guides gives points P, Q, R, S; join them to obtain rectangle PQRS.
5. Verify: $$PQ = 8\text{ units},\; QR = 6\text{ units},\; PQ \times QR = 48\text{ square units} = \dfrac{1}{2}A_{\text{outer}}.$$

Rectangle PQRS is completely inside ABCD, does not touch it anywhere, and its area is exactly half of the area of ABCD, as required.

Step 1 – Draw and name the square

Let the square piece of paper have side $$s\,(\text{cm})$$. Its perimeter is therefore

$$P_{\text{square}} = 4s$$

and its area is

$$A_{\text{square}} = s^2$$

Step 2 – Fold the square exactly in half

Because the fold must divide the square into two rectangles, the only possible folds are the lines that join the mid–points of opposite sides (horizontal or vertical).
The fold line is therefore parallel to one pair of sides and passes through the mid-points of the other pair.

After cutting along the fold we obtain two equal rectangles, each having

• length $$= s$$ (the full side of the square)
• breadth $$= \dfrac{s}{2}$$ (half the side, because the fold is at the mid-point)

Step 3 – Area of each rectangle

$$A_{\text{one rectangle}} = s \times \frac{s}{2} = \frac{s^2}{2}$$

Step 4 – Perimeter of each rectangle

$$P_{\text{one rectangle}} = 2\bigl(\text{length} + \text{breadth}\bigr) \\ = 2\Bigl(s + \frac{s}{2}\Bigr) = 2\Bigl(\frac{3s}{2}\Bigr) = 3s$$

Step 5 – Compare the required quantities

1. Statement (a): $$A_{\text{one rectangle}} = \dfrac{s^2}{2} < s^2 = A_{\text{square}}$$, so (a) is false.
2. Statement (b): Combined perimeter of the two rectangles is $$2\times 3s = 6s$$, while $$P_{\text{square}} = 4s$$. Hence $$6s > 4s$$, so (b) is false (it says the opposite).
3. Statement (c): \[ \frac{P_{\text{both rectangles}}}{P_{\text{square}}} = \frac{6s}{4s} = \frac{3}{2} = 1\frac12 \] Therefore the added perimeters are always $$1\frac12$$ times the perimeter of the square. Statement (c) is true.
4. Statement (d): Total area of the two rectangles is $$2\times \dfrac{s^2}{2} = s^2$$, exactly equal to the square’s area, not three times. So (d) is false.

Step 6 – Conclusion

The only statement that is always true for any square folded as described is statement (c).