Chapter 6: Measuring Space: Perimeter and Area

Known facts from athletics

• The width of one lane is practically the same for every standard track: we write it as $$w$$ (the International norm is $$w = 1.22\,\text{m}$$).
• Each lane is a concentric circle (in the curved part of the track). If the inner-most lane has radius $$R_1$$, the next lane has radius $$R_2 = R_1 + w$$, the third $$R_3 = R_1 + 2w$$, and so on.
• When the race is run in lanes (as in the $$4 \times 100\,\text{m}$$ relay) every team must cover exactly the same total distance; therefore an outer-lane runner must start a little farther ahead so that the extra length of his circular path is compensated. That forward shift is called the stagger.

1. How much extra distance appears in one complete circle?

The arc length of a full circle is its circumference. Hence, for two consecutive lanes

$$\text{extra distance for one full circle} \;=\; 2\pi R_2\; -\; 2\pi R_1\; =\; 2\pi (R_2-R_1)\; =\; 2\pi w.$$

Notice that this number does not depend on the actual radius of the track: it depends only on the lane width $$w$$.

2. The situation on the standard (400 m) track

• The baton travels one complete lap (one circle) in every lane, because the total race length is $$400\,\text{m}$$ and the lap itself measures $$400\,\text{m}$$.
• Therefore the distance that the outer lane would otherwise have to run in excess is $$2\pi w$$.
• To equalise the paths, the outer lane’s starting line is moved forward by exactly that amount: the usual figure is
  $$2\pi w \approx 2\times 3.14 \times 1.22 \approx 7.66\,\text{m}.$$

3. What changes when the track itself is only 200 m long?

• One lap is now $$200\,\text{m}$$;
• but the relay is still a $$4 \times 100\,\text{m} = 400\,\text{m}$$ event, so the baton must complete two laps.

Over two complete circles the outer lane inherits the extra length twice:

$$\text{total extra distance (two laps)} = 2\,\bigl(2\pi w\bigr) = 4\pi w.$$

4. Numerical comparison

5. Conclusion

The outer lanes on a 200 m track actually need twice as much forward shift as on a 400 m track, because the baton travels twice around the curves. Hence the stagger must be larger, not smaller.

Step 1 : Recall what “perimeter” means
The perimeter of any closed figure is the total length of its boundary. For a circle we give this special name circumference.

Step 2 : The key fact discovered from measurement
If we measure the circumference $$C$$ of several circles and divide it by their diameter $$d$$, the quotient is always the same number. We denote this constant by the Greek letter $$\pi$$ (pi).

In symbols

$$\frac{C}{d}=\pi \quad\Longrightarrow\quad C = \pi d$$

Step 3 : Write the diameter in terms of the radius
Because the diameter is twice the radius, $$d = 2r$$.

Step 4 : Substitute

$$C = \pi (2r) = 2\pi r$$

The circumference therefore depends only on the radius. Collecting the result in display form,

This is the perimeter of the given circle.

Step 5 : Connection with the 400 m athletics track
A standard 400 m track is not a full circle; it consists of

• two straight parallel stretches (usually 2 × 100 m), and
• two semicircular ends of equal radius.

The two semicircles together form one complete circle, so their combined length is exactly the circumference $$2\pi r$$ derived above. Track designers use the same formula to choose the radius of the bends once the lengths of the straight portions have been fixed, ensuring that the total distance around one lap is 400 m.

Part A : A quick hands-on estimate

1. Measure the diameter $$D$$ of the cotton reel as accurately as possible (use a ruler or, if available, a vernier calliper).
2. Wrap a very thin thread tightly round the reel exactly 20 times, then unwind it and measure its total length $$L$$.
3. The length that goes once round the reel is the circumference, so the experimental value of the circumference is $$\dfrac{L}{20}$$.
4. The required ratio is therefore $$\dfrac{C}{D}=\dfrac{L}{20D}.$$
5. A careful measurement usually gives $$\dfrac{L}{20D}\approx 3.14$$, which lies between $$3$$ and $$4$$ and more precisely between $$3.1$$ and $$3.2$$.

Part B : Obtaining the same bounds by pure geometry

We now show, without any measurement, that for every circle

1. Lower bound 3 — inscribed regular hexagon.

Draw a circle of centre $$O$$ and radius $$R$$, and inscribe a regular hexagon $$A_1A_2A_3A_4A_5A_6$$ in it. Each side of a regular hexagon inscribed in a circle equals the radius, so $$A_1A_2=A_2A_3=\dots =A_6A_1=R$$ and the perimeter of the hexagon is $$6R$$. A chord is always shorter than the arc it subtends, so the circumference $$C$$ exceeds the hexagon’s perimeter:

Since $$D=2R$$,

2. Upper bound 4 — circumscribed square.

Draw the smallest square that completely contains the circle; each side of this square equals the diameter $$D$$, so the perimeter of the square is $$4D$$. The circle lies entirely inside the square, and the circle’s circumference is shorter than the perimeter of any polygon that circumscribes it. Hence

Combining the two bounds obtained so far,

3. Sharper lower bound 3.105… — inscribed regular dodecagon (12-gon).

The side of an inscribed regular $$n$$-gon is $$2R\sin\dfrac{180^{\circ}}{n}$$, so its perimeter is $$P_{\text{in}}=2nR\sin\dfrac{180^{\circ}}{n}$$. For $$n=12$$,

Using $$\sin 15^{\circ}\approx 0.25882$$,

Hence

4. Sharper upper bound 3.215… — circumscribed regular dodecagon.

The side of a regular $$n$$-gon circumscribing a circle of radius $$R$$ is $$2R\tan\dfrac{180^{\circ}}{n}$$, so its perimeter is $$P_{\text{out}}=2nR\tan\dfrac{180^{\circ}}{n}$$. For $$n=12$$,

Using $$\tan 15^{\circ}\approx 0.26795$$,

Hence

5. Combining the refined bounds.

which, rounded to one decimal place, gives

Hence pure geometry, without any actual measurement, yields the same conclusion as the simple thread-and-reel experiment.

Data that can be read off Fig. 6.11

• The track consists of two straight parts, each $$100\,\text{m}$$ long.
• Each lane is a ring-shaped strip of width $$1\,\text{m}$$ (the figure shows four such concentric lanes).
• The centre line of every lane is a circle; its radius increases by exactly the lane–width from one lane to the next.

1. Radius difference between the first and second lanes

Let the radius of the centre-line of the first lane be $$r_1\,(\text{m}).$$
Because the width of each lane is $$1\,\text{m},$$ the radius of the centre-line of the second lane is

$$r_2 \,=\, r_1 + 1.$$ Therefore the required difference is simply

$$r_2 - r_1 = 1\,\text{m}.$$

2. Extra distance that the runner in lane 2 would cover on the bends

The curved part of the track consists of two semicircles. The length of one semicircle of radius $$r$$ is $$\pi r.$$ Hence, in the first lane, total curved distance = $$2\pi r_1,$$ while in the second lane it is $$2\pi r_2.$$ The extra distance that the lane-2 runner would have to run on the two bends is therefore

$$\begin{aligned} \bigl(2\pi r_2\bigr) - \bigl(2\pi r_1\bigr) &= 2\pi\,(r_2-r_1) \\ &= 2\pi\times1 \\ &= 2\pi\,\text{m}\;\approx\;6.28\,\text{m}. \end{aligned}$$

3. How the stagger solves the problem

During a race every athlete must run exactly the same total distance. The extra $$2\pi\,\text{m}$$ appearing in the curved part for lane 2 is compensated by placing the lane-2 starting line $$2\pi\,\text{m}$$ ahead of the lane-1 starting line along the straight. This forward shift of about $$6.28\,\text{m}$$ is the required stagger for lane 2.

4. Will the stagger between the third and second lanes be the same?

Yes. The radius of the centre-line of lane 3 is $$r_3 = r_2 + 1.$$ Hence $$r_3 - r_2 = 1\,\text{m}$$ once again, and the extra curved distance for lane 3 over lane 2 is also $$2\pi\times1 = 2\pi\,\text{m}.$$ Therefore the stagger required between lanes 3 and 2 is the same $$6.28\,\text{m}.$$

Conclusion

• Difference in radius (lane 2 − lane 1): $$1\,\text{m}.$$
• Stagger for lane 2: $$2\pi\,\text{m}\;\approx\;6.28\,\text{m}.$$
• The same stagger is needed between lanes 3 and 2, because every consecutive pair of lanes differs in radius by the same $$1\,\text{m}.$$

The difficulty. In Fig. 6.18 the proof that a parallelogram can be cut and rearranged to a rectangle starts with the perpendicular from vertex $$C$$ to the base $$AD$$. When the parallelogram is very “thin”, the perpendicular meets the produced line of $$AD$$, not the segment itself; the little right-triangle to be cut is then no longer inside the figure, so at first sight the construction collapses.

How we repair it.

1. Produce the base. Extend $$AD$$ beyond $$D$$ to meet the perpendicular from $$C$$ at $$H$$ (so $$CH\perp AD$$).
2. Cut the same triangle as before. Join $$A$$ to $$H$$; the triangle $$\triangle ADH$$ that we now see outside the parallelogram is precisely the triangle that was inside the figure when the perpendicular fell on $$AD$$.
3. Translate the triangle. Because $$ABCD$$ is a parallelogram, $$AD=BC$$ and $$AD\parallel BC$$. Cutting off $$\triangle ADH$$ and sliding it leftwards so that $$D$$ falls on $$B$$ fits the triangle exactly on the other side, turning the whole figure into the rectangle $$ADHC$$.

Result of the repair.

Thus the cut-and-paste argument is still valid; we simply allow the “foot” of the perpendicular to lie on the line containing the base. (Alternatively one may avoid any extension by taking $$AB$$ instead of $$AD$$ as the base; for at least one pair of opposite sides the perpendicular from the other vertices always falls inside.) Hence the apparent gap is completely filled.

Step 1 : Recall the rectangle case
For a rectangle of length $$l$$ and breadth $$b$$, the angle between each pair of adjacent sides is $$90^\circ$$, so its height is exactly the other side: $$\text{Area}=l\times b$$. Therefore knowing the two side-lengths fixes one and only one area.

Step 2 : Write the general area formula for a parallelogram
Take any parallelogram with adjacent sides of lengths $$a$$ and $$b$$. Let $$\theta$$ be the interior angle between them (say the one at the bottom left corner). Drop a perpendicular from the opposite vertex to the base $$a$$. The perpendicular length (the height) is $$h=b\sin\theta$$, because that perpendicular, the side $$b$$ and the projection $$b\cos\theta$$ form a right-angled triangle.
Hence $$\text{Area}=\text{base}\times\text{height}=a\,h=a\,(b\sin\theta)=ab\sin\theta.$$

Step 3 : Why the two side-lengths are not enough
The formula $$ab\sin\theta$$ contains the extra factor $$\sin\theta$$. If we keep $$a$$ and $$b$$ fixed but change $$\theta$$, the numerical value of $$\sin\theta$$ changes and therefore the area changes.

• When $$\theta=90^\circ$$ (parallelogram becomes a rectangle), $$\sin\theta=1$$ and $$\text{Area}=ab$$.
• If we make the parallelogram “flatter”, say $$\theta=30^\circ$$, then $$\sin30^\circ=\tfrac12$$ and the area becomes $$\tfrac12 ab$$ — only half of the rectangular case.
• As we keep decreasing $$\theta$$ towards $$0^\circ$$, $$\sin\theta$$ tends to $$0$$ and the area can be made as small as we like, without altering $$a$$ and $$b$$.

Step 4 : Conclusion
Because the same pair of side-lengths $$a,b$$ can correspond to infinitely many different areas (one for each possible interior angle $$\theta$$ between $$0^\circ$$ and $$180^\circ$$), the lengths alone do not determine the area of a parallelogram. We must also know either the height, the included angle $$\theta$$, or some equivalent information.

Problem restated
Triangles $$\triangle ABD$$ and $$\triangle ACD$$ have the same area (they stand on the same base $$AD$$ and vertices $$B,\,C$$ lie on the line $$BC\parallel AD$$). We want to know whether $$\triangle ABD$$ can be cut into a few straight-edged pieces that can be re-assembled to cover $$\triangle ACD$$ exactly.

Key idea
Any two polygons of equal area are equidecomposable; a finite set of straight cuts always exists that turns one into the other (Bolyai–Gerwien theorem). For two equal-area triangles a very small number of cuts suffices. In this particular arrangement only two cuts are needed.

Construction of the cuts

1. Let $$P$$ be the mid-point of the common base; that is, $$AP = PD$$ on $$AD$$.
2. Inside $$\triangle ABD$$ draw the two straight segments
   $$PB \quad\text{and}\quad PD.$$
   Both lie wholly in $$\triangle ABD$$, so these are legitimate cuts.

What pieces are obtained?
The two segments divide $$\triangle ABD$$ into exactly two smaller triangles:

• Piece 1 : $$\triangle ABP$$
• Piece 2 : $$\triangle BPD$$

Why do these two pieces fit in $$\triangle ACD$$?

• Because $$AP = PD$$ and $$BC\parallel AD$$, corresponding angles are equal: $$\angle BAP = \angle DCP$$ and $$\angle BPA = \angle CPD$$ (vertical). Hence
  $$\triangle ABP \cong \triangle PCD\quad(\text{ASA criterion}).$$
• Similarly, $$\angle BDP = \angle PCA$$ (corresponding) and again $$AP = PD$$, giving
  $$\triangle BPD \cong \triangle PAC\quad(\text{AAS criterion}).$$

Thus each of the two pieces cut from $$\triangle ABD$$ is congruent to exactly one of the two complementary regions inside $$\triangle ACD$$ (obtained by the same straight lines $$PC$$ and $$PD$$ in that triangle). If we lift the pieces and place them on their congruent partners, they cover the whole of $$\triangle ACD$$ with neither gap nor overlap.

Conclusion
Yes, it is possible: two straight cuts (along $$PB$$ and $$PD$$ with $$P$$ the mid-point of $$AD$$) are enough to dissect $$\triangle ABD$$ into parts that can be rearranged to fill $$\triangle ACD$$ exactly.

Remark The argument is a special case of a general fact: any two polygons of equal area can always be cut into finitely many polygonal pieces that can be reassembled to form one another.

What is being asked. We are allowed to make straight-line cuts on one polygon, move the pieces as rigid tiles (no bending, no scaling) and try to re-assemble them so that they coincide exactly with the other polygon. Such an operation is called a dissection or equidecomposition.

A classical theorem (Wallace–Bolyai–Gerwien) states:

Theorem. Any two simple polygons of equal area can be cut by finitely many straight cuts into pieces that can be rearranged to form one another.

We verify the statement in three familiar cases.

1. Square ↔ non-square rectangle of equal area.

1. Let rectangle $$ABCD$$ have $$AB=l$$ (longer side), $$AD=b$$ (shorter side). The square of equal area has side $$s=\sqrt{lb}$$; note $$b\le s\le l$$.
2. On $$AB$$ mark $$E$$ with $$AE=s$$ and on $$DC$$ mark $$F$$ with $$DF=s$$. Join $$EF$$; this segment is perpendicular to $$AB$$, so the rectangle is divided into rectangle $$AEFD$$ (of sides $$s$$ and $$b$$) and rectangle $$EBCF$$ (of sides $$l-s$$ and $$b$$).
3. We must now turn rectangle $$AEFD$$ (sides $$s,b$$ with $$s\ge b$$) into a square of side $$s$$. Extend $$DA$$ upward to $$A'$$ with $$AA'=s-b$$ (so $$DA'=s$$). Draw the diagonal $$A'E$$. This diagonal cuts off a small triangle from the strip $$EBCF$$ which, slid into place above $$AEFD$$, completes the square. The cuts needed are $$EF$$ and $$A'E$$ (extended); the three pieces rearrange to a square of side $$s$$.

(If $$l>2s$$ the rectangle is first halved and the halves stacked, then the procedure is applied. Either way, only finitely many straight cuts are used.)

2. Two triangles of different shapes but equal area.

1. Take any triangle $$\triangle PQR$$ with area $$\Delta$$. Let $$M$$ be the midpoint of $$PR$$ and $$N$$ the midpoint of $$QR$$. Cut along $$MN$$, separating the small triangle $$\triangle MNR$$ from the trapezium $$PQNM$$.
2. Rotate $$\triangle MNR$$ through $$180^{\circ}$$ about the midpoint $$N$$. Because $$N$$ is the midpoint of $$QR$$, the point $$R$$ goes to $$Q$$, and because $$MN\parallel PQ$$ (mid-segment theorem) and $$MN=\tfrac12 PQ$$, the rotated piece fits exactly along $$QN$$. The result is a parallelogram $$PQNM'$$ (where $$M'$$ is the image of $$M$$) with base $$PQ$$ and the same area $$\Delta$$ as the triangle.
3. Apply the same recipe to $$\triangle XYZ$$ to obtain a second parallelogram of area $$\Delta$$.
4. It remains to show that any two equal-area parallelograms can be dissected into one another. Place them so that one pair of equal sides lies along the same straight line, with both parallelograms on the same side.
   • If their bases happen to be equal in length, both have the same base and the same height (since the areas are equal). A single straight cut from the top vertex of one perpendicular to the base, together with sliding the resulting right triangle to the other end, converts it into a rectangle; the same procedure converts the second parallelogram into a rectangle of the same dimensions. The two rectangles coincide.
   • If the bases differ, cut the parallelogram with the longer base into vertical strips whose widths add up to the shorter base; stacking the strips gives a parallelogram with the shorter base and the same height (a shear move). Now apply the previous case.

3. Triangle ↔ square of equal area.

1. By Part 2, the triangle can be turned (with two straight cuts) into a parallelogram of the same area.
2. Drop a perpendicular from one top vertex of this parallelogram to the base, slide the resulting right triangle to the other end, and obtain a rectangle of the same area (one further cut).
3. Apply the rectangle-to-square dissection of Part 1 to that rectangle (a small bounded number of additional cuts).

The triangle is thus cut, in finitely many straight cuts, into pieces that re-assemble to a square of equal area.

Conjecture. Any two plane polygons of equal area can be dissected, with finitely many straight cuts, into pieces that rearrange to form one another.

This is exactly what the Wallace–Bolyai–Gerwien theorem confirms. Hence the answer to the question is yes.

Known facts

• If a rectangle has length $$l$$ and breadth $$b$$, its perimeter is $$2(l+b)$$ and its area is $$A=l\times b$$.
• Every side length must be strictly positive.

The question fixes the perimeter at $$40$$ units, so

Because $$l \gt 0$$ and $$b \gt 0$$, the relation $$b=20-l$$ forces $$0 \lt l \lt 20$$ (and consequently $$0 \lt b \lt 20$$).

(1) How many such rectangles are there?

The only restriction on $$l$$ is $$0 \lt l \lt 20$$, and in that open interval there are infinitely many real numbers (for example $$6$$, $$6.1$$, $$6.01$$, $$6.001$$, …). Hence there are infinitely many different rectangles with perimeter $$40$$.

(2) Which one has the largest area?

Write the area in terms of one variable:

Complete the square:

The term $$-(l-10)^{2}$$ is never positive, and equals $$0$$ only when $$l=10$$. Hence the maximum value of $$A$$ is $$100$$, attained at $$l=10$$. The corresponding breadth is $$b=20-10=10$$.

The rectangle of greatest area is therefore the square of side $$10$$ units.

(3) Is there a rectangle of smallest area?

As $$l$$ approaches $$0$$ from the positive side, $$A=l(20-l)$$ also approaches $$0$$. For example:

• $$l=0.1 \Rightarrow b=19.9 \Rightarrow A=1.99$$,
• $$l=0.01 \Rightarrow b=19.99 \Rightarrow A=0.1999$$,
• $$l=0.001 \Rightarrow b=19.999 \Rightarrow A=0.019999$$.

So the area can be made arbitrarily small, but it cannot reach $$0$$ (because $$l$$ and $$b$$ must remain strictly positive). Hence

(4) Any surprises?

• The maximum-area answer (a square) is satisfyingly symmetric.
• The non-existence of a smallest area can be surprising. It arises because the side lengths can be chosen as small positive numbers, so the area can be squeezed as close to $$0$$ as we like without ever reaching it.

Problem restatement. Given a triangle $$\triangle ABC$$, construct with ruler and compasses a square whose area equals the area of the triangle.

Idea behind the construction. The area of a triangle equals half the product of one side and the corresponding altitude. We first convert $$\triangle ABC$$ into a rectangle of the same area (whose sides are the base of the triangle and half its altitude). A classical mean-proportional construction then turns that rectangle into a square of equal area.

1. Draw the altitude of the triangle. From vertex $$A$$ drop the perpendicular to the line $$BC$$, meeting it at $$D$$. Thus $$AD\perp BC$$ and $$AD$$ is the altitude $$h$$.
2. Halve the altitude. Construct the midpoint $$M$$ of segment $$AD$$ (perpendicular-bisector method). Hence $$MD=\dfrac{1}{2}AD=\dfrac{h}{2}$$.
3. Form a rectangle of area equal to $$\triangle ABC$$. Erect perpendiculars to $$BC$$ at $$B$$ and at $$C$$. On these perpendiculars (both on the same side of $$BC$$) mark points $$F$$ and $$G$$ such that $$BF=CG=\dfrac{h}{2}$$. Join $$FG$$. Then $$BCGF$$ is a rectangle of base $$BC$$ and height $$\dfrac{h}{2}$$, so \[\text{Area}(BCGF)=BC\times\frac{h}{2}=\frac{1}{2}\,BC\times AD=\text{Area}(\triangle ABC).\]
4. Lay off the two side lengths on one straight line. Draw an auxiliary line and mark successive points $$P,\,Q,\,R$$ such that $$PQ=BC$$ (length of the rectangle) and $$QR=BF=\dfrac{h}{2}$$ (breadth of the rectangle). Then $$PR=PQ+QR=BC+\dfrac{h}{2}$$.
5. Construct the mean proportional of $$BC$$ and $$\dfrac{h}{2}$$.
   (a) Construct the midpoint $$O$$ of $$PR$$ (perpendicular-bisector method).
   (b) With centre $$O$$ and radius $$OP$$, draw a semicircle on $$PR$$ as diameter.
   (c) At $$Q$$ (the junction point on $$PR$$) erect a perpendicular to $$PR$$; let it meet the semicircle at $$S$$. Then by the geometric-mean property, \[QS^{2}=PQ\times QR=BC\times\frac{h}{2}.\] Thus $$QS=\sqrt{BC\times\dfrac{h}{2}}$$.
6. Construct the required square. With $$QS$$ as one side, construct square $$QSTU$$. Then \[\text{Area}(QSTU)=QS^{2}=BC\times\frac{h}{2}=\text{Area}(BCGF)=\text{Area}(\triangle ABC).\]

Conclusion. The square $$QSTU$$ has area equal to the area of the given triangle $$\triangle ABC$$; the triangle has therefore been “squared”.

Understanding the preference for the circle

A circle is defined in mathematics as the set (locus) of all points in a plane that are at a fixed distance $$r$$ (called the radius) from a fixed point O (called the centre). This single geometric fact gives the circle several properties that have attracted human beings since prehistoric times.

1. Simplicity of construction
   Starting with a taut vine or a piece of string tied to a peg, early artists and engineers could trace an almost perfect circle simply by turning the free end through one full turn. No other closed curve of equal smoothness is as easy to draw without modern tools.
2. Perfect rotational symmetry
   Because every radius is the same, rotating a circle about its centre through any angle brings it exactly onto itself. This property makes circular objects roll smoothly, an advantage noticed long before people could articulate the idea of symmetry.
3. Constant width
   The distance across a circle measured through the centre (the diameter) is everywhere the same: $$d = 2r$$. This constancy means a circular wheel bears weight uniformly and turns without bumping.
4. Efficient enclosure
   For a given perimeter, the circle encloses the maximum possible area: if the perimeter is $$P$$, the enclosed area is greatest when the shape is a circle with radius $$r = \dfrac{P}{2\pi}$$, giving area $$A = \pi r^2 = \dfrac{P^2}{4\pi}$$. Early builders of huts, forts and city walls could surround more land with the same length of fence by choosing a round plan.
5. Aesthetic and symbolic appeal
   Cultures around the world have associated the circle with the Sun, the Moon, eternity (no beginning or end) and perfection. Temples, altars and ornaments therefore often adopted circular motifs for religious or artistic reasons that go beyond mere practicality.

Was it only for practical reasons?

Clearly not. While the mechanical benefits of the circle (rolling, even pressure distribution, efficient enclosure) are practical, the circle’s perfect symmetry also appeals psychologically and spiritually. Architecture, pottery and jewellery from every ancient civilisation show circular patterns even where no mechanical advantage exists—for example, mandalas painted on flat walls or round halo designs in sculpture. Hence both utility and aesthetic / symbolic meaning explain humanity’s fondness for circles.

Major human uses of the circular shape

Conclusion

The circle combines unmatched mechanical advantages with cultural, artistic and symbolic richness. That is why, throughout history, human beings have been—and still are—so fond of using circular shapes.

Given data
Two equal circles have radius $$r$$. Each centre lies on the circumference of the other, so the distance between the centres is also $$r$$.

Step 1 – Name the important points
Let the centres be $$O_1$$ and $$O_2$$. The circles meet at two points; mark the upper one $$A$$ and the lower one $$B$$.

Step 2 – Identify the triangle formed by the centres and an intersection point
In $$\triangle O_1AO_2$$ we have
$$O_1O_2 = r, \; O_1A = r, \; O_2A = r$$ (all are radii).
Therefore $$\triangle O_1AO_2$$ is equilateral  ⇒  each interior angle is $$60^{\circ}$$.

Step 3 – Central angle that lies inside the overlap
At centre $$O_1$$ the angle subtended by arc $$AB$$ is $$60^{\circ}$$. This minor arc of the circle centred at $$O_1$$ is the part inside the other circle, so it will not appear on the required boundary. The same is true for the circle centred at $$O_2$$.

Step 4 – Central angle that lies on the required perimeter
Total angle of a full circle is $$360^{\circ}$$. Hence, on each circle the arc that does form the outer boundary measures
$$360^{\circ} - 60^{\circ} = 300^{\circ}.$$

Step 5 – Length of one such arc
Arc length formula: $$L = \frac{\theta}{360^{\circ}} \times 2\pi r.$$ Substituting $$\theta = 300^{\circ}$$ gives
$$L = \frac{300}{360}\times 2\pi r = \frac{5}{6}\,\times 2\pi r = \frac{5\pi r}{3}.$$

Step 6 – Total perimeter of the required shape
There are two identical outer arcs (one from each circle), so
$$P = 2 \times \frac{5\pi r}{3} = \frac{10\pi r}{3}.$$

Result

\[P = \frac{10\pi r}{3}\;\text{units}\]

Given.  Points $$P$$ and $$Q$$ are joined in two different ways.

• Path $$a$$ is a single semicircle drawn on $$\overline{PQ}$$ as diameter.
• Path $$b+c+d$$ consists of three consecutive semicircles whose diameters exactly cover $$\overline{PQ}$$ one after another.

Let the length of the straight segment $$\overline{PQ}$$ be $$D\;\text{cm}$$.
Suppose the three smaller diameters are

$$PR = d_1,\; RS = d_2,\; SQ = d_3$$  so that

$$d_1 + d_2 + d_3 = D.$$

1. Length of path $$a$$

The circumference of a full circle with diameter $$D$$ is $$\pi D$$, so the length of its semicircle is half of that:

\[L_a = \dfrac{\pi D}{2}.\]

2. Length of path $$b+c+d$$

Each of the small semicircles contributes

• $$b:\; \dfrac{\pi d_1}{2}$$
• $$c:\; \dfrac{\pi d_2}{2}$$
• $$d:\; \dfrac{\pi d_3}{2}$$

Adding them,

$$L_{b+c+d} = \dfrac{\pi}{2}\,(d_1 + d_2 + d_3).$$

But $$d_1 + d_2 + d_3 = D$$, hence

\[L_{b+c+d} = \dfrac{\pi D}{2}.\]

3. Comparison

$$L_a = \dfrac{\pi D}{2}=L_{b+c+d}.$$

Therefore the two paths have exactly the same length.

Choice : (iii) The two paths have equal length.

We are told that the perimeter of a circle is 44 cm. For a circle, the word “perimeter” means the circumference.

The formula for circumference is

$$\text{Circumference} = 2\pi r$$

where $$r$$ is the radius of the circle and $$\pi$$ is the constant pi. The question asks us to use the approximation $$\pi \approx \tfrac{22}{7}$$.

Step 1: Substitute the given circumference and the given value of $$\pi$$ into the formula:

$$44 = 2 \times \frac{22}{7} \times r$$

Step 2: Multiply the constants on the right‐hand side:

$$2 \times \frac{22}{7} = \frac{44}{7}$$

So the equation becomes

$$44 = \frac{44}{7} \times r$$

Step 3: To isolate $$r$$, divide both sides of the equation by $$\tfrac{44}{7}$$:

$$r = \frac{44}{\tfrac{44}{7}}$$

Step 4: Dividing by a fraction is equivalent to multiplying by its reciprocal:

$$r = 44 \times \frac{7}{44}$$

Step 5: The factor 44 cancels:

$$r = 7$$

Therefore, the radius of the circle is 7 cm.

Given: Radius $$r = 14\,\text{cm}$$ and sector angle $$\theta = 75^\circ$$.

For a sector of a circle:

• Arc length $$l = \dfrac{\theta}{360^\circ}\times 2\pi r$$,
• Perimeter $$P = l + 2r$$ (curved edge + two radii).

1. Arc length

$$l = \frac{75^\circ}{360^\circ}\times 2\pi \times 14$$

$$l = \frac{75}{360}\times 28\pi = \frac{5}{24}\times 28\pi$$

$$l = \frac{5\times28}{24}\,\pi = \frac{140}{24}\,\pi = \frac{35}{6}\,\pi\;\text{cm}$$

Taking $$\pi = \dfrac{22}{7}$$ (NCERT convention),

$$l = \frac{35}{6}\times\frac{22}{7}\;\text{cm} = \frac{5\times22}{6}\;\text{cm} = \frac{110}{6}\;\text{cm} = \frac{55}{3}\;\text{cm} \approx 18.33\,\text{cm}$$

2. Perimeter of the sector

$$P = l + 2r$$

$$P = \frac{55}{3} + 2\times14$$

$$P = \frac{55}{3} + 28 = \frac{55}{3} + \frac{84}{3} = \frac{139}{3}\,\text{cm}$$

$$P = 46\frac{1}{3}\,\text{cm} \;(\text{approximately } 46.33\,\text{cm}).$$

Therefore, the perimeter of the sector is $$46\frac{1}{3}\,\text{cm}\,(\approx 46.33\,\text{cm}).$$

Let the radii of the two circles be $$r_1$$ and $$r_2$$, and let their perimeters (circumferences) be $$P_1$$ and $$P_2$$ respectively.

The perimeter of a circle is given by

So,

$$P_1 = 2\pi r_1 \quad\text{and}\quad P_2 = 2\pi r_2.$$

We are told that the perimeters are in the ratio 5 : 4, i.e.

$$\dfrac{P_1}{P_2} = \dfrac{5}{4}.$$

Substituting the expressions for $$P_1$$ and $$P_2$$:

$$\dfrac{2\pi r_1}{2\pi r_2} = \dfrac{5}{4}.$$

The common factor $$2\pi$$ cancels out:

$$\dfrac{r_1}{r_2} = \dfrac{5}{4}.$$

Therefore, the ratio of their radii is 5 : 4.

Given : An equilateral triangle $$\triangle ABC$$ whose three sides are equal, i.e. $$AB = BC = CA = a\text{ units}$$.

We have to determine the measure of each interior angle of the triangle and its perimeter.

1. Let each interior angle be $$\theta$$
Because the triangle is equilateral, its three angles are congruent, so call each of them $$\theta$$.

2. Use the Angle–Sum Property of a triangle
For every triangle, the sum of the three interior angles is $$180^{\circ}$$. Therefore

$$\theta + \theta + \theta = 180^{\circ}$$

3. Solve for $$\theta$$

$$3\theta = 180^{\circ}$$
$$\theta = \dfrac{180^{\circ}}{3}$$
$$\theta = 60^{\circ}$$

4. Find the perimeter
The perimeter $$P$$ is the sum of the lengths of the three sides:

$$P = AB + BC + CA = a + a + a = 3a$$

Result

• Each interior angle of the equilateral triangle measures $$60^{\circ}$$.
• Its perimeter is $$3a$$ units.

Given : $$\triangle ABC$$ is isosceles with $$AB = AC = a$$ and base $$BC = 2b$$.

To prove : The altitude from the vertex $$A$$ to the base $$BC$$ is $$\sqrt{a^{2}-b^{2}}$$ (and hence the area of the triangle is $$b\sqrt{a^{2}-b^{2}}$$).

Construction : Draw altitude $$AD$$ from $$A$$ to $$BC$$ so that $$AD \perp BC$$ and $$D$$ lies on $$BC$$. (In a neat free-hand sketch show $$\triangle ABC$$ with $$AB = AC$$ and a perpendicular from $$A$$ to $$BC$$.)

Reasoning and calculation

1. Because $$\triangle ABC$$ is isosceles with $$AB = AC$$, the altitude from the vertex to the base is also the median and the angle-bisector. Hence
   $$BD = DC = \dfrac{BC}{2} = \dfrac{2b}{2} = b.$$
2. Right-angled $$\triangle ABD$$ has
   $$AB = a\; ,\; BD = b\; ,\; AD = h \; (\text{say}).$$
3. Apply Pythagoras’ theorem to $$\triangle ABD$$:
   $$AB^{2} = AD^{2} + BD^{2}.$$
   Substitute the known lengths:
   $$a^{2} = h^{2} + b^{2}.$$
4. Solve for $$h$$:
   $$h^{2} = a^{2} - b^{2} \;\Longrightarrow\; h = \sqrt{a^{2} - b^{2}}.$$
   (We take the positive root because a length is always positive.)
5. Therefore the altitude drawn to the base is
   \[AD = \sqrt{a^{2}-b^{2}}.\]
6. Optional — Area of the triangle
   $$\text{Area} = \dfrac{1}{2}\times \text{base}\times \text{height} = \dfrac{1}{2}\times 2b \times \sqrt{a^{2}-b^{2}} = b\sqrt{a^{2}-b^{2}}.$$

Hence proved.

Given data

• The three line segments that are intended to form the sides of a triangle have lengths $$3\text{ units}$$, $$4\text{ units}$$ and $$5\text{ units}$$.

We have to examine whether these three segments can indeed form a triangle and, if they do, to decide what kind of triangle they form (acute-angled, obtuse-angled or right-angled).

Step 1 : Verifying the Triangle Inequality

For any three positive numbers to be the lengths of the sides of a triangle, each one of them must be less than the sum of the other two (Triangle Inequality).

Since all three inequalities hold, the segments can be joined end-to-end to form a triangle.

Step 2 : Identifying the Type of Triangle (Converse of Pythagoras’ Theorem)

• The largest side is $$5\text{ units}$$, so if the triangle turns out to be right-angled, that side would have to be the hypotenuse.

Compute the squares of the three side lengths:

$$3^{2} = 9, \quad 4^{2} = 16, \quad 5^{2} = 25$$

Now compare the sum of the squares of the two shorter sides with the square of the longest side:

$$3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$$

Because $$3^{2} + 4^{2} = 5^{2}$$ holds exactly, the converse of the Pythagoras Theorem tells us that the angle opposite the side of length $$5\text{ units}$$ is a right angle (i.e. $$90^{\circ}$$).

Conclusion

The three given segments do form a triangle, and that triangle is a right-angled triangle whose right angle lies opposite the side of length $$5\text{ units}$$.

Given: A rectangle with consecutive sides $$a$$ and $$b$$.

Recall Brahmagupta’s formula (for a cyclic quadrilateral with sides $$a,\;b,\;c,\;d$$ and semi-perimeter $$s$$):

\[ A = \sqrt{(s-a)(s-b)(s-c)(s-d)} \quad(1) \]

where $$s = \dfrac{a+b+c+d}{2}$$.

A rectangle is a cyclic quadrilateral (each interior angle is $$90^\circ$$, so opposite angles sum to $$180^\circ$$). Its opposite sides are equal, hence

• $$a = c$$
• $$b = d$$

Step 1 : Find the semi-perimeter

$$\begin{aligned} s &= \frac{a + b + c + d}{2} \\ &= \frac{a + b + a + b}{2} \\ &= \frac{2a + 2b}{2} \\ &= a + b. \end{aligned}$$

Step 2 : Compute each factor in (1)

$$\begin{aligned} s - a &= (a + b) - a = b, \\ s - b &= (a + b) - b = a, \\ s - c &= (a + b) - c = (a + b) - a = b, \\ s - d &= (a + b) - d = (a + b) - b = a. \end{aligned}$$

Step 3 : Substitute into (1)

$$\begin{aligned} A &= \sqrt{(s-a)(s-b)(s-c)(s-d)} \\ &= \sqrt{\bigl(b\bigr)\bigl(a\bigr)\bigl(b\bigr)\bigl(a\bigr)} \\ &= \sqrt{a^2 b^2}. \end{aligned}$$

Step 4 : Simplify the square root

Because $$a>0$$ and $$b>0$$ (lengths), $$\sqrt{a^2 b^2}=ab$$.

Step 5 : Compare with the known area of a rectangle

The usual formula for the area of a rectangle is also $$A = ab$$. Thus Brahmagupta’s formula gives exactly the same area.

Conclusion: Brahmagupta’s formula is verified for the case of a rectangle.

What we have to prove
For a cyclic quadrilateral whose consecutive sides are $$a,\;b,\;c,\;d$$ Brahmagupta’s formula gives its area as

The question asks us to verify this formula when the quadrilateral is an isosceles trapezium.

1. Naming the trapezium and writing its sides

Draw an isosceles trapezium $$ABCD$$ such that

• $$AB \parallel CD$$ (these are the two bases),
• $$BC = AD = b$$ (the two equal non-parallel sides, or legs),
• Let $$AB = a$$ and $$CD = c$$ with $$c > a$$ (we could also choose the reverse, it does not matter).

Because the legs are equal and the pair of base angles are equal, the four vertices are concyclic. Hence Brahmagupta’s formula is applicable.

2. The usual (trapezium) formula for the area

Drop perpendiculars from $$C$$ and $$D$$ on $$AB$$; they meet $$AB$$ at $$P$$ and $$Q$$ respectively. Since $$AB \parallel CD$$ and the trapezium is isosceles, $$P$$ and $$Q$$ divide $$AB$$ symmetrically:

• $$AP = BQ = \dfrac{c-a}{2}$$,
• The height of the trapezium is $$h = CP = DQ$$.

In right $$\triangle DQA$$, by Pythagoras

$$h^2 = AD^2 - AQ^2 = b^2 - \left(\dfrac{c-a}{2}\right)^2 \;\;\Longrightarrow\;\; h = \sqrt{\,b^2-\left(\dfrac{c-a}{2}\right)^2\,}.$$

Hence the familiar area is

$$\text{Area}_{\text{trapezium}} = \dfrac{AB + CD}{2}\;h = \dfrac{a+c}{2}\;\sqrt{\,b^2-\left(\dfrac{c-a}{2}\right)^2\,}. \quad(1)$$

3. Brahmagupta’s formula for the same trapezium

For $$ABCD$$ the four sides are $$a,\;b,\;c,\;b$$ (in that order). Hence the semi-perimeter is

$$s = \dfrac{a + b + c + b}{2} = \dfrac{a + c + 2b}{2}.$$

Plugging these values in Brahmagupta’s formula:

\[\Delta_B = \sqrt{(s-a)(s-b)(s-c)(s-b)} = (s-b)\,\sqrt{(s-a)(s-c)}.\]

But

$$s-b = \frac{a + c + 2b}{2} - b = \frac{a + c}{2}. \quad(2)$$

Also

\(\begin{aligned} (s-a) &= \frac{a + c + 2b}{2} - a = \frac{c - a + 2b}{2},\\[2pt] (s-c) &= \frac{a + c + 2b}{2} - c = \frac{a - c + 2b}{2}. \end{aligned}\)

Therefore

\(\begin{aligned} (s-a)(s-c) &= \frac{c - a + 2b}{2}\;\frac{a - c + 2b}{2}\\ &= \frac{(2b + (c-a))(2b - (c-a))}{4}\\ &= \frac{4b^2 - (c-a)^2}{4}\\ &= b^2 - \left(\dfrac{c-a}{2}\right)^2.\quad(3) \end{aligned}\)

Taking the square root of (3) gives

$$\sqrt{(s-a)(s-c)} = \sqrt{\,b^2-\left(\dfrac{c-a}{2}\right)^2\,} = h.$$

Substituting (2) and this value of $$h$$ in Brahmagupta’s formula:

$$\Delta_B = \frac{a + c}{2}\;h. \quad(4)$$

4. Comparing the two expressions

Equation (4) obtained from Brahmagupta is exactly the same as the trapezium-area expression (1). Hence

$$\boxed{\;\Delta_B = \text{Area}_{\text{trapezium}}\;}.$$

Thus Brahmagupta’s formula is verified for an isosceles trapezium.

Let the rectangle be such that the vertices are taken in order A (top-left), B (top-right), C (bottom-right) and D (bottom-left). Thus

• AB is horizontal and $$AB = 10\text{ cm}$$,
• BC is vertical and $$BC = 8\text{ cm}$$, so $$AD = 8\text{ cm}$$ (opposite sides of a rectangle are equal).

E is any point on BC. Join A E and D E to obtain $$\triangle ADE$$.

To compute its area, drop a perpendicular from E to the base AD. Let the foot of this perpendicular be F, so that $$EF \perp AD$$.

Because AD and BC are vertical lines, EF is horizontal and therefore parallel to AB. In a rectangle opposite sides are equal; hence

$$EF = AB = 10\text{ cm}.$$

In $$\triangle ADE$$ take

• base $$AD = 8\text{ cm},$$
• corresponding height $$EF = 10\text{ cm}.$$

Therefore

\[ \text{Area of } \triangle ADE = \tfrac12 \times AD \times EF = \tfrac12 \times 8 \times 10 = 40\,\text{cm}^2 \]

Thus the required area is $$40\,\text{cm}^2$$.

Given data

• Trapezium $$ABCD$$ with $$AB\parallel CD$$.
• $$AB=40\,\text{cm},\;CD=20\,\text{cm}$$ (parallel sides).
• $$AD=BC=26\,\text{cm}$$ (equal non-parallel sides).

Since the two non-parallel sides are equal, $$ABCD$$ is an isosceles trapezium.

Drop perpendiculars from $$D$$ and from $$C$$ to the longer base $$AB$$. Let the foot of the perpendicular from $$D$$ be $$N$$, and the foot of the perpendicular from $$C$$ be $$M$$. Then $$DN\perp AB$$, $$CM\perp AB$$, and $$DN=CM=h$$, the height of the trapezium.

Step 1 : Split the longer base into three parts.

Because $$CD\parallel AB$$ and $$DN,CM$$ are both perpendicular to $$AB$$, the figure $$DNMC$$ is a rectangle, so $$NM=DC=20\,\text{cm}$$.

By the symmetry of an isosceles trapezium, the two remaining segments $$AN$$ and $$MB$$ are equal in length; call each of them $$x\,\text{cm}$$. Hence

$$AB=AN+NM+MB=x+20+x=40,$$

so $$2x=20$$ and $$x=10\,\text{cm}.$$

Step 2 : Find the height $$h$$.

Consider right $$\triangle DAN$$ (right-angled at $$N$$):

$$AN=x=10\,\text{cm},\quad DN=h,\quad AD=26\,\text{cm}.$$

By the Pythagoras theorem,

$$AD^{2}=AN^{2}+DN^{2},$$

$$26^{2}=10^{2}+h^{2}\;\Longrightarrow\;676=100+h^{2}\;\Longrightarrow\;h^{2}=576\;\Longrightarrow\;h=24\,\text{cm}.$$

Step 3 : Area of the trapezium.

$$A=\dfrac{1}{2}\times(\text{sum of parallel sides})\times h=\dfrac{1}{2}\times(40+20)\times 24=30\times 24=720\,\text{cm}^{2}.$$

\[\boxed{\,\text{Area of the trapezium}=720\,\text{cm}^{2}\,}\]

Step 1 : Identify the known data

Two sides of a triangle are given as
$$a = 8\;\text{cm}, \; b = 11\;\text{cm}$$
and its perimeter is given as
$$P = 32\;\text{cm}.$$

Step 2 : Find the third side

Let the third side be $$c\;\text{cm}$$. By definition, the perimeter of a triangle is the sum of the lengths of its three sides, so

$$a + b + c = P.$$

Substituting the known values,

$$8 + 11 + c = 32.$$

Solve for $$c$$:

$$c = 32 - (8 + 11) = 32 - 19 = 13.$$

Thus the three sides are $$8\text{ cm},\; 11\text{ cm},\; 13\text{ cm}.$$

Step 3 : Verify the triangle inequality (quick check)

• $$8 + 11 = 19 > 13\;\;(\text{True})$$
• $$11 + 13 = 24 > 8\;\;(\text{True})$$
• $$13 + 8 = 21 > 11\;\;(\text{True})$$

The three lengths can indeed form a triangle.

Step 4 : Use Heron’s formula to find the area

First compute the semi-perimeter $$s$$:

$$s = \frac{a + b + c}{2} = \frac{8 + 11 + 13}{2} = \frac{32}{2} = 16\;\text{cm}.$$

Heron’s formula states

$$\text{Area} = \sqrt{\,s(s-a)(s-b)(s-c)\,}.$$

Substitute $$s = 16\text{ cm},\; a = 8\text{ cm},\; b = 11\text{ cm},\; c = 13\text{ cm}:$$

$$\text{Area} = \sqrt{16\,\bigl(16-8\bigr)\bigl(16-11\bigr)\bigl(16-13\bigr)}$$

$$= \sqrt{16 \times 8 \times 5 \times 3}$$

Group the factors conveniently:

$$= \sqrt{(16) \times (8) \times (15)} = \sqrt{128 \times 15} = \sqrt{1920}.$$

Factor inside the radical to isolate a perfect square:

$$1920 = 64 \times 30 \quad\Longrightarrow\quad \sqrt{1920} = \sqrt{64} \times \sqrt{30} = 8\sqrt{30}.$$

\[ \boxed{\text{Area} = 8\sqrt{30}\;\text{cm}^2} \]

Therefore, the required area of the triangle is $$8\sqrt{30}\;\text{cm}^2$$ (approximately $$43.8\;\text{cm}^2$$).

Step 1 – Express the three sides in one unknown.
The ratio of the sides is given as $$3:5:7$$. Let the common multiplying factor be $$k$$ metres.
\[\text{Sides }=\;3k,\;5k,\;7k\]

Step 2 – Determine the value of $$k$$ from the perimeter.
Perimeter $$=300\text{ m}$$.

$$3k+5k+7k=300$$

$$15k=300$$

$$k=\dfrac{300}{15}=20$$

Step 3 – Write the actual lengths of the sides.

• First side: $$3k=3\times20=60\text{ m}$$
• Second side: $$5k=5\times20=100\text{ m}$$
• Third side: $$7k=7\times20=140\text{ m}$$

Step 4 – Find the semi-perimeter $$s$$ needed for Heron’s formula.

$$s=\dfrac{\text{perimeter}}{2}=\dfrac{300}{2}=150\text{ m}$$

Step 5 – Apply Heron’s formula for the area.

Heron’s formula: $$\text{Area}=\sqrt{s\,(s-a)\,(s-b)\,(s-c)}$$.

Substitute $$a=60$$, $$b=100$$, $$c=140$$, $$s=150$$.

$$\text{Area}=\sqrt{150\,(150-60)\,(150-100)\,(150-140)}$$

$$=\sqrt{150\times90\times50\times10}$$

Group the numbers conveniently:

$$150\times90=13\,500\quad\text{and}\quad50\times10=500$$

$$\text{Area}=\sqrt{13\,500\times500}=\sqrt{6\,750\,000}$$

Factor inside the radical:

$$6\,750\,000=100\times100\times(9\times25\times3)$$

Since $$\sqrt{100}=10$$, $$\sqrt{9}=3$$ and $$\sqrt{25}=5$$,

$$\text{Area}=10\times10\times3\times5\times\sqrt3$$

$$=1500\sqrt3\;\text{m}^2$$

Step 6 – Numerical approximation (if required).
Using $$\sqrt3\approx1.732$$,

$$\text{Area}\approx1500\times1.732\approx2598\text{ m}^2$$

Therefore, the area of the triangular plot is \(1500\sqrt3\,\text{m}^2\), which is approximately 2598 m².

Let the shorter diagonal be $$d_1$$ cm.

The other diagonal is twice as long, so $$d_2 = 2d_1\;\text{cm}.$$

The area $$A$$ of a rhombus in terms of its diagonals is

$$A = \frac{1}{2} \times d_1 \times d_2.$$

Given $$A = 128\,\mathrm{cm^2}$$, substitute $$d_2 = 2d_1$$:

$$128 = \tfrac{1}{2} \times d_1 \times 2d_1.$$

$$128 = d_1^2.$$

Taking the positive square root (because a length is positive):

$$d_1 = \sqrt{128} = \sqrt{64\times 2} = 8\sqrt{2}.$$

Therefore, the length of the shorter diagonal is

\[8\sqrt{2}\;\text{cm}.\]

Given : ABCD is a parallelogram and P, Q are any two points on side AB.

To prove : The ratio area $$ ( \triangle PCD ) $$ : area $$ ( \triangle QCD ) $$ equals 1 : 1.

Step 1 — Recall properties of a parallelogram

• Opposite sides are parallel; hence $$ AB \parallel CD $$ and $$ AD \parallel BC $$.

Step 2 — Identify the common base

• Both triangles $$ \triangle PCD $$ and $$ \triangle QCD $$ have the same base $$ CD $$.
  Therefore $$ \text{base}_{\triangle PCD}=\text{base}_{\triangle QCD}=CD $$.

Step 3 — Show that the heights are equal

• The height of a triangle is the perpendicular distance from the third vertex to the line containing the base.
• For $$ \triangle PCD $$, drop a perpendicular from P to the line CD. For $$ \triangle QCD $$, drop a perpendicular from Q to the line CD.
• Since P and Q both lie on AB and $$ AB \parallel CD $$, the distance of every point of AB from the line CD is the same fixed number. Consequently, the perpendiculars from P and from Q to CD are equal in length. Hence $$ \text{height}_{\triangle PCD}=\text{height}_{\triangle QCD} $$.

Step 4 — Compare the areas

The area of any triangle is

Because the two triangles share both the same base and the same height, we have

$$ \text{area}(\triangle PCD) = \tfrac12 \times CD \times h $$

and

$$ \text{area}(\triangle QCD) = \tfrac12 \times CD \times h, $$

where $$ h $$ is the common height.

Thus

$$ \text{area}(\triangle PCD) = \text{area}(\triangle QCD). $$

Step 5 — Write the required ratio

\[ \frac{\text{area}(\triangle PCD)}{\text{area}(\triangle QCD)} = \frac{\text{area}(\triangle QCD)}{\text{area}(\triangle QCD)} = 1. \]

Therefore

$$ \text{area}(\triangle PCD) : \text{area}(\triangle QCD) = 1 : 1. $$

Conclusion. No matter where P and Q are chosen on AB, the two triangles are equal in area.

Given data

• $$PQRS$$ is a parallelogram.
• $$PR$$ is one of its diagonals.
• $$O$$ is any point on the diagonal $$PR$$.

To prove. The areas of triangles $$PSO$$ and $$PQO$$ are equal, i.e.

Step 1 : The diagonal $$PR$$ divides the parallelogram into two equal-area triangles.

In $$\triangle PSR$$ and $$\triangle RQP$$:

• $$PS=RQ$$ (opposite sides of the parallelogram are equal),
• $$SR=QP$$ (opposite sides of the parallelogram are equal),
• $$PR=RP$$ (common side).

Hence $$\triangle PSR\cong\triangle RQP$$ by the SSS congruence criterion, and so

(This is the well-known result that a diagonal of a parallelogram divides it into two triangles of equal area.)

Step 2 : Relate each smaller triangle to one of the two big triangles.

(i) Consider $$\triangle PSO$$ and $$\triangle PSR$$. Both have vertex $$S$$, and their bases $$PO$$ and $$PR$$ lie on the same straight line $$PR$$. Hence the perpendicular distance from $$S$$ to that line is common to both triangles, i.e. they share the same height.

For two triangles with the same altitude, the ratio of their areas equals the ratio of their bases:

$$\dfrac{\text{ar}(\triangle PSO)}{\text{ar}(\triangle PSR)}=\dfrac{PO}{PR}.\quad(2)$$

(ii) Similarly, $$\triangle PQO$$ and $$\triangle PQR$$ share vertex $$Q$$, and their bases $$PO$$ and $$PR$$ lie on the same line $$PR$$. Hence

$$\dfrac{\text{ar}(\triangle PQO)}{\text{ar}(\triangle PQR)}=\dfrac{PO}{PR}.\quad(3)$$

Step 3 : Combine the three relations.

From (2): $$\text{ar}(\triangle PSO)=\dfrac{PO}{PR}\,\text{ar}(\triangle PSR).$$

From (3): $$\text{ar}(\triangle PQO)=\dfrac{PO}{PR}\,\text{ar}(\triangle PQR).$$

By (1), $$\text{ar}(\triangle PSR)=\text{ar}(\triangle PQR)$$, so the right-hand sides are equal. Therefore

Conclusion. The two triangles $$PSO$$ and $$PQO$$ have equal area for every position of $$O$$ on the diagonal $$PR$$ of the parallelogram $$PQRS$$. Proved.

Let $$ABCD$$ be the given 4-gon, and let the mid-points of its sides $$AB,BC,CD$$ and $$DA$$ be $$P,Q,R$$ and $$S$$ respectively. Join the points in the order $$P,Q,R,S$$. We have to show two things.

1. PQRS is a parallelogram.
2. Area$$\,(PQRS)=\dfrac12\,\text{Area}\,(ABCD).$$

Draw the two diagonals $$AC$$ and $$BD$$ of $$ABCD$$; they intersect (not necessarily at right angles) at $$O$$.

1. PQRS is a parallelogram

• In $$\triangle ABC$$, $$P$$ and $$Q$$ are the mid-points of $$AB$$ and $$BC$$, so by the Mid-point Theorem
  $$PQ\;\parallel\;AC.\quad(1)$$
• In $$\triangle CDA$$, $$R$$ and $$S$$ are the mid-points of $$CD$$ and $$DA$$, so again by the Mid-point Theorem
  $$RS\;\parallel\;AC.\quad(2)$$
• Because of (1) and (2) we have $$PQ\parallel RS.$$
• Likewise, in $$\triangle ABD$$ the mid-points $$P$$ and $$S$$ give $$PS\parallel BD,$$ and in $$\triangle BCD$$ the mid-points $$Q$$ and $$R$$ give $$QR\parallel BD.$$ Hence $$PS\parallel QR.$$

Both pairs of opposite sides are parallel, so $$PQRS$$ is a parallelogram.

2. The area of PQRS

With the help of the two diagonals the 4-gon is split up into four triangles that sit at its vertices:

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

Together with $$PQRS$$ they fill the whole 4-gon, so

(a) The pair \(\triangle BPQ\) and \(\triangle DRS\)

Consider $$\triangle ABC$$. Because $$P$$ and $$Q$$ are mid-points, $$PQ\parallel AC$$ (already used in step 1) and

$$\angle BPQ=\angle BCA,\;\angle BQP=\angle BAC;$$

hence $$\triangle BPQ\sim\triangle BCA$$ (both are right way round at $$B$$). The similarity ratio is

$$\dfrac{BP}{BA}=\dfrac12,$$

so the ratio of their areas is the square of this number:

\[\dfrac{\text{Area}(\triangle BPQ)}{\text{Area}(\triangle BCA)}=\left(\dfrac12\right)^2=\dfrac14.\quad(4)\]

Exactly the same argument in $$\triangle CDA$$ (mid-points $$R,S$$, segment $$RS\parallel AC$$) gives

Add (4) and (5):

(b) The pair \(\triangle APS\) and \(\triangle CRQ\)

Work this time with the other diagonal $$BD$$.

• In $$\triangle ABD$$ the points $$P$$ and $$S$$ are mid-points; hence $$PS\parallel BD$$, and a similarity argument exactly like that above gives

• In $$\triangle BCD$$, $$Q$$ and $$R$$ are the mid-points and $$QR\parallel BD$$, so

Add (7) and (8):

(c) Total area of the four corner triangles

Combine (6) and (9):

\[\sum_{\text{four corner }\triangle}s=\dfrac14\,\text{Area}(ABCD)+\dfrac14\,\text{Area}(ABCD)=\dfrac12\,\text{Area}(ABCD).\quad(10)\]

(d) Finish

Insert (10) into (3):

Rearrange:

This completes the proof: the parallelogram obtained by joining the mid-points of the sides of any 4-gon has exactly half the area of the original 4-gon.

Given data
In $$\triangle ABC$$, point $$D$$ is the midpoint of $$BC$$, so $$BD = DC$$. Segment $$AD$$ is therefore a median. A variable point $$P$$ is taken on $$AD$$.

To prove
$$\text{ar}(\triangle ABP)=\text{ar}(\triangle ACP).$$

Proof

1. Because $$D$$ is the midpoint of $$BC$$, we have equal bases $$BD=DC$$.
  Triangles $$\triangle ABD$$ and $$\triangle ACD$$ lie between the same parallels $$BC$$ (as base) and vertex $$A$$ (common height). Hence

$$\text{ar}(\triangle ABD)=\text{ar}(\triangle ACD)\;\;\cdots(1)$$

2. Points $$A,\,P,\,D$$ are collinear. Take $$AD$$ as the reference line (base).

• In $$\triangle ABP$$ the base is $$AP$$; its altitude is the perpendicular distance from $$B$$ to line $$AD$$.
• In $$\triangle ABD$$ the base is $$AD$$; the altitude is the same perpendicular distance from $$B$$ to line $$AD$$ (because the vertex is still $$B$$ and the base line is still $$AD$$).

Therefore, with equal heights, the areas are proportional to their bases:

$$\frac{\text{ar}(\triangle ABP)}{\text{ar}(\triangle ABD)}=\frac{AP}{AD}\;\;\cdots(2)$$

3. Exactly the same reasoning with vertex $$C$$ instead of $$B$$ gives

$$\frac{\text{ar}(\triangle ACP)}{\text{ar}(\triangle ACD)}=\frac{AP}{AD}\;\;\cdots(3)$$

4. From (2) and (3) we have

$$\frac{\text{ar}(\triangle ABP)}{\text{ar}(\triangle ABD)}=\frac{\text{ar}(\triangle ACP)}{\text{ar}(\triangle ACD)}.$$

Using (1), $$\text{ar}(\triangle ABD)=\text{ar}(\triangle ACD)$$, so the denominators in the last equality are equal. Consequently, the numerators must also be equal:

Hence, for every point $$P$$ on the median $$AD$$, the two triangles $$\triangle ABP$$ and $$\triangle ACP$$ always have equal area. The statement is proved.

Step 1 : Label the data

• Let the length of each side of the square $$ABCD$$ be $$a$$.
• Drop perpendiculars from the interior point $$P$$ to the four sides.
      • Let the distance of $$P$$ from side $$AB$$ be $$h_1$$.
      • Let the distance of $$P$$ from side $$BC$$ be $$h_2$$.
      • Let the distance of $$P$$ from side $$CD$$ be $$h_3$$.
      • Let the distance of $$P$$ from side $$DA$$ be $$h_4$$.

Step 2 : Observe two key relations

Because $$AB$$ and $$CD$$ are parallel and the distance between them is exactly the side of the square, we have

$$h_1+h_3=a$$      (1)

Similarly, since $$BC$$ || $$DA$$, we get

$$h_2+h_4=a$$      (2)

Step 3 : Write the four triangular areas

• $$\triangle PAB$$ has base $$AB=a$$ and height $$h_1$$, hence $$\text{ar}(\triangle PAB)=\tfrac12\,a\,h_1$$.
• $$\triangle PBC$$ has base $$BC=a$$ and height $$h_2$$, hence $$\text{ar}(\triangle PBC)=\tfrac12\,a\,h_2$$.
• $$\triangle PCD$$ has base $$CD=a$$ and height $$h_3$$, hence $$\text{ar}(\triangle PCD)=\tfrac12\,a\,h_3$$.
• $$\triangle PDA$$ has base $$DA=a$$ and height $$h_4$$, hence $$\text{ar}(\triangle PDA)=\tfrac12\,a\,h_4$$.

Step 4 : Form the required sums

Red region (two opposite triangles):
$$\text{Red}=\tfrac12\,a\,h_1+\tfrac12\,a\,h_3=\tfrac12\,a\,(h_1+h_3)$$

Green region (the other two triangles):
$$\text{Green}=\tfrac12\,a\,h_2+\tfrac12\,a\,h_4=\tfrac12\,a\,(h_2+h_4)$$

Step 5 : Substitute from (1) and (2)

Using $$h_1+h_3=a$$ and $$h_2+h_4=a$$, we get

$$\text{Red}=\tfrac12\,a\,a=\tfrac12\,a^2$$

$$\text{Green}=\tfrac12\,a\,a=\tfrac12\,a^2$$

Step 6 : State the ratio

Thus the two coloured regions always have equal area.

Given: In $$\triangle ABC$$ let $$D$$ be the midpoint of $$AB$$, $$P$$ any point on $$BC$$ and $$Q$$ a point on the line $$AB$$ such that $$CQ\parallel PD$$. Join $$PQ$$.

Goal

Step 1  Setting up convenient co-ordinates

• Put $$A(0,0)$$ and $$B(2b,0)\;(b>0)$$ on the x-axis, so that $$AB=2b$$.
• Because $$D$$ is the midpoint of $$AB$$, we have $$D(b,0)$$.
• Choose $$C(0,2c)\;(c>0)$$; the triangle is still perfectly general.

Step 2  Co-ordinates of the variable point P

Let $$t\,(0 $$ P\bigl(x_P,y_P\bigr)=\bigl(2b(1-t),\,2c\,t\bigr). $$

Step 3  Finding the co-ordinates of Q

The slope of $$PD$$ is

The line through $$C(0,2c)$$ parallel to $$PD$$ is $$y-2c=m x$$. Putting $$y=0$$ (because $$Q$$ lies on the x-axis) gives

Step 4  Area of $$\triangle ABC$$

With base $$AB=2b$$ and height $$2c$$ above the axis,

Step 5  Area of $$\triangle BPQ$$

Using the determinant formula for the area of a triangle whose vertices are $$B(2b,0)$$, $$P(2b(1-t),2c t)$$ and $$Q(x_Q,0)$$,

Since $$y_B=y_Q=0$$ this becomes

Substituting
$$x_B-x_Q=2b-\frac{b(2t-1)}{t}=\frac{b}{t},\qquad y_P=2c\,t,$$ we obtain

Step 6  Comparing the two areas

The result does not involve the parameter $$t$$, so it is true for every position of $$P$$ on $$BC$$. The required relation is therefore proved.

Step 1 : Formula for the area of a sector. For a sector of central angle $$\theta$$ (in degrees) in a circle of radius $$r$$,

Step 2 : Given data. $$r=7\,\text{cm},\;\theta=60^{\circ}.$$

Step 3 : Substitute and simplify.

$$A=\dfrac{60^{\circ}}{360^{\circ}}\,\pi\,(7)^{2}=\dfrac{1}{6}\times\pi\times 49=\dfrac{49\pi}{6}\,\text{cm}^{2}.$$

Step 4 : Numerical value. Taking $$\pi\approx\dfrac{22}{7}$$,

$$A\approx\dfrac{49}{6}\times\dfrac{22}{7}=\dfrac{49\times 22}{6\times 7}=\dfrac{1078}{42}=\dfrac{77}{3}\approx 25.67\,\text{cm}^{2}.$$

Therefore, the exact area is $$\dfrac{49\pi}{6}\,\text{cm}^{2}$$, which is approximately $$25.7\,\text{cm}^{2}$$.

Step 1 : Express the radius in terms of the given circumference
Let the radius of the circle be $$r\text{ cm}$$. For any circle, the circumference is given by $$C = 2\pi r$$.

The problem states that the circumference is $$44\text{ cm}$$, i.e. $$2\pi r = 44$$.

NCERT ordinarily uses the value $$\pi = \dfrac{22}{7}$$. Substituting this value, we have $$2 \times \dfrac{22}{7} \times r = 44.$$

Solve for $$r$$: $$\frac{44}{7}\,r = 44 \;\;\Rightarrow\;\; r = 44 \times \frac{7}{44} = 7.$$ Thus, $$r = 7\text{ cm}$$.

Step 2 : Find the area of the whole circle
The area of a circle is $$A = \pi r^{2}$$. Hence $$A = \frac{22}{7} \times 7^{2} = \frac{22}{7} \times 49 = 154\text{ cm}^{2}.$$

Step 3 : Compute the area of one quadrant
A quadrant is exactly one–fourth of a full circle, so $$\text{Area of quadrant} = \frac{1}{4} \times 154 = 38.5\text{ cm}^{2}.$$

Therefore, the area of the required quadrant is $$38.5 \text{ cm}^{2}.$$

The tip of the minute-hand traces a circle of radius equal to the length of the hand.

Radius $$r = 7\text{ cm}$$.

Step 1 – Central angle for 10 minutes

In 60 minutes the hand turns $$360^{\circ}$$.
In 1 minute it turns $$\dfrac{360^{\circ}}{60}=6^{\circ}$$.
Therefore, in 10 minutes it turns
$$\theta = 10 \times 6^{\circ}=60^{\circ}$$.

Step 2 – Area of the corresponding sector

Area of a sector: $$A = \dfrac{\theta}{360^{\circ}}\,\pi r^{2}$$.

Substitute $$\theta = 60^{\circ}$$ and $$r = 7\text{ cm}$$:

$$A = \dfrac{60^{\circ}}{360^{\circ}} \times \pi \times 7^{2} = \dfrac{1}{6}\,\pi\,49 = \dfrac{49\pi}{6}\text{ cm}^{2}$$.

Step 3 – Approximate value

Using $$\pi = \dfrac{22}{7}$$,

$$A = \dfrac{49 \times \tfrac{22}{7}}{6} = \dfrac{154}{6} = \dfrac{77}{3} \approx 25.7\text{ cm}^{2}$$.

Hence, the minute hand sweeps an area of $$\dfrac{49\pi}{6}\text{ cm}^{2}\,(\approx 25.7\text{ cm}^{2})$$ in 10 minutes.

Step 1 : Draw and label

Draw a circle with centre $$O$$ and radius $$r = 15\,\text{cm}$$. Mark two radii $$OA$$ and $$OB$$ such that $$\angle AOB = 60^{\circ}$$. Chord $$AB$$ subtends this angle. Shade the region bounded by arc $$AB$$ and chord $$AB$$ — that is the minor segment. The remaining part of the circle is the major segment.

Step 2 : Area of the sector $$AOB$$

The sector subtends $$60^{\circ}$$ at the centre, so

$$ \text{Area of sector } AOB = \frac{60^{\circ}}{360^{\circ}} \times \pi r^{2} $$

Substituting $$r = 15\,\text{cm}$$ and $$\pi \approx 3.14$$,

$$ \text{Area of sector } AOB = \frac16 \times 3.14 \times 15^{2} $$

$$ = \frac16 \times 3.14 \times 225 $$

$$ = \frac{706.5}{6} $$

$$ = 117.75\,\text{cm}^2 $$

Step 3 : Area of triangle $$\triangle AOB$$

Triangle $$AOB$$ is isosceles with sides $$OA = OB = 15\,\text{cm}$$ and included angle $$60^{\circ}$$.

Using $$\text{Area} = \tfrac12 ab \sin C$$,

$$ \text{Area}(\triangle AOB) = \tfrac12 \times 15 \times 15 \times \sin 60^{\circ} $$

Given $$\sqrt3 \approx 1.73$$, so $$\sin 60^{\circ} = \dfrac{\sqrt3}{2} \approx \dfrac{1.73}{2} = 0.865$$.

$$ \text{Area}(\triangle AOB) = \tfrac12 \times 225 \times 0.865 $$

$$ = 112.5 \times 0.865 $$

$$ = 97.3125 \text{ cm}^2 $$

We shall keep two–decimal accuracy: $$97.31\,\text{cm}^2$$.

Step 4 : Area of the minor segment

Minor segment area = (sector area) – (triangle area)

$$ \text{Minor segment} = 117.75\,\text{cm}^2 - 97.31\,\text{cm}^2 $$

$$ = 20.44\,\text{cm}^2 $$

Step 5 : Area of the major segment

Total area of the circle

$$ \pi r^{2} = 3.14 \times 15^{2} = 3.14 \times 225 = 706.5\,\text{cm}^2 $$

Major segment area = (circle area) – (minor segment area)

$$ \text{Major segment} = 706.5\,\text{cm}^2 - 20.44\,\text{cm}^2 $$

$$ = 686.06\,\text{cm}^2 $$

Answer :

• Area of the minor segment ≈ $$20.44\,\text{cm}^2$$
• Area of the major segment ≈ $$686.06\,\text{cm}^2$$

Step 1 – Each sweep is a sector. When a wiper moves once, the tip of its blade traces an arc of a circle whose radius equals the blade length and whose central angle equals the sweep angle. For one wiper:

• Radius $$r=28\,\text{cm}$$,
• Central angle $$\theta=120^{\circ}$$.

Step 2 – Area cleaned by one wiper. The area of a sector is

$$\text{Sector area}=\dfrac{\theta}{360^{\circ}}\,\pi r^{2}=\dfrac{120^{\circ}}{360^{\circ}}\,\pi(28)^{2}=\dfrac{1}{3}\,\pi\,(784)=\dfrac{784\pi}{3}\,\text{cm}^{2}.$$

Step 3 – Total area cleaned by both wipers. The two blades do not overlap, so the total area is twice the area cleaned by one:

$$\text{Total area}=2\times\dfrac{784\pi}{3}=\dfrac{1568\pi}{3}\,\text{cm}^{2}.$$

Step 4 – Numerical value. Using $$\pi=\dfrac{22}{7}$$,

$$\text{Total area}=\dfrac{1568}{3}\times\dfrac{22}{7}=\dfrac{1568\times 22}{3\times 7}=\dfrac{1568\times 22}{21}=\dfrac{4928}{3}\,\text{cm}^{2}\approx 1642.67\,\text{cm}^{2}.$$

(If $$\pi\approx 3.14$$ is used instead, $$\dfrac{1568\pi}{3}\approx \dfrac{1568\times 3.14}{3}\approx 1641.2\,\text{cm}^{2}$$.)

Hence the two wipers together clean an area of $$\dfrac{1568\pi}{3}\,\text{cm}^{2}\approx 1642.67\,\text{cm}^{2}$$ at each sweep.

Given. A circle of radius $$r$$ has a chord $$AB$$ that subtends $$\angle AOB=60^{\circ}$$ at the centre $$O$$.

To show. The area of the minor segment (the region bounded by the chord $$AB$$ and the minor arc $$\widehat{AB}$$) is

which is the meaning of the expression $$\pi r^{2}\left(\dfrac{1}{6}-\dfrac{\sqrt{3}}{4}\right)$$ stated in the question.

Step 1 : Area of the sector $$AOB$$. For a sector of central angle $$\theta^{\circ}$$ in a circle of radius $$r$$,

$$\text{Area of sector}=\dfrac{\theta^{\circ}}{360^{\circ}}\,\pi r^{2}.$$

For $$\theta=60^{\circ}$$,

$$\text{Area of sector }AOB=\dfrac{60^{\circ}}{360^{\circ}}\,\pi r^{2}=\dfrac{\pi r^{2}}{6}.\quad(1)$$

Step 2 : Area of $$\triangle AOB$$. Two sides of the triangle are radii ($$OA=OB=r$$) and the included angle is $$60^{\circ}$$. Hence $$\triangle AOB$$ is equilateral with side $$r$$, and its area is

$$\text{Area of }\triangle AOB=\dfrac{1}{2}\cdot r\cdot r\cdot\sin 60^{\circ}=\dfrac{1}{2}r^{2}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{4}r^{2}.\quad(2)$$

Step 3 : Area of the minor segment. The segment is what remains of the sector after the triangle is removed, so

$$\text{Area of minor segment}=\text{Area of sector}-\text{Area of }\triangle AOB=\dfrac{\pi r^{2}}{6}-\dfrac{\sqrt{3}}{4}r^{2}.$$

Factor out $$r^{2}$$:

\[\text{Area of minor segment}=r^{2}\left(\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\right),\]

which is the required result.

Note. The expression $$\pi r^{2}\!\left(\dfrac{1}{6}-\dfrac{\sqrt{3}}{4}\right)$$ in the question is intended to be read as $$r^{2}\!\left(\dfrac{\pi}{6}-\dfrac{\sqrt{3}}{4}\right)$$, i.e. the $$\pi$$ multiplies only the first term; one cannot literally pull a common factor of $$\pi$$ out of both $$\dfrac{\pi r^{2}}{6}$$ and $$\dfrac{\sqrt{3}}{4}r^{2}$$.

Hence proved.

Step 1 : Relate the side of the equilateral triangle to the radius of the circle

Let the equilateral triangle be $$\triangle ABC$$ and let the centre of the circle be $$O$$.
Because all three vertices lie on the circle, $$OA = OB = OC = r$$.

In a circle, equal chords subtend equal angles at the centre. Since $$AB = BC = CA$$, the three central angles are equal and together make one full revolution $$360^\circ$$:

$$\angle AOB = \angle BOC = \angle COA = \frac{360^\circ}{3} = 120^\circ.$$

Thus each of the three identical isosceles triangles $$\triangle AOB,\,\triangle BOC,\,\triangle COA$$ has two sides $$r,\,r$$ enclosing an angle of $$120^\circ$$.

Step 2 : Area of one isosceles triangle

The area of a triangle whose two sides are $$a$$ and $$b$$ with included angle $$C$$ is
$$\text{Area} = \tfrac12 ab \sin C.$$

Here $$a = b = r$$ and $$C = 120^\circ$$, so the area of $$\triangle AOB$$ is

$$\text{Area}(\triangle AOB) = \tfrac12 r\,r\,\sin 120^\circ = \tfrac12 r^2 \left( \tfrac{\sqrt3}{2} \right) = \tfrac{\sqrt3}{4} r^2.$$

Step 3 : Area of the whole equilateral triangle

Adding the three congruent parts,

$$\text{Area}(\triangle ABC) = 3 \times \tfrac{\sqrt3}{4} r^2 = \tfrac{3\sqrt3}{4} r^2.$$

Step 4 : Area of the circle

The circle’s area is $$\pi r^2.$$

Step 5 : Required ratio

\[ \dfrac{\text{Area of }\triangle ABC}{\text{Area of the circle}} = \dfrac{\tfrac{3\sqrt3}{4} r^2}{\pi r^2} = \dfrac{3\sqrt3}{4\pi}. \]

Step 6 : Numerical value

Using $$\sqrt3 \approx 1.732$$ and $$\pi \approx 3.142$$:

$$\dfrac{3\times 1.732}{4\times 3.142} \approx \dfrac{5.196}{12.568} \approx 0.413.$$

Hence the ratio of the area of the equilateral triangle to the area of the circle is $$\dfrac{3\sqrt3}{4\pi} \approx 0.413$$, as required.

Given: A square is inscribed in a circle whose radius is $$r$$.

Objective: Find the ratio of the area of the square to the area of the circle.

Step 1 — Relate the side of the square to the radius of the circle.

• In an inscribed (circumscribed) square, the four vertices lie on the circle.
• The diagonal of the square therefore coincides with the diameter of the circle.

If the side of the square is $$s$$, then by the Pythagoras theorem applied to one right-angled triangle formed by two sides and the diagonal, we have

$$s^2 + s^2 = (\text{diagonal})^2$$

$$2s^2 = (2r)^2$$ (because the diagonal = diameter = $$2r$$)

$$s^2 = \dfrac{(2r)^2}{2} = 2r^2$$

Therefore,

\[s = r\sqrt{2}\]

Step 2 — Compute the area of the square.

Area of square = $$s^2 = 2r^2$$.

Step 3 — Compute the area of the circle.

Area of circle = $$\pi r^2$$.

Step 4 — Form the required ratio.

$$\text{Ratio} = \dfrac{\text{Area of square}}{\text{Area of circle}} = \dfrac{2r^2}{\pi r^2} = \dfrac{2}{\pi}$$

Numerically,

$$\dfrac{2}{\pi} \approx 0.637$$.

Hence proved.

Step 1 – Draw the figure
Draw a circle with centre $$O$$ and radius $$r$$. Mark six points $$A,\,B,\,C,\,D,\,E,\,F$$ on the circumference such that the arcs AB, BC, … , FA are equal. Join adjacent points to obtain the regular hexagon $$ABCDEF$$ and join every vertex to the centre.

Step 2 – Show that each central angle is $$60^\circ$$
Because the whole circle is $$360^\circ$$ and the six arcs are equal, each arc subtends a central angle of

$$ \dfrac{360^\circ}{6}=60^\circ. $$

Step 3 – Show that the six triangles are equilateral
Consider one of the six triangles, say $$\triangle OAB$$. Two of its sides are radii, so

$$ OA = OB = r. $$

The included angle is $$\angle AOB = 60^\circ$$. In an isosceles triangle whose two equal sides enclose $$60^\circ$$, the third side is equal to the two equal sides; hence

$$ AB = r. $$

Thus $$\triangle OAB$$ is equilateral. The same argument works for the other five triangles, so the hexagon is made up of six congruent equilateral triangles.

Step 4 – Find the area of one small triangle
For an equilateral triangle of side $$s$$, the area is $$\dfrac{\sqrt3}{4}s^2$$. Taking $$s=r$$ gives

$$ \text{Area of }\triangle OAB = \dfrac{\sqrt3}{4}r^2. $$

Step 5 – Area of the hexagon
Because there are six such triangles,

$$ \text{Area of hexagon} = 6\times\left( \dfrac{\sqrt3}{4}r^2 \right) = \dfrac{6\sqrt3}{4}r^2 = \dfrac{3\sqrt3}{2}r^2. $$

Step 6 – Area of the circle
$$ \text{Area of circle}= \pi r^2. $$

Step 7 – Required ratio

\[\frac{\text{Area of hexagon}}{\text{Area of circle}} = \frac{ \dfrac{3\sqrt3}{2}r^2 }{ \pi r^2 } = \frac{3\sqrt3}{2\pi} \approx 0.827.\]

Step 8 – Why is this exactly twice the answer to Question 8?
In Question 8 you inscribed an equilateral triangle in the same circle. That triangle is obtained by taking every alternate vertex of the regular hexagon, so it is made up of three of the six small equilateral triangles found above. Hence

$$ \frac{\text{Area of triangle}}{\text{Area of circle}} = \frac{3\times\left(\dfrac{\sqrt3}{4}r^2\right)}{\pi r^2} = \frac{3\sqrt3}{4\pi}, $$

and the hexagon/circle ratio is twice this value:

$$ 2\times\frac{3\sqrt3}{4\pi} = \frac{3\sqrt3}{2\pi}. $$

Hence proved.

Recall The textbook has already shown that when we draw a square of side $$a + b$$ and divide it into convenient rectangles, the total area breaks up as

\[(a + b)^2 = a^2 + 2ab + b^2\]

The same “area‐as‐algebra” idea can be used for the two new identities. Because we cannot insert actual pictures here, the construction of each diagram is explained step–by-step so that you can reproduce it accurately in your notebook or on graph paper.

1. Diagram for $$ (a + b)(a - b) = a^2 - b^2 $$

1. Along a horizontal line, mark a segment $$AB$$ of length $$a + b$$. At the left end, mark a perpendicular segment $$AD$$ of length $$a - b$$ (assume $$a > b>0$$ so that $$a - b$$ is positive).

2. Through points $$B$$ and $$D$$ draw lines parallel to $$AD$$ and $$AB$$ respectively; their intersection point is $$C$$. Thus, $$ABCD$$ is a rectangle.

3. On $$AB$$, locate point $$E$$ so that $$AE = a$$ and $$EB = b$$. Draw a perpendicular from $$E$$ to the opposite side meeting $$DC$$ at $$F$$. You have now drawn rectangle $$AEFD$$ (area $$a(a-b) = a^2 - ab$$) and rectangle $$EBFC$$ (area $$b(a-b) = ab - b^2$$).

4. Shade rectangle $$EBFC$$ and write its area algebraically as $$b(a - b) = ab - b^2$$.

The large rectangle $$ABCD$$ represents $$ (a+b)(a-b) $$. Removing the shaded rectangle removes an area of $$ab - b^2$$ and leaves exactly a square of side $$a$$ (area $$a^2$$) minus a square of side $$b$$ (area $$b^2$$). Hence the picture visually shows

\[(a + b)(a - b) = a^2 - b^2\]

2. Diagram for $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca $$

1. Draw a large square $$PQRS$$ whose side is $$a + b + c$$.

2. On the base $$PQ$$ mark off $$PA = a$$, $$AB = b$$ and $$BQ = c$$. Through points $$A$$ and $$B$$ draw lines parallel to $$PS$$, meeting the top side $$RS$$ at $$A'$$ and $$B'$$. Similarly, on the left side $$PS$$ mark $$PC = a$$, $$CD = b$$ and $$DS = c$$. Draw lines $$C C'$$ and $$D D'$$ parallel to $$PQ$$ to meet $$RS$$.

3. The square is now cut by two vertical and two horizontal lines into nine regions. Label every smaller rectangle or square with its dimensions:

   • Region at the bottom-left corner: $$a \times a$$ (area $$a^2$$).
   • Immediately to its right: $$a \times b$$ (area $$ab$$).
   • Extreme bottom-right: $$a \times c$$ (area $$ac$$).
   • Second row, first column: $$b \times a$$ (area $$ba = ab$$).
   • Middle square: $$b \times b$$ (area $$b^2$$).
   • Second row, third column: $$b \times c$$ (area $$bc$$).
   • Top-left: $$c \times a$$ (area $$ca$$).
   • Top-middle: $$c \times b$$ (area $$cb = bc$$).
   • Top-right: $$c \times c$$ (area $$c^2$$).

4. If you now collect like terms:

   \[a^2 + b^2 + c^2 + (ab + ba) + (bc + cb) + (ca + ac)\]

   which simplifies to

   \[a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\]

Because the nine parts are a partition of the whole square, their areas must add exactly to the area of the big square; hence the diagram proves

\[(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\]

Summary of construction steps

• For $$ (a + b)(a - b) $$: draw rectangle $$ (a+b) \times (a-b) $$, cut off a smaller rectangle of width $$b$$ to reveal the difference of two squares.
• For $$ (a + b + c)^2 $$: draw a square of side $$a+b+c$$, erect vertical and horizontal segments at distances $$a$$ and $$b$$ from the edges to dissect it into nine smaller rectangles/squares.

Once the figures have been drawn, merely listing the dimensions of each small region and adding their areas yields the two required identities.

Given an isosceles triangle $$\triangle ABC$$ with equal sides

• $$AB = AC = 15\,\text{cm}$$
• Perimeter $$P = 40\,\text{cm}$$

1. Find the length of the base

The perimeter of a triangle is the sum of its three sides, so

$$AB + AC + BC = P$$

Substituting the known values,

$$15 + 15 + BC = 40$$

$$30 + BC = 40$$

$$BC = 40 - 30 = 10\,\text{cm}$$

2. Draw the altitude to obtain the height

Because $$AB = AC$$, the altitude from vertex $$A$$ to base $$BC$$ is also a median and angle-bisector. Let the foot of the altitude be $$D$$.

Then $$BD = DC = \dfrac{BC}{2} = \dfrac{10}{2} = 5\,\text{cm}$$.

Right triangle $$\triangle ABD$$ has

• Hypotenuse $$AB = 15\,\text{cm}$$
• One leg $$BD = 5\,\text{cm}$$
• Other leg $$AD$$ (the required height)

Using Pythagoras’ theorem,

$$AB^{2} = AD^{2} + BD^{2}$$

$$15^{2} = AD^{2} + 5^{2}$$

$$225 = AD^{2} + 25$$

$$AD^{2} = 225 - 25 = 200$$

$$AD = \sqrt{200} = 10\sqrt{2}\,\text{cm}$$

3. Calculate the area

The area of a triangle is

$$\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}$$

Substituting $$BC = 10\,\text{cm}$$ and $$AD = 10\sqrt{2}\,\text{cm}$$,

$$\text{Area} = \dfrac{1}{2} \times 10 \times 10\sqrt{2} = 50\sqrt{2}\,\text{cm}^{2}$$

💡 Therefore, the area of the given isosceles triangle is

\[\boxed{50\sqrt{2}\,\text{cm}^{2}}\]

Given data and notation

• The triangle is isosceles, so two of its sides are equal. Let the common length of these two equal sides be $$a\,\text{cm}$$.
• The unequal side (the base) is $$10\,\text{cm}$$ long.
• The area of the triangle is $$60\,\mathrm{cm^2}$$.

Step 1: Draw an altitude to the base

From the vertex opposite the base, draw the altitude that meets the base at a right angle. (In a diagram, this altitude would land at the midpoint of the base because the triangle is isosceles.)

• This altitude splits the base into two equal segments, each measuring $$5\,\text{cm}$$.
• The altitude itself is the height of the triangle. Let that height be $$h\,\text{cm}$$.
• Each half‐triangle formed is a right triangle whose hypotenuse is $$a$$ and whose base is $$5$$.

Step 2: Express the height $$h$$ in terms of $$a$$

Using the Pythagoras Theorem in one of the right triangles:

$$a^2 = 5^2 + h^2$$

\[\Rightarrow\; h^2 = a^2 - 25\]

\[\Rightarrow\; h = \sqrt{a^2 - 25}\quad(\text{height is positive})\]

Step 3: Relate the height to the given area

The area $$A$$ of any triangle is

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

Substituting the known base and the given area:

$$60 = \tfrac12 \times 10 \times h$$

\[\Rightarrow\; 60 = 5h\]

\[\Rightarrow\; h = 12\]

Step 4: Find $$a$$ using $$h=12$$

Insert $$h = 12$$ into $$h^2 = a^2 - 25$$:

$$12^2 = a^2 - 25$$

\[\Rightarrow\; 144 = a^2 - 25\]

\[\Rightarrow\; a^2 = 169\]

\[\Rightarrow\; a = 13\;(\because a > 0)\]

Conclusion

The two equal sides of the isosceles triangle are each $$13\,\text{cm}$$ long.

Given data

• Area of the right-angled triangle  =  $$54\;\text{cm}^2$$
• One perpendicular side (leg)  =  $$12\;\text{cm}$$

Label the triangle $$\triangle ABC$$ with the right angle at $$B$$ so that

• $$AB = 12\;\text{cm}$$ (given leg)
• $$BC = x\;\text{cm}$$ (other leg to be found)
• $$AC$$ is the hypotenuse.

Step 1 – Use the area formula

The area of a right-angled triangle is one-half the product of its legs:

$$\text{Area} = \frac12 \times AB \times BC$$

Substitute the known values:

$$54 = \frac12 \times 12 \times x$$

Simplify:

$$54 = 6x$$

$$x = \frac{54}{6} = 9$$

So, $$BC = 9\;\text{cm}$$.

Step 2 – Find the hypotenuse using Pythagoras’ theorem

The theorem gives:

$$AC^2 = AB^2 + BC^2$$

$$AC^2 = 12^2 + 9^2$$

$$AC^2 = 144 + 81 = 225$$

$$AC = \sqrt{225} = 15\;\text{cm}$$

Step 3 – Calculate the perimeter

Perimeter $$P$$ of $$\triangle ABC$$ is the sum of all three sides:

$$P = AB + BC + AC$$

$$P = 12 + 9 + 15$$

$$P = 36\;\text{cm}$$

Conclusion

The perimeter of the given right-angled triangle is $$36\;\text{cm}$$.

Step 1 ― Express the sides by a single variable

The ratio of the sides is $$2:3:4$$. Let the common multiplying factor be $$x$$. Therefore the three sides are $$2x$$, $$3x$$ and $$4x$$.

Step 2 ― Use the perimeter to find $$x$$

Perimeter $$=45\,\text{cm}$$, so

$$2x+3x+4x=45$$

$$9x=45$$

$$x=5$$

Hence the actual side-lengths are

• $$a=2x=2\times5=10\,\text{cm}$$
• $$b=3x=3\times5=15\,\text{cm}$$
• $$c=4x=4\times5=20\,\text{cm}$$

Step 3 ― Find the semi-perimeter $$s$$

$$s=\dfrac{a+b+c}{2}=\dfrac{10+15+20}{2}=\dfrac{45}{2}=22.5\,\text{cm}$$

Step 4 ― Apply Heron’s formula

Heron’s formula: $$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$$

\[\text{Area}=\sqrt{22.5\,(22.5-10)\,(22.5-15)\,(22.5-20)}\]

Compute the factors:

$$22.5-10=12.5,\;22.5-15=7.5,\;22.5-20=2.5$$

So

$$\text{Area}=\sqrt{22.5\times12.5\times7.5\times2.5}$$

Step 5 ― Exact simplification

Write every decimal as a fraction with denominator $$2$$:

$$22.5=\dfrac{45}{2},\;12.5=\dfrac{25}{2},\;7.5=\dfrac{15}{2},\;2.5=\dfrac{5}{2}$$

$$\text{Area}=\sqrt{\dfrac{45\times25\times15\times5}{2^4}}=\dfrac{1}{4}\,\sqrt{84375}$$

Prime–factorising $$84375=3^3\,5^5$$ gives

$$\sqrt{84375}=75\sqrt{15}$$

Therefore

\[\boxed{\text{Area}=\dfrac{75}{4}\sqrt{15}\;\text{cm}^2}\]

Step 6 ― Numerical value (optional)

$$\sqrt{15}\approx3.873$$, so

$$\text{Area}\approx18.75\times3.873\approx72.5\,\text{cm}^2$$

We are given a triangle whose three sides measure $$a = 7 \text{ cm},\; b = 24 \text{ cm},\; c = 25 \text{ cm}.$$

Method 1  :  Heron’s formula

Step 1 — Semi-perimeter.

$$s = \frac{a + b + c}{2} = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28 \text{ cm}.$$

Step 2 — Substitute in Heron’s formula.

Heron’s formula for the area (denoted by $$\Delta$$) is $$\Delta = \sqrt{s\,(s - a)(s - b)(s - c)}.$$

Insert the numerical values:

$$\Delta = \sqrt{28\,(28 - 7)\,(28 - 24)\,(28 - 25)} = \sqrt{28 \times 21 \times 4 \times 3}.$$

Compute step by step so that nothing is skipped:

• $$28 \times 21 = 588$$
• $$4 \times 3 = 12$$
• $$588 \times 12 = 7056$$

Therefore $$\Delta = \sqrt{7056} = 84 \text{ cm}^2.$$

Method 2  :  Using the right angle (Pythagoras)

First confirm whether the triangle is right-angled:

$$7^2 + 24^2 = 49 + 576 = 625 = 25^2.$$

Because the square of the longest side equals the sum of the squares of the other two sides, the triangle is right-angled, with the right angle opposite the 25 cm side. Hence the sides 7 cm and 24 cm are perpendicular.

The area of a right-angled triangle is

$$\Delta = \frac12 \times (\text{product of the perpendicular sides})$$

$$\;\;\;\;\;\; = \frac12 \times 7 \times 24 = \frac12 \times 168 = 84 \text{ cm}^2.$$

Both methods give exactly the same result.

Given data

• Diameter of the bicycle wheel: $$d = 60\,\text{cm}$$
• Number of complete revolutions: $$N = 100$$

Step 1 · Find the radius

Radius is half of the diameter:

$$r = \frac{d}{2} = \frac{60\,\text{cm}}{2} = 30\,\text{cm}$$

Step 2 · Circumference of the wheel

The distance travelled in one complete revolution equals the circumference of the wheel. Using $$C = 2\pi r$$ (or $$\pi d$$):

$$C = 2 \pi r = 2 \pi (30\,\text{cm}) = 60 \pi \,\text{cm}$$

Step 3 · Total distance for 100 revolutions

Distance = circumference × number of revolutions

$$\text{Distance} = C \times N = (60 \pi) \times 100 = 6000 \pi \,\text{cm}$$

Step 4 · Convert to metres and evaluate

1 metre = 100 cm, so

$$6000 \pi \,\text{cm} = \frac{6000 \pi}{100} \,\text{m} = 60 \pi \,\text{m}$$

Using $$\pi \approx \tfrac{22}{7}$$:

$$60 \pi \text{ m} \approx 60 \times \frac{22}{7} \text{ m} = \frac{1320}{7} \text{ m} \approx 188.57 \text{ m}$$

Therefore, the cyclist travels about 188.6 metres (to one decimal place) after 100 rotations of the wheel.

Step 1 : Relate the given circumference to the radius

The circumference C of a circle and its radius r are connected by the formula

$$C = 2\pi r$$

We are told that the circumference is 66 cm, so

$$2\pi r = 66$$

Step 2 : Solve for r

Divide both sides by $$2\pi$$:

$$r = \dfrac{66}{2\pi}$$

The NCERT textbook works with $$\pi = \dfrac{22}{7}$$. Substituting this value, we get

$$r = \dfrac{66}{2 \times \dfrac{22}{7}} = \dfrac{66}{\dfrac{44}{7}}$$

To divide by a fraction, multiply by its reciprocal:

$$r = 66 \times \dfrac{7}{44}$$

First reduce the fraction:

$$\dfrac{66}{44} = \dfrac{3}{2}$$  (because 22 is a common factor)

Hence

$$r = \dfrac{3}{2} \times 7 = \dfrac{21}{2} = 10.5\text{ cm}$$

Step 3 : Find the area of the whole circle

The area A of a circle is

$$A = \pi r^2$$

Substitute $$r = 10.5\text{ cm}$$ and $$\pi = \dfrac{22}{7}$$:

$$A = \dfrac{22}{7}\,(10.5)^2$$

Compute $$ (10.5)^2 $$ first:

$$ (10.5)^2 = 10.5 \times 10.5 = 110.25 $$

Now multiply:

$$ A = \dfrac{22}{7} \times 110.25 $$

$$\phantom{A} = \dfrac{22 \times 110.25}{7}$$

$$\phantom{A} = \dfrac{2425.5}{7} = 346.5\text{ cm}^2$$

Step 4 : Extract the area of one quadrant

A quadrant is one–fourth of a circle, so

$$\text{Area of quadrant} = \dfrac{1}{4} \times \text{Area of circle}$$

$$\text{Area of quadrant} = \dfrac{1}{4} \times 346.5$$

$$\text{Area of quadrant} = 86.625\text{ cm}^2$$

Step 5 : State the result

The area of the required quadrant is therefore

\[ \boxed{86.625\text{ cm}^2} \]

Step 1 : Understand the quantities given
The outer radius of the wheel is given; hence this is the radius of the circle traced by every point on the tyre.
Given radius : $$r = 28\,\text{cm}$$

Step 2 : Distance covered in one full turn (the circumference)
For a circle of radius $$r$$, the circumference is $$C = 2\pi r$$.
Using $$\pi = \dfrac{22}{7}$$ (the value recommended in NCERT), we have
\[C = 2 \times \frac{22}{7} \times 28\]

Carry out the arithmetic step by step:

• First simplify the fraction: $$\dfrac{28}{7} = 4$$.
• Thus $$C = 2 \times 22 \times 4$$.
• Multiplying: $$2 \times 22 = 44$$, and $$44 \times 4 = 176$$.

Hence the distance the car travels in one complete revolution of its wheel is
\[C = 176\,\text{cm}\]

Because 100 cm = 1 m, convert into metres for everyday use:
$$176\,\text{cm} = \dfrac{176}{100}\,\text{m} = 1.76\,\text{m}$$.

Step 3 : Convert the journey length into the same unit
1 kilometre = 1000 metres = 1000 × 100 cm = 100 000 cm.

Step 4 : How many turns cover 1 km?
The wheel turns once whenever it covers 176 cm. Therefore the required number of turns $$n$$ is
$$n = \frac{\text{total distance}}{\text{distance per turn}} = \frac{100\,000\,\text{cm}}{176\,\text{cm}}$$

Calculate the quotient:
\[n = \frac{100\,000}{176} \approx 568.18\]

• Exactly, $$n = 568\dfrac{32}{176} = 568\dfrac{2}{11}$$ turns.
• In practical terms, we say the wheel makes about $$568$$ complete turns (a little more than 568).

Conclusion
Distance per complete revolution : 1.76 m (or 176 cm).
Number of revolutions in 1 km : approximately 568.

Step 1 · Name the two rectangles

Let  Rectangle I have length $$l_1$$ and breadth $$b_1$$.
Let  Rectangle II have length $$l_2$$ and breadth $$b_2$$.
(All lengths are positive.)

Step 2 · Translate the given information into equations

• Same area  ⇒  $$l_1 b_1 = l_2 b_2$$  …(1)
• Same perimeter  ⇒  $$2\,(l_1 + b_1) = 2\,(l_2 + b_2)\;\Rightarrow\; l_1 + b_1 = l_2 + b_2$$  …(2)

Step 3 · Form the quadratic whose roots are the two side-lengths

Denote the common sum and product by

$$S = l_1 + b_1 = l_2 + b_2, \qquad P = l_1 b_1 = l_2 b_2.$$

Consider the quadratic equation

$$x^2 - Sx + P = 0.$$

Because $$l_1$$ and $$b_1$$ satisfy it, they are its roots. For the same reason, $$l_2$$ and $$b_2$$ are also roots.

Step 4 · Use uniqueness of the roots

A quadratic equation with real coefficients has at most two real roots. Hence the unordered pairs of roots must be identical:

• Either $$l_1 = l_2$$ and $$b_1 = b_2$$,
• or  $$l_1 = b_2$$ and $$b_1 = l_2$$.

Step 5 · Interpret geometrically

In the second possibility the lengths are merely interchanged; rotating one rectangle through $$90^{\circ}$$ makes the correspondence exact. Thus, in either case the two rectangles have the same pair of side-lengths.

Step 6 · Conclusion

Having equal and correspondingly matching sides, the rectangles are congruent.

Given. A trapezium $$ABCD$$ with $$AB \parallel CD$$, $$AB = a$$, $$CD = b$$ and the perpendicular distance between the parallel sides equal to $$h$$.

Step 1 — Create a copy. Make an exact copy of $$ABCD$$, name it $$A'B'C'D'$$. Rotate (turn) this copy through $$180^{\circ}$$ and place it so that

• side $$AB$$ of the first trapezium continues straight on to $$C'D'$$ of the copy,
• side $$CD$$ of the first continues straight on to $$A'B'$$ of the copy.

The two trapeziums now form a single quadrilateral $$ABCD'A'B'$$.

Step 2 — Why is the new quadrilateral a parallelogram?

• In each trapezium $$AB \parallel CD$$ and $$A'B' \parallel C'D'$$.
• Because we joined $$AB$$ in a straight line with $$C'D'$$, the whole edge $$ABB'C'D'$$ is one straight segment. Similarly $$CD$$ joins $$A'B'$$ in a straight line.
• Thus the opposite edges of the new quadrilateral are pairs of straight, parallel lines, so the figure is a parallelogram.

Step 3 — Dimensions of the parallelogram.

• Base. One long side is made of segments $$AB$$ and $$C'D'$$, i.e. length $$a+b$$.
• Height. Rotation has not changed the vertical separation of the parallel sides, so the height is still $$h$$.

Step 4 — Area of the parallelogram.
Using the known formula,

Area $$(\text{parallelogram}) = (a+b) \times h = (a+b)h.$$

Step 5 — Area of one trapezium.
The parallelogram is composed of two congruent trapeziums, therefore

Area $$(\text{one trapezium}) = \dfrac{1}{2}(a+b)h.$$

Thus we have shown

as required.

Given. A trapezium $$ABCD$$ in which $$AB \parallel CD$$.

Let

• the length of the upper parallel side be $$AB = a$$,
• the length of the lower parallel side be $$CD = b$$,
• the perpendicular distance between the two parallel sides (the height) be $$h$$.

Construction : Join the non-parallel vertices with a diagonal $$AC$$. The diagonal divides the trapezium into two triangles Δ₁ $$ABC$$ and Δ₂ $$ADC$$.

You may illustrate the construction by drawing trapezium $$ABCD$$ with $$AB$$ the shorter parallel side on top, $$CD$$ the longer one below, then draw diagonal $$AC$$.

1. Area of ΔABC
   In Δ$$ABC$$, the base is $$AB$$ and its altitude is the height of the trapezium, $$h$$ (because a perpendicular dropped from any point on $$AB$$ to the line $$CD$$ has length $$h$$).
   Therefore
   $$ \text{Area}(\triangle ABC)=\frac12 \times AB \times h=\frac12 a h. $$

2. Area of ΔADC
   In Δ$$ADC$$, the base is $$CD$$ and its altitude is again the same perpendicular distance $$h$$ (the height of the trapezium).
   Thus
   $$ \text{Area}(\triangle ADC)=\frac12 \times CD \times h=\frac12 b h. $$

3. Area of trapezium $$ABCD$$
   The trapezium is exactly the union of the two triangles formed, so
   $$ \text{Area}(ABCD)=\text{Area}(\triangle ABC)+\text{Area}(\triangle ADC) \;.$$ Substituting from steps 1 and 2, we get $$ \text{Area}(ABCD)=\frac12 a h+\frac12 b h=\frac12(a+b)h. $$

Conclusion. A trapezium’s area equals half the sum of the lengths of its parallel sides multiplied by the height, i.e.

\[ \boxed{\text{Area of trapezium}=\dfrac{(\text{sum of parallel sides})\times(\text{height})}{2}} \]

Step 1 — Recall the parts of a trapezium

• In a trapezium (also called a trapezoid) only one pair of opposite sides is parallel.
• Call the parallel sides $$a$$ and $$b$$ with $$a \gt b$$ (their order is immaterial, we only want to distinguish the two lengths).
• Let the distance between the parallel sides be the height $$h$$ of the trapezium.

Step 2 — Make a copy and place it upside-down

1. Draw the given trapezium ABCD in which AB ∥ CD, $$AB = a$$, $$CD = b$$ and the perpendicular distance between the lines $$AB$$ and $$CD$$ is $$h$$.
2. Make another copy of the same trapezium. Call the copy A^\*B^\*C^\*D^\*. This copy is congruent to the first one, so its corresponding sides are also $$a, b$$ and its height is $$h$$.
3. Rotate the copy through 180° in the plane (turn it upside-down). The side that was on the top will now be at the bottom.
4. Now slide the rotated copy so that its non-parallel side that was next to $$a$$ lies exactly along the non-parallel side of the original trapezium that is next to $$b$$.

Diagram to draw: Draw the original trapezium with its longer base on the top. Next to it, attach the rotated copy so that together they look like a slanted parallelogram. The two long bases form one continuous slanted side, and the two short bases form the opposite side.

Step 3 — Why the new figure is a parallelogram

• The two copies are congruent, so each pair of corresponding non-parallel sides is equal in length.
• When we place one non-parallel side of the original exactly on the matching non-parallel side of the copy, the remaining free sides are:
  • the two long bases, each of length $$a$$, lying one after the other, making a single straight line of length $$a + a$$;
  • the two short bases, each of length $$b$$, lying one after the other, making the opposite straight side of length $$b + b$$.
• Thus we have a quadrilateral whose opposite sides are equal and parallel; hence the whole figure is a parallelogram. Its two different side-lengths are $$a + b$$ and $$h$$.

Step 4 — Area of the parallelogram obtained

The area of any parallelogram equals its base multiplied by its corresponding height.

Take the longer side $$a + b$$ as the base; its perpendicular distance to the opposite side is still $$h$$ (exactly the same height of the original trapezium, because the two copies were just moved, not stretched).

Therefore

Step 5 — Area of one trapezium

• The parallelogram is made of two congruent copies of the original trapezium.
• Hence the area of one copy is exactly half the area of the parallelogram.

Step 6 — Write the standard formula

Since $$a$$ and $$b$$ are the two parallel sides and $$h$$ is the distance between them, the result can be quoted as:

Thus, by assembling two identical trapezia into a parallelogram we have proved the usual area formula for a trapezium.

Fig. 6.43 — the triangle

Let the vertices of the triangle be $$A(0,0),\;B(1,0)\text{ and }C(0,1)$$. Each side is divided into three equal parts and we join the point that is one-third of the way along every side in the same cyclic order. Thus

• on $$AB$$ we take $$P(1/3,0)$$ (so that $$AP:PB=1:2$$);
• on $$BC$$ we take $$Q\bigl(2/3,1/3\bigr)$$ (so that $$BQ:QC=1:2$$);
• on $$CA$$ we take $$R(0,2/3)$$ (so that $$CR:RA=1:2$$).

The shaded triangle is $$\triangle PQR$$. Using the co-ordinate (shoelace) formula, its area is

The area of the whole triangle $$ABC$$ is

Hence

Exactly one-third of the triangle is shaded.

Fig. 6.44 — the square

Let the square be $$ABCD$$ with side 1 and vertices

$$A(0,0),\;B(1,0),\;C(1,1),\;D(0,1).$$

The mid-points of the opposite sides are

• $$M_1\bigl(\tfrac12,1\bigr)$$ on $$CD$$,
• $$M_2\bigl(0,\tfrac12\bigr)$$ on $$AD$$,
• $$M_3\bigl(\tfrac12,0\bigr)$$ on $$AB$$,
• $$M_4\bigl(1,\tfrac12\bigr)$$ on $$BC$$.

Join each vertex to the mid-point of the opposite side:

$$A\!M_1,\;B\!M_2,\;C\!M_3,\;D\!M_4.$$

The four straight lines just written meet pair-wise in a smaller central square. We now find its vertices.

1. Line $$A\!M_1$$ has parametric form $$(0,0)+t\,(\tfrac12,1)=\bigl(\tfrac12t,\,t\bigr).$$

2. Line $$B\!M_2$$ has parametric form $$(1,0)+s\,(-1,\tfrac12)=\bigl(1-s,\,\tfrac12s\bigr).$$

Equating coordinates gives $$\tfrac12t=1-s,\;t=\tfrac12s\;\Rightarrow\;s=\tfrac45,\;t=\tfrac25.$$ Thus the first corner is $$P\bigl(\tfrac15,\,\tfrac25\bigr).$$ Working round the figure in exactly the same way we obtain

• $$Q\bigl(\tfrac35,\,\tfrac15\bigr)$$ from $$B\!M_2\cap C\!M_3,$$
• $$R\bigl(\tfrac45,\,\tfrac35\bigr)$$ from $$C\!M_3\cap D\!M_4,$$
• $$S\bigl(\tfrac25,\,\tfrac45\bigr)$$ from $$D\!M_4\cap A\!M_1.$$

Using the shoelace formula for the quadrilateral $$PQRS$$ in that order,

The original square has area $$1^2=1$$, so

Exactly one-fifth of the square is shaded.

Given figures

• Fig. 6.45 shows a rectangle in which three congruent circles are arranged side–by-side so that each touches its neighbours and both the longer as well as the shorter sides of the rectangle.
• Fig. 6.46 shows the same situation, but with four congruent circles in a row.

Let the (common) radius of every circle be $$r$$. All the required dimensions of the two rectangles can be expressed in terms of this single unknown.

1. Dimensions of the rectangles

Height (common to both figures)

Because each circle touches the top and the bottom of the rectangle, the height equals the diameter:

$$\text{height}=2r$$

Width of Fig. 6.45

Three circles placed in a row touch one another successively, so the width equals three diameters:

$$\text{width}_{(3\text{ circles})}=3\times 2r=6r$$

Width of Fig. 6.46

Similarly, with four circles,

$$\text{width}_{(4\text{ circles})}=4\times 2r=8r$$

2. Areas of the rectangles

Fig. 6.45

$$\text{Area}_{\text{rect.,3}}=\text{height}\times\text{width}=\bigl(2r\bigr)\bigl(6r\bigr)=12r^{2}$$

Fig. 6.46

$$\text{Area}_{\text{rect.,4}}=\bigl(2r\bigr)\bigl(8r\bigr)=16r^{2}$$

3. Combined area of the circles

Fig. 6.45

There are three identical circles:

$$\text{Area}_{\text{circles,3}}=3\times(\pi r^{2})=3\pi r^{2}$$

Fig. 6.46

There are four identical circles:

$$\text{Area}_{\text{circles,4}}=4\times(\pi r^{2})=4\pi r^{2}$$

4. Required fractions

Fig. 6.45

\[ \text{Fraction covered}=\frac{\text{area of 3 circles}}{\text{area of rectangle}}=\frac{3\pi r^{2}}{12r^{2}}=\frac{\pi}{4} \]

Fig. 6.46

\[ \text{Fraction covered}=\frac{4\pi r^{2}}{16r^{2}}=\frac{\pi}{4} \]

5. Numerical approximation

Using $$\pi\approx3.14$$,

$$\dfrac{\pi}{4}\approx0.785\;\text{(or}\;78.5\%\text{)}$$

Hence, in both figures the circles cover the same fraction of the rectangle.

Given activity (recall)
On a sheet we first draw a long rectangle. We then draw a grid of equal squares inside the rectangle and in each square we draw a circle that just touches all four sides of the square. Thus every square contains exactly one circle, and the squares fill the rectangle completely without gaps or overlaps.

Suppose the rectangle contains n such circles. We want to guess (conjecture) what fraction of the area of the rectangle is actually covered by the circles, then test the guess for some particular values of n, and finally prove our guess.

Step 1 – Notation for one circle & its square

• Let the radius of every circle be $$r\;\text{cm}$$.
• Therefore the diameter of every circle is $$2r\;\text{cm}$$.
• Each circle is inscribed in a square whose side is also $$2r\;\text{cm}$$ (because the circle just touches all four sides).

Step 2 – Area of one pair “circle + its square”

• Area of the circle
  \[A_{\text{circle}} = \pi r^{2}\]
• Area of the square
  \[A_{\text{square}} = (2r)^{2} = 4r^{2}\]
• Fraction of the square that is covered by its circle
  \[\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi r^{2}}{4r^{2}} = \frac{\pi}{4}\;\;\approx\;0.7854\,(78.54\%)\]

Step 3 – Extending to n circles

Because all the squares are congruent, a rectangle containing n such squares has

• Total area of circles
  \[A_{\text{circles}} = n \times \pi r^{2} = n\pi r^{2}\]
• Total area of rectangle (which is the same as the combined area of the n squares)
  \[A_{\text{rect}} = n \times 4r^{2} = 4nr^{2}\]
• Fraction of the rectangle that is covered by all the circles
  \[\frac{A_{\text{circles}}}{A_{\text{rect}}} = \frac{n\pi r^{2}}{4nr^{2}} = \frac{\pi}{4}\]

Conjecture
No matter how many equal circles we arrange in this way, the circles will always occupy exactly $$\dfrac{\pi}{4}\;(≈78.5\%)$$ of the area of the containing rectangle.

Step 4 – Testing the conjecture

Let us take $$r = 1\;\text{cm}$$ to keep the numbers small.

In every case the ratio is practically $$0.7854$$, i.e. $$\dfrac{\pi}{4}$$, exactly as predicted.

Step 5 – Formal proof

Let the rectangle be completely subdivided into n congruent squares, each of side $$2r$$, with a circle of radius $$r$$ inscribed in every square. By simple multiplication the total area of the rectangle is $$4nr^{2}$$ and the total area of all the circles is $$n\pi r^{2}$$. Because both totals contain the common factor $$nr^{2}$$, their ratio simplifies to $$\tfrac{\pi}{4}$$, a constant independent of n.

Hence our conjecture is proved: