NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles — Exercise 11.3

NCERT Exemplar Problems, Class 10 Mathematics — Areas Related to Circles, Exercise 11.3. All 16 questions solved.

Sixteen questions, and nine of them are shaded-region problems: a composite figure where the answer is one area minus, or plus, another. These have a reputation for being fiddly, and the reputation is undeserved — every one of them yields to the same three steps.

1. Decompose. Name the shapes the figure is built from. Almost always a rectangle, square, triangle or trapezium, together with circles, semicircles, quadrants or sectors.
2. Find each piece separately, writing down which is being added and which subtracted before computing anything.
3. Combine, and sanity-check against the whole. The shaded part must be smaller than the figure containing it, and usually a recognisable fraction of it.

The formulae are few:

$$\text{sector} = \frac{\theta}{360^\circ}\pi r^2, \qquad \text{arc} = \frac{\theta}{360^\circ} \times 2\pi r, \qquad \text{minor segment} = \text{sector} – \text{triangle}$$

And one fact does a surprising amount of work: the angles of any triangle sum to $180^\circ$, and of any quadrilateral to $360^\circ$. Questions 13 and 15 are nothing but that, applied to sectors drawn at the corners.

A note on the current syllabus. “Areas of Combinations of Plane Figures” has been removed from the Class 10 textbook — the chapter is now a single section on sectors and segments, and the phrase does not occur in it. Nine of these sixteen questions are combination problems, so they are exemplar-only material now. They remain the most useful thing in the chapter, and the decomposition habit is worth having whatever the syllabus says.

Unless a question states otherwise, $\pi = \dfrac{22}{7}$ — the chapter’s own note. Questions 7, 11 and 12 say to use $3.14$.

Key insight. When sectors are drawn at every corner of a polygon with the same radius, their angles add to the polygon’s angle sum. Three corners of a triangle give $180^\circ$ — half a circle. Four corners of a quadrilateral give $360^\circ$ — a whole one. You never need to know the individual angles, or even the shape, which is why Questions 13 and 15 give you neither.

Question 1

Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii $15$ cm and $18$ cm.

Solution. Circumference is proportional to radius, so adding circumferences adds radii:

$$2\pi R = 2\pi(15) + 2\pi(18)$$

Dividing by $2\pi$:

$$R = 15 + 18 = 33 \text{ cm}$$

No value of $\pi$ is needed, because it cancels — a sign that the relation is purely one of lengths.

Contrast this with Questions 1 and 10 of Exercise 11.1, where areas were added and the radii combined as $R^2 = R_1^2 + R_2^2$. Here that would have given $\sqrt{225 + 324} = \sqrt{549} \approx 23.4$ cm, which is not the answer to this question.

Question 2

In Fig. 11.5, a square of diagonal $8$ cm is inscribed in a circle. Find the area of the shaded region.

Solution. The shaded region is the circle with the square removed, so find each separately.

The circle. The square’s corners lie on the circle, so the square’s diagonal is a diameter:

$$\text{diameter} = 8 \text{ cm} \implies r = 4 \text{ cm} \implies \text{area} = \pi(4)^2 = 16\pi \text{ cm}^2$$

The square. With both diagonals equal to $8$ and perpendicular:

$$\text{area} = \frac12 d_1 d_2 = \frac12 \times 8 \times 8 = 32 \text{ cm}^2$$

(Or: side $= \frac{8}{\sqrt2} = 4\sqrt2$, and $(4\sqrt2)^2 = 32$.)

The shaded region:

$$16\pi – 32 \text{ cm}^2 \approx 50.27 – 32 = 18.27 \text{ cm}^2$$

The answer is left in terms of $\pi$ because the question does not name a value for it.

A check on plausibility: the square occupies $\frac{32}{16\pi} \approx 64\%$ of the circle, so the four corner pieces make up the other $36\%$ — which looks right in the figure.

Question 3

Find the area of a sector of a circle of radius $28$ cm and central angle $45^\circ$.

Solution. Straight substitution into the sector formula:

$$\text{area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{45}{360} \times \frac{22}{7} \times 28^2$$

Simplify before multiplying — $\frac{45}{360} = \frac18$, and $28^2 = 784$, with $784 \div 7 = 112$:

$$= \frac18 \times 22 \times 112 = \frac{2464}{8} = 308 \text{ cm}^2$$

A check: the whole circle has area $\frac{22}{7} \times 784 = 2464$ cm², and a $45^\circ$ sector is one-eighth of it. Indeed $\frac{2464}{8} = 308$ ✓.

Choosing a radius that is a multiple of $7$ is the book’s way of keeping $\frac{22}{7}$ exact; whenever you see $7$, $14$, $21$, $28$ or $35$ in one of these questions, the cancellation is intended and the answer will be a whole number.

Question 4

The wheel of a motor cycle is of radius $35$ cm. How many revolutions per minute must the wheel make so as to keep a speed of $66$ km/h?

Solution. The units have to be made compatible before anything else — this is a unit-conversion question with a circle attached.

Distance per minute. At $66$ km/h:

$$\frac{66 \text{ km}}{1 \text{ hour}} = \frac{66\,000 \text{ m}}{60 \text{ min}} = 1100 \text{ m/min} = 110\,000 \text{ cm/min}$$

Distance per revolution. One revolution lays down one circumference:

$$2\pi r = 2 \times \frac{22}{7} \times 35 = 220 \text{ cm}$$

Revolutions per minute:

$$\frac{110\,000}{220} = 500$$

So the wheel turns $500$ times a minute.

Work in centimetres throughout, since the radius is given in centimetres. Mixing kilometres with centimetres is where this question goes wrong, and the error is invisible in the final number — $500$ looks as plausible as $5$ or $50\,000$.

Question 5

A cow is tied with a rope of length $14$ m at the corner of a rectangular field of dimensions $20$ m $\times$ $16$ m. Find the area of the field in which the cow can graze.

Solution. The cow can reach any point within $14$ m of the corner that is inside the field. At a corner of a rectangle the interior angle is $90^\circ$, so the accessible region is a quarter circle of radius $14$ m.

Before using that, check the rope does not overshoot either side. The two sides meeting at the corner are $20$ m and $16$ m, and the rope is $14$ m — shorter than both. So the quarter circle lies entirely within the field and no part of it is cut off.

$$\text{area} = \frac{90}{360} \times \pi r^2 = \frac14 \times \frac{22}{7} \times 196 = \frac14 \times 616 = 154 \text{ m}^2$$

The check matters. Had the rope been $18$ m, it would have exceeded the $16$ m side, and the quarter circle would have run off the end — the grazed region would then be the quarter circle minus the overhang, and the answer would be quite different. Always compare the rope against the sides before assuming a clean quarter circle.

Question 6

Find the area of the flower bed (with semi-circular ends) shown in Fig. 11.6.

Solution. Read the figure carefully: the $38$ cm arrow spans the rectangular part only, not the whole bed. The dimension line stops where each semicircle begins, as the dashed lines above show. Getting this wrong changes the answer, and the next question sets the same shape the other way round.

The semicircular ends. Each has the bed’s width as its diameter:

$$\text{diameter} = 10 \text{ cm} \implies r = 5 \text{ cm}$$

Two semicircles of the same radius make one full circle:

$$\text{area of both ends} = \pi r^2 = 25\pi \text{ cm}^2$$

The rectangle.

$$38 \times 10 = 380 \text{ cm}^2$$

Total:

$$380 + 25\pi \text{ cm}^2 \approx 380 + 78.5 = 458.5 \text{ cm}^2$$

Combining the two semicircles into one circle before computing is worth doing every time: it removes a halving step and the error that goes with it.

Question 7

In Fig. 11.7, $AB$ is a diameter of the circle, $AC = 6$ cm and $BC = 8$ cm. Find the area of the shaded region (Use $\pi = 3.14$).

Solution. The shaded region is the circle minus the triangle. Two facts unlock it.

The angle at $C$ is a right angle. $AB$ is a diameter and $C$ is on the circle, so $\angle ACB = 90^\circ$ — the angle in a semicircle.

So $AB$ is the hypotenuse, and Pythagoras gives it:

$$AB = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10 \text{ cm}$$

the $6$–$8$–$10$ triangle. Since $AB$ is a diameter, the radius is $5$ cm.

The circle:

$$\pi r^2 = 3.14 \times 25 = 78.5 \text{ cm}^2$$

The triangle. Right angled at $C$, so the two given sides are its legs:

$$\frac12 \times 6 \times 8 = 24 \text{ cm}^2$$

The shaded region:

$$78.5 – 24 = 54.5 \text{ cm}^2$$

The right angle at $C$ is doing double duty — it makes $AB$ the hypotenuse and makes $AC$, $BC$ the base and height. Without it neither step works.

Question 8

Find the area of the shaded field shown in Fig. 11.8.

Solution. The field is a rectangle with a semicircle attached below part of it. The figure gives three numbers and only two of them are needed directly — the third fixes the semicircle.

The rectangle is $8$ m by $4$ m:

$$8 \times 4 = 32 \text{ m}^2$$

The semicircle. The total height on the left is $6$ m, of which the rectangle accounts for $4$ m. So the semicircle projects

$$6 – 4 = 2 \text{ m}$$

below the rectangle. That projection is the semicircle’s radius, so $r = 2$ m and its diameter is $4$ m — which is why it occupies only part of the rectangle’s lower edge.

$$\text{semicircle} = \frac12 \pi r^2 = \frac12 \times \pi \times 4 = 2\pi \text{ m}^2$$

Total:

$$32 + 2\pi \text{ m}^2 \approx 32 + 6.28 = 38.28 \text{ m}^2$$

The $6$ m is the measurement most easily misused: it is the overall height, not a dimension of either piece, and the semicircle’s radius has to be extracted from it by subtraction.

Question 9

Find the area of the shaded region in Fig. 11.9.

Solution. A rectangle with a stadium-shaped hole. The hole’s dimensions are not given directly — they have to be recovered from the margins, and here the margins are measured to the extreme ends of the hole, which is the opposite convention to Question 6.

The hole’s overall size.

$$\text{length} = 26 – 3 – 3 = 20 \text{ m}, \qquad \text{width} = 12 – 4 – 4 = 4 \text{ m}$$

Decompose the hole. Its width is $4$ m, so the semicircular ends have diameter $4$ m and radius $2$ m. Those ends account for $2 + 2 = 4$ m of the overall length, leaving

$$\text{rectangular part of the hole} = 20 – 4 = 16 \text{ m by } 4 \text{ m}$$

The hole’s area:

$$16 \times 4 + \pi(2)^2 = 64 + 4\pi \text{ m}^2$$

again combining the two semicircles into one circle.

The outer rectangle:

$$26 \times 12 = 312 \text{ m}^2$$

The shaded region:

$$312 – (64 + 4\pi) = 248 – 4\pi \text{ m}^2 \approx 248 – 12.57 = 235.4 \text{ m}^2$$

Compare with Question 6. Both figures are stadium shapes, and the given length means something different in each — the rectangle there, the whole shape here. Neither question says so; the figures do, and that is why they have to be read rather than skimmed.

Question 10

Find the area of the minor segment of a circle of radius $14$ cm, when the angle of the corresponding sector is $60^\circ$.

Solution. For a minor segment,

$$\text{segment} = \text{sector} – \text{triangle}$$

The sector, with $\theta = 60^\circ$ and $r = 14$:

$$\frac{60}{360} \times \frac{22}{7} \times 196 = \frac16 \times 616 = \frac{308}{3} \text{ cm}^2 \approx 102.67$$

The triangle. It is formed by the two radii and the chord, so two of its sides are $14$ cm with $60^\circ$ between them — which makes it equilateral, since the base angles must then be $60^\circ$ too. For an equilateral triangle of side $a$:

$$\text{area} = \frac{\sqrt3}{4}a^2 = \frac{\sqrt3}{4} \times 196 = 49\sqrt3 \text{ cm}^2 \approx 84.87$$

The segment:

$$\frac{308}{3} – 49\sqrt3 \text{ cm}^2 \approx 102.67 – 84.87 = 17.80 \text{ cm}^2$$

The $60^\circ$ is what makes this clean: any other angle would need $\frac12 r^2\sin\theta$ for the triangle, but at $60^\circ$ the equilateral formula applies and the answer stays exact.

Note how small the segment is — under $18$ cm² against a sector of nearly $103$ cm². A narrow sector is almost entirely filled by its triangle, which is a useful check: if your segment comes out close to the sector’s size, the triangle has gone missing.

Question 11

Find the area of the shaded region in Fig. 11.10, where arcs drawn with centres $A$, $B$, $C$ and $D$ intersect in pairs at mid-points $P$, $Q$, $R$ and $S$ of the sides $AB$, $BC$, $CD$ and $DA$, respectively of a square $ABCD$ (Use $\pi = 3.14$).

Solution. Each arc is centred at a corner and passes through the mid-points of the two sides meeting there, so its radius is half the side:

$$r = \frac{12}{2} = 6 \text{ cm}$$

and each arc turns through the square’s corner angle, $90^\circ$ — a quadrant.

The four quadrants together. Four quarter circles of the same radius make one whole circle:

$$4 \times \frac14 \pi r^2 = \pi r^2 = 3.14 \times 36 = 113.04 \text{ cm}^2$$

They do not overlap, because each pair meets exactly at a mid-point and goes no further.

The square:

$$12 \times 12 = 144 \text{ cm}^2$$

The shaded region is what is left of the square after the four quadrants are removed:

$$144 – 113.04 = 30.96 \text{ cm}^2$$

The four quadrants combining into a single circle is the shortcut worth taking. It works whenever equal-radius sectors are drawn at every vertex of a polygon and their angles sum to $360^\circ$ — here, four right angles.

Question 12

In Fig. 11.11, arcs are drawn by taking vertices $A$, $B$ and $C$ of an equilateral triangle of side $10$ cm to intersect the sides $BC$, $CA$ and $AB$ at their respective mid-points $D$, $E$ and $F$. Find the area of the shaded region (Use $\pi = 3.14$).

Solution. The shaded region is the three sectors themselves — the corner pieces — not the triangle with them removed. Read the figure before computing.

Each arc is centred at a vertex and reaches the mid-points of the two sides there, so

$$r = \frac{10}{2} = 5 \text{ cm}$$

and each sector’s angle is the triangle’s interior angle, $60^\circ$.

The three sectors together. Their angles sum to $60^\circ \times 3 = 180^\circ$, the angle sum of a triangle — so they amount to exactly half a circle of radius $5$:

$$\frac{180}{360} \times \pi r^2 = \frac12 \times 3.14 \times 25 = 39.25 \text{ cm}^2$$

No individual angle was needed, only their total.

A check against the triangle. The equilateral triangle of side $10$ has area

$$\frac{\sqrt3}{4} \times 100 \approx 43.30 \text{ cm}^2$$

so the three sectors fill all but about $4.05$ cm² of it — the small curved region in the middle, which is exactly what the figure shows. If your answer exceeds $43.30$, the sectors have been over-counted.

Question 13

In Fig. 11.12, arcs have been drawn with radii $14$ cm each and with centres $P$, $Q$ and $R$. Find the area of the shaded region.

Solution. The triangle’s shape and size are never given, and are never needed. That is the point of the question.

Three sectors, all of radius $14$ cm, are drawn at the three vertices of a triangle, each cut off by the two sides meeting there. So each sector’s angle is one of the triangle’s interior angles, and those sum to $180^\circ$ whatever the triangle looks like:

$$\theta_P + \theta_Q + \theta_R = 180^\circ$$

Since all three have the same radius, their areas add directly:

$$\text{total} = \frac{\theta_P}{360}\pi r^2 + \frac{\theta_Q}{360}\pi r^2 + \frac{\theta_R}{360}\pi r^2 = \frac{\theta_P + \theta_Q + \theta_R}{360}\pi r^2 = \frac{180}{360}\pi r^2$$

So the three sectors together make exactly half a circle of radius $14$:

$$\frac12 \times \frac{22}{7} \times 196 = \frac12 \times 616 = 308 \text{ cm}^2$$

This is the same argument as Question 12, freed of the assumption that the triangle is equilateral. There the angles were $60^\circ$ each; here they are unknown and it makes no difference, because only the total enters.

Question 14

A circular park is surrounded by a road $21$ m wide. If the radius of the park is $105$ m, find the area of the road.

Solution. The road is the region between two concentric circles — an annulus. Its area is the big circle minus the small one.

$$\text{inner radius} = 105 \text{ m}, \qquad \text{outer radius} = 105 + 21 = 126 \text{ m}$$

$$\text{area} = \pi R^2 – \pi r^2 = \pi\left(R^2 – r^2\right)$$

Factor the difference of squares rather than squaring both numbers — it is far less work:

$$R^2 – r^2 = (R + r)(R – r) = (126 + 105)(126 – 105) = 231 \times 21$$

Note that $R – r$ is just the width of the road, $21$ m. So

$$\text{area} = \frac{22}{7} \times 231 \times 21 = 22 \times 33 \times 21 = 15\,246 \text{ m}^2$$

using $231 \div 7 = 33$.

The factorised form $\pi(R+r)(R-r)$ is worth using for every annulus question. Here it turns a subtraction of five-digit squares into a product of two small numbers, and it is how Question 5 of Exercise 11.4 is best done too.

Question 15

In Fig. 11.13, arcs have been drawn of radius $21$ cm each with vertices $A$, $B$, $C$ and $D$ of quadrilateral $ABCD$ as centres. Find the area of the shaded region.

Solution. The same argument as Question 13, with one more corner.

Four sectors, all of radius $21$ cm, are drawn at the four vertices of a quadrilateral, each cut off by the sides meeting there. Their angles are the quadrilateral’s interior angles, and those sum to $360^\circ$ — for any quadrilateral whatsoever:

$$\theta_A + \theta_B + \theta_C + \theta_D = 360^\circ$$

Equal radii again let the areas add directly:

$$\text{total} = \frac{360}{360} \times \pi r^2 = \pi r^2$$

The four sectors make exactly one whole circle of radius $21$:

$$\frac{22}{7} \times 441 = 22 \times 63 = 1386 \text{ cm}^2$$

The progression across Questions 11, 12, 13 and 15 is the thing to take away. Equal-radius sectors at every vertex of a polygon always combine into a single sector whose angle is the polygon’s angle sum:

• triangle: $180^\circ$, so half a circle (Questions 12 and 13)
• quadrilateral: $360^\circ$, so a whole circle (Questions 11 and 15)

Nothing about the polygon’s shape or side lengths ever enters, provided the sectors do not overlap.

Question 16

A piece of wire $20$ cm long is bent into the form of an arc of a circle subtending an angle of $60^\circ$ at its centre. Find the radius of the circle.

Solution. The wire becomes the arc, so its length is the arc length:

$$\ell = \frac{\theta}{360^\circ} \times 2\pi r$$

$$20 = \frac{60}{360} \times 2\pi r = \frac16 \times 2\pi r = \frac{\pi r}{3}$$

Solving for $r$:

$$r = \frac{60}{\pi} \text{ cm} \approx 19.1 \text{ cm}$$

The answer is left in terms of $\pi$ deliberately, and the key prints it that way too. Substituting $\frac{22}{7}$ would give $\frac{60 \times 7}{22} = \frac{210}{11} \approx 19.09$ cm — correct, but less exact and less tidy.

A quick check on size: a $60^\circ$ arc is one-sixth of the circumference, so the full circumference is $6 \times 20 = 120$ cm, and $r = \frac{120}{2\pi} = \frac{60}{\pi}$ ✓. Going via the whole circumference is often the quickest route for these.

Common mistakes

• Questions 6 and 9 — assuming a labelled length means the same thing in both. In Question 6 the $38$ cm spans only the rectangle, so the bed is $38 + 10$ long overall; in Question 9 the margins locate the ends of the hole, so its rectangular part is $4$ m shorter than the $20$ m span. The figures settle it and the words do not.
• Question 12 — shading the wrong region. The shaded part is the three sectors ($39.25$ cm²), not the triangle minus them ($4.05$ cm²). Identify which side of the boundary is hatched before calculating.
• Questions 11, 12, 13 and 15 — computing individual sector angles. You never need them. Equal radii plus the polygon’s angle sum give the total in one step: half a circle for a triangle, a whole circle for a quadrilateral.
• Question 8 — treating the $6$ m as a dimension of a piece. It is the overall height; the semicircle’s radius is $6 – 4 = 2$ m, and its diameter is therefore $4$ m, not $8$ m.
• Question 5 — using a quarter circle without checking the rope. The rope must be shorter than both sides meeting at the corner, or part of the circle falls outside the field. Here $14 < 16 < 20$, so it is safe.
• Question 4 — mixing units. The radius is in centimetres and the speed in kilometres per hour. Convert to centimetres per minute first; an unconverted answer is wrong by a factor of $1000$ or $60$ and looks perfectly reasonable.
• Question 10 — forgetting the triangle. A segment is a sector minus the triangle on the two radii, and at $60^\circ$ that triangle is equilateral, hence $\frac{\sqrt3}{4}a^2$. A segment that comes out nearly as large as its sector has lost it.
• Question 14 — squaring $126$ and $105$ separately. Use $\pi(R+r)(R-r)$ instead, where $R – r$ is just the road’s width. It turns the arithmetic into $22 \times 33 \times 21$.

Practise next

• Exercise 11.4 — twenty long-answer questions, where these decompositions get one layer more involved and the sector-angle-sum trick reappears for triangles and hexagons.
• Exercise 11.2 — worth revisiting after Question 10, since its Questions 4, 9 and 10 are the segment and sector facts this exercise applies.
• Exercise 9.3 — the tangent and chord results that fix the lengths inside these composite figures, particularly for questions where circles touch one another.