NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles — Exercise 11.4

NCERT Exemplar Problems, Class 10 Mathematics — Areas Related to Circles, Exercise 11.4. All 20 questions solved.

Twenty questions, and the hardest in the unit — though not because any single step is difficult. What makes them long is that most need two or three areas found separately and then combined, with a unit conversion or a cost calculation on the end.

Three patterns account for almost all of them.

The annulus. A path round a pond, a road round a park, a ring on a target. Always use the factorised form:

$$\pi R^2 – \pi r^2 = \pi (R + r)(R – r)$$

where $R – r$ is simply the width of the ring. Questions 5 and 13 both fall to this.

Sectors at the corners of a polygon. Equal radii at every vertex means the angles add to the polygon’s angle sum, so the sectors combine into one: half a circle for a triangle, a whole circle for a quadrilateral or a hexagon’s worth of corners. Questions 3, 6, 9 and 10 are this.

Circles packed inside a rectangle. Count them, find one, multiply. Question 11 is the archetype and the counting is the only place it goes wrong.

A note on $\pi$. The chapter states that “unless stated otherwise, the value of $\pi$ is to be taken as $\frac{22}{7}$”. Questions 4, 11 and 12 say to use $3.14$; everything else uses $\frac{22}{7}$. This matters in Question 5, where the printed answer turns out to have used $3.14$ without saying so — the page gives both figures and says which is which.

A note on the current syllabus. “Areas of Combinations of Plane Figures” has been cut from the Class 10 textbook, which now runs to one section on sectors and segments. Eight of these twenty questions are combination problems, so they are exemplar-only material.

Key insight. Before computing anything, ask whether the pieces you are about to add have the same radius. If they do, add their angles first and evaluate one sector at the end. Questions 3, 6, 9, 10 and 17 each collapse from four or six calculations to one that way.

Question 1

The area of a circular playground is $22176$ m². Find the cost of fencing this ground at the rate of ₹$50$ per metre.

Solution. Fencing runs round the boundary, so the quantity needed is the circumference — but the area is what is given, so the radius has to be recovered first.

$$\pi r^2 = 22176 \implies \frac{22}{7}r^2 = 22176 \implies r^2 = \frac{22176 \times 7}{22} = 1008 \times 7 = 7056$$

$$r = \sqrt{7056} = 84 \text{ m}$$

Now the circumference:

$$2\pi r = 2 \times \frac{22}{7} \times 84 = 2 \times 22 \times 12 = 528 \text{ m}$$

And the cost:

$$528 \times 50 = ₹26\,400$$

Three steps, and the middle one is where the question can be lost: it is easy to compute the area-based quantity and multiply the rate by that. Fencing is priced per metre, so it is a length that gets multiplied — a cost in ₹ per m² would have meant the area instead.

Question 2

The diameters of front and rear wheels of a tractor are $80$ cm and $2$ m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes $1400$ revolutions.

Solution. The two wheels travel the same distance — they are on the same tractor. So compute that distance from the front wheel and then divide by the rear wheel’s circumference.

Convert first: the rear diameter is $2$ m $= 200$ cm.

Distance covered, from the front wheel:

$$1400 \times \pi \times 80 = 112\,000\pi \text{ cm}$$

Revolutions of the rear wheel:

$$\frac{112\,000\pi}{\pi \times 200} = \frac{112\,000}{200} = 560$$

The $\pi$ cancels, so no value for it is needed — which is the sign that this is really a ratio question. In fact the whole thing is one line:

$$\text{rear revolutions} = 1400 \times \frac{80}{200} = 1400 \times \frac{2}{5} = 560$$

because the number of revolutions is inversely proportional to the diameter. The bigger wheel turns fewer times, which is the check: $560 < 1400$, as it must be.

Question 3

Sides of a triangular field are $15$ m, $16$ m and $17$ m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length $7$ m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Solution. The ungrazed area is the field minus the three grazed sectors.

The field. Three sides and no height, so use Heron’s formula. The semi-perimeter is

$$s = \frac{15 + 16 + 17}{2} = 24$$

$$\text{area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{24 \times 9 \times 8 \times 7}$$

$$= \sqrt{12096} = \sqrt{576 \times 21} = 24\sqrt{21} \text{ m}^2 \approx 109.98$$

The grazed region. Each animal sweeps a sector of radius $7$ m centred at a corner, with the corner’s own angle. All three radii are equal and the angles sum to $180^\circ$, so the three sectors make exactly half a circle of radius $7$:

$$\frac12 \times \frac{22}{7} \times 49 = \frac{154}{2}\cdot\frac{7}{7} = 77 \text{ m}^2$$

$$\left(\text{that is } \tfrac12 \times \tfrac{22}{7} \times 49 = \tfrac{1078}{14} = 77\right)$$

No individual angle is ever needed — the same argument as Questions 12, 13 and 15 of Exercise 11.3.

The ungrazed area:

$$24\sqrt{21} – 77 \text{ m}^2 \approx 109.98 – 77 = 32.98 \text{ m}^2$$

A check that the sectors fit: the shortest side is $15$ m and the two ropes at its ends are $7$ m each, totalling $14 < 15$, so no two grazed regions overlap.

Question 4

Find the area of the segment of a circle of radius $12$ cm whose corresponding sector has a central angle of $60^\circ$ (Use $\pi = 3.14$).

Solution. For a minor segment,

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

The sector, with $\theta = 60^\circ$ and $r = 12$:

$$\frac{60}{360} \times 3.14 \times 144 = \frac16 \times 452.16 = 75.36 \text{ cm}^2$$

The triangle. Two radii of $12$ cm with $60^\circ$ between them force the third angle to $60^\circ$ as well, so the triangle is equilateral of side $12$:

$$\frac{\sqrt3}{4} \times 144 = 36\sqrt3 \text{ cm}^2 \approx 62.35$$

The segment:

$$75.36 – 36\sqrt3 \text{ cm}^2 \approx 75.36 – 62.35 = 13.01 \text{ cm}^2$$

The answer is left with $\sqrt3$ unevaluated because the question gives a value for $\pi$ but not for $\sqrt3$ — a signal that the surd is meant to stay. Had it wanted a decimal it would have supplied $\sqrt3 = 1.73$, as the textbook’s own exercises do.

This is Question 10 of Exercise 11.3 with a different radius and $\pi = 3.14$ instead of $\frac{22}{7}$; the structure is identical.

Question 5

A circular pond is $17.5$ m is of diameter. It is surrounded by a $2$ m wide path. Find the cost of constructing the path at the rate of ₹$25$ per m².

Solution. The path is an annulus — the ring between the pond’s edge and the outer edge of the path.

$$\text{inner radius} = \frac{17.5}{2} = 8.75 \text{ m}, \qquad \text{outer radius} = 8.75 + 2 = 10.75 \text{ m}$$

Use the factorised form, since the width of the ring is exactly $R – r$:

$$\text{area} = \pi\left(R^2 – r^2\right) = \pi(R+r)(R-r) = \pi \times 19.5 \times 2 = 39\pi \text{ m}^2$$

Now the cost, at ₹$25$ per m²:

$$\text{cost} = 25 \times 39\pi = 975\pi$$

With the chapter’s $\pi = \frac{22}{7}$, which is what this question’s silence calls for:

$$975 \times \frac{22}{7} = \frac{21450}{7} = ₹3064.29 \text{ (to the nearest paisa)}$$

A note on the printed answer. The key gives ₹$3061.50$, which is exactly $975 \times 3.14$. So the key has used $\pi = 3.14$ even though this question — unlike Questions 4, 11 and 12 — does not say to, and the chapter’s stated convention is $\frac{22}{7}$.

Neither figure is a mistake in the arithmetic; they differ only in the value of $\pi$, and the gap is about ₹$2.79$ on ₹$3000$, under a tenth of a percent. If you are working to the question as printed, ₹$3064.29$ is the answer; if you are matching the book, ₹$3061.50$ is. Say which convention you used and either will earn the marks.

Question 6

In Fig. 11.17, $ABCD$ is a trapezium with $AB \parallel DC$, $AB = 18$ cm, $DC = 32$ cm and distance between $AB$ and $DC$ $= 14$ cm. If arcs of equal radii $7$ cm with centres $A$, $B$, $C$ and $D$ have been drawn, then find the area of the shaded region of the figure.

Solution. The shaded region is the trapezium minus the four sectors.

The trapezium. With parallel sides $18$ and $32$ and distance $14$ between them:

$$\text{area} = \frac12 \times (\text{sum of parallel sides}) \times \text{height} = \frac12 \times (18 + 32) \times 14 = \frac12 \times 50 \times 14 = 350 \text{ cm}^2$$

The four sectors. All have radius $7$ cm, and their angles are the four interior angles of the trapezium — which sum to $360^\circ$ for any quadrilateral. So they combine into exactly one whole circle of radius $7$:

$$\pi r^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2$$

The trapezium’s shape never enters. This is Question 15 of Exercise 11.3 with a specific quadrilateral and its area supplied.

The shaded region:

$$350 – 154 = 196 \text{ cm}^2$$

A check: the sectors take up $44\%$ of the trapezium, leaving $56\%$ shaded — which matches the figure.

Question 7

Three circles each of radius $3.5$ cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Solution. “Enclosed between” means the small curved triangle in the middle, bounded by three arcs.

The triangle of centres. Two circles touching externally have their centres exactly $r + r = 7$ cm apart, so joining the three centres gives an equilateral triangle of side $7$ cm:

$$\text{area} = \frac{\sqrt3}{4} \times 49 \approx 21.22 \text{ cm}^2$$

The three sectors inside it. Each corner of that triangle is the centre of a circle of radius $3.5$ cm, and the triangle’s angle there is $60^\circ$. So the three sectors together make

$$\frac{3 \times 60}{360} \times \pi r^2 = \frac12 \times \frac{22}{7} \times (3.5)^2 = \frac12 \times \frac{22}{7} \times 12.25 = 19.25 \text{ cm}^2$$

— half a circle of radius $3.5$, by the usual angle-sum argument.

The enclosed region:

$$21.22 – 19.25 = 1.97 \text{ cm}^2 \text{ (approximately)}$$

The answer is small, and it should be: three touching circles leave only a sliver between them. If your figure exceeds a few square centimetres, the sectors have been under-counted.

The key prints $1.967$ cm² (approx), which is $\frac{49\sqrt3}{4} – \frac{539}{28}$ evaluated.

Question 8

Find the area of the sector of a circle of radius $5$ cm, if the corresponding arc length is $3.5$ cm.

Solution. The angle is not given, so use the form of the sector area that does not need it:

$$\text{sector area} = \frac{\ell r}{2}$$

where $\ell$ is the arc length. Then

$$\text{area} = \frac{3.5 \times 5}{2} = \frac{17.5}{2} = 8.75 \text{ cm}^2$$

Where that formula comes from, since it is worth being able to rebuild:

$$\text{area} = \frac{\theta}{360}\pi r^2 = \frac{r}{2} \times \underbrace{\frac{\theta}{360} \times 2\pi r}_{= \ \ell} = \frac{\ell r}{2}$$

The longer route works too — find $\theta$ from $3.5 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 5$, giving $\theta \approx 40.09^\circ$, then substitute back — but it takes three times as long and introduces a rounding error that the direct formula avoids entirely.

This is the same formula that settled Questions 9 and 10 of Exercise 11.2, and it is the single most useful thing in the chapter.

Question 9

Four circular cardboard pieces of radii $7$ cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

Solution. Exactly Question 7 with four circles instead of three.

The quadrilateral of centres. Each pair of touching circles has centres $7 + 7 = 14$ cm apart, and with four circles each touching two others the centres form a square of side $14$ cm:

$$\text{area} = 14^2 = 196 \text{ cm}^2$$

The four sectors inside it. Each corner of the square is a circle’s centre, the radius is $7$ cm, and the corner angle is $90^\circ$. Four quadrants make one whole circle:

$$\pi r^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2$$

The enclosed region:

$$196 – 154 = 42 \text{ cm}^2$$

Compare with Question 7. Three circles gave a triangle of centres and half a circle of sectors; four give a square and a whole circle. The gap left in the middle is proportionally much bigger here — $21\%$ of the square against $9\%$ of the triangle — because four circles round a square leave a wider hole than three round a triangle.

Question 10

On a square cardboard sheet of area $784$ cm², four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.

Solution. Find the side of the square first, since the area is what is given:

$$\text{side} = \sqrt{784} = 28 \text{ cm}$$

The radius of each plate. The four plates sit in a $2 \times 2$ arrangement, each touching its two neighbours and the two sides of the square nearest it. So along any side of the square, two diameters fit exactly:

$$4r = 28 \implies r = 7 \text{ cm}$$

The four plates together. Four full circles of radius $7$:

$$4 \times \frac{22}{7} \times 49 = 4 \times 154 = 616 \text{ cm}^2$$

Uncovered:

$$784 – 616 = 168 \text{ cm}^2$$

A useful invariant. Four circles packed into a square cover $4 \times \pi r^2$ out of $(4r)^2 = 16r^2$, a fraction of $\frac{4\pi r^2}{16r^2} = \frac{\pi}{4} \approx 78.5\%$ — the same fraction as a single inscribed circle in a square. Packing more circles never improves the coverage in this arrangement, which is why the uncovered $168$ cm² is $21.4\%$ of $784$ whatever the number of plates.

Note the contrast with Question 9, where the circles were the same size but the region asked for was the gap between them, not the whole uncovered part.

Question 11

Floor of a room is of dimensions $5$ m $\times$ $4$ m and it is covered with circular tiles of diameters $50$ cm each as shown in Fig. 11.18. Find the area of floor that remains uncovered with tiles. (Use $\pi = 3.14$)

Solution. Convert to a single unit first. The floor is in metres and the tiles in centimetres, so work in metres: each tile has diameter $50$ cm $= 0.5$ m, so $r = 0.25$ m.

Count the tiles. The tiles are laid in rows and columns, touching:

$$\text{along the 5 m side: } \frac{5}{0.5} = 10, \qquad \text{along the 4 m side: } \frac{4}{0.5} = 8$$

$$\text{total} = 10 \times 8 = 80 \text{ tiles}$$

Area covered. Each tile covers $\pi r^2 = 3.14 \times 0.0625 = 0.19625$ m², so

$$80 \times 0.19625 = 15.7 \text{ m}^2$$

Floor area:

$$5 \times 4 = 20 \text{ m}^2$$

Uncovered:

$$20 – 15.7 = 4.3 \text{ m}^2$$

A cross-check without counting: each tile sits in its own $0.5 \times 0.5$ m square, and a circle inscribed in a square covers $\frac{\pi}{4} = 78.5\%$ of it. So the uncovered fraction is $21.5\%$ of $20$ m², which is $4.3$ m² ✓. That check works however many tiles there are, and catches a miscount instantly.

Question 12

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is $1256$ cm². (Use $\pi = 3.14$)

Solution. A rhombus inscribed in a circle is a square. That is the fact the question turns on, and it is worth seeing why: a rhombus has all four sides equal, and any quadrilateral inscribed in a circle has opposite angles summing to $180^\circ$; a parallelogram has opposite angles equal, so each must be $90^\circ$. Equal sides and right angles make a square.

The radius.

$$\pi r^2 = 1256 \implies 3.14 r^2 = 1256 \implies r^2 = 400 \implies r = 20 \text{ cm}$$

The diagonals. The square’s corners lie on the circle, so both its diagonals are diameters:

$$d_1 = d_2 = 2r = 40 \text{ cm}$$

The area. For a rhombus (and so for a square), the area is half the product of the diagonals:

$$\frac12 d_1 d_2 = \frac12 \times 40 \times 40 = 800 \text{ cm}^2$$

Answering without spotting that the rhombus must be a square leaves the problem underdetermined — a general rhombus has two independent diagonals, and nothing here would fix their ratio.

Question 13

An archery target has three regions formed by three concentric circles as shown in Fig. 11.19. If the diameters of the concentric circles are in the ratio $1 : 2 : 3$, then find the ratio of the areas of three regions.

Solution. Let the radii be $r$, $2r$ and $3r$ — the same ratio as the diameters, since halving all three preserves it.

The three circles’ areas:

$$\pi r^2, \qquad \pi(2r)^2 = 4\pi r^2, \qquad \pi(3r)^2 = 9\pi r^2$$

The three regions are not the circles. Only the innermost region is a full circle; the other two are rings, each found by subtracting the circle inside it:

$$\text{inner region} = \pi r^2$$

$$\text{middle ring} = 4\pi r^2 – \pi r^2 = 3\pi r^2$$

$$\text{outer ring} = 9\pi r^2 – 4\pi r^2 = 5\pi r^2$$

So the ratio is

$$\pi r^2 : 3\pi r^2 : 5\pi r^2 = 1 : 3 : 5$$

The pattern is worth remembering. Concentric circles with radii in the ratio $1 : 2 : 3 : \ldots$ divide the disc into rings whose areas are in the ratio $1 : 3 : 5 : \ldots$ — the odd numbers, because $n^2 – (n-1)^2 = 2n – 1$. It is why archery and darts targets have their rings drawn at equally spaced radii and yet the outer rings are much larger in area.

Answering $1 : 4 : 9$ is the error the question is set to catch: those are the areas of the three circles, not of the three regions.

Question 14

The length of the minute hand of a clock is $5$ cm. Find the area swept by the minute hand during the time period $6:05$ a.m. and $6:40$ a.m.

Solution. Find the angle first. The minute hand sweeps a full $360^\circ$ in $60$ minutes, so

$$\text{per minute} = \frac{360^\circ}{60} = 6^\circ$$

From $6{:}05$ to $6{:}40$ is $35$ minutes, so

$$\theta = 35 \times 6^\circ = 210^\circ$$

The area swept is a sector of radius $5$ cm and angle $210^\circ$:

$$\frac{210}{360} \times \frac{22}{7} \times 25 = \frac{7}{12} \times \frac{550}{7} = \frac{550}{12} = \frac{275}{6} = 45\frac56 \text{ cm}^2$$

using $\frac{210}{360} = \frac{7}{12}$ and $\frac{22}{7} \times 25 = \frac{550}{7}$, after which the sevens cancel.

$45\frac56 \approx 45.83$ cm².

A check on size: the whole circle has area $\frac{22}{7} \times 25 \approx 78.6$ cm², and $210^\circ$ is a bit over half of $360^\circ$. Indeed $45.83$ is about $58\%$ of $78.6$ ✓.

The common slip is to use the clock times as though the hour mattered — it does not. Only the elapsed minutes matter, because the minute hand’s position depends on nothing else.

Question 15

Area of a sector of central angle $200^\circ$ of a circle is $770$ cm². Find the length of the corresponding arc of this sector.

Solution. Two routes, and the second is much shorter.

The direct route. Recover the radius from the area, then find the arc.

$$\frac{200}{360} \times \frac{22}{7} \times r^2 = 770 \implies \frac59 \times \frac{22}{7} \times r^2 = 770$$

$$r^2 = 770 \times \frac{9}{5} \times \frac{7}{22} = 770 \times \frac{63}{110} = 441 \implies r = 21 \text{ cm}$$

Then

$$\ell = \frac{200}{360} \times 2 \times \frac{22}{7} \times 21 = \frac59 \times 132 = \frac{660}{9} = \frac{220}{3} = 73\frac13 \text{ cm}$$

The short route. Use $\text{area} = \frac{\ell r}{2}$, which needs only the radius:

$$770 = \frac{\ell \times 21}{2} \implies \ell = \frac{1540}{21} = \frac{220}{3} = 73\frac13 \text{ cm}$$

Either way the radius has to be found, but the second avoids a second pass through the angle.

$73\frac13 \approx 73.33$ cm. A check: the full circumference is $2 \times \frac{22}{7} \times 21 = 132$ cm, and $200^\circ$ is $\frac59$ of the circle — and $\frac59 \times 132 = 73.33$ ✓.

Question 16

The central angles of two sectors of circles of radii $7$ cm and $21$ cm are respectively $120^\circ$ and $40^\circ$. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

Solution. Four quantities, then the observation — which is the real point of the question.

First sector, $r = 7$ cm, $\theta = 120^\circ$:

$$\text{area} = \frac{120}{360} \times \frac{22}{7} \times 49 = \frac13 \times 154 = \frac{154}{3} \text{ cm}^2 \approx 51.33$$

$$\text{arc} = \frac{120}{360} \times 2 \times \frac{22}{7} \times 7 = \frac13 \times 44 = \frac{44}{3} \text{ cm} \approx 14.67$$

Second sector, $r = 21$ cm, $\theta = 40^\circ$:

$$\text{area} = \frac{40}{360} \times \frac{22}{7} \times 441 = \frac19 \times 1386 = 154 \text{ cm}^2$$

$$\text{arc} = \frac{40}{360} \times 2 \times \frac{22}{7} \times 21 = \frac19 \times 132 = \frac{44}{3} \text{ cm} \approx 14.67$$

The observation. The two arc lengths are equal — both $\frac{44}{3}$ cm — while the areas are not: $\frac{154}{3}$ cm² against $154$ cm², a ratio of $1 : 3$.

So equal arcs on different circles do not give equal sector areas. The reason is visible in $\text{area} = \frac{\ell r}{2}$: with $\ell$ the same, the areas are in the ratio of the radii, and $7 : 21 = 1 : 3$ ✓.

This is Question 9 of Exercise 11.2 as a calculation. There the claim “the areas of two sectors of two different circles with equal corresponding arc lengths are equal” had to be judged false; here the counterexample is constructed explicitly, with these very numbers.

Question 17

Find the area of the shaded region given in Fig. 11.20.

Solution. A square with a four-petalled flower removed. The flower’s size has to be recovered from the $3$ cm margins.

The flower’s span. It stands $3$ cm clear of each side, so across the square:

$$14 – 3 – 3 = 8 \text{ cm}$$

in both directions.

Decompose the flower. It is a central square with a semicircle erected on each of its four sides. If the central square has side $a$ and each semicircle has radius $\frac a2$, then the full span is

$$\frac a2 + a + \frac a2 = 2a = 8 \implies a = 4 \text{ cm}$$

So the central square is $4 \times 4$ cm and each semicircle has radius $2$ cm.

The flower’s area. The four semicircles have equal radius, so together they make two full circles of radius $2$:

$$\text{flower} = \underbrace{4 \times 4}_{\text{central square}} + \underbrace{2 \times \pi(2)^2}_{\text{four semicircles}} = 16 + 8\pi \text{ cm}^2$$

The square:

$$14 \times 14 = 196 \text{ cm}^2$$

The shaded region:

$$196 – \left(16 + 8\pi\right) = 180 – 8\pi \text{ cm}^2 \approx 180 – 25.13 = 154.87 \text{ cm}^2$$

The check that the decomposition is right: the flower spans $8$ cm, so it fits inside an $8 \times 8$ square of area $64$, and $16 + 8\pi \approx 41.1$ is comfortably less — about $64\%$ of it, which is what a four-petal shape should occupy.

Question 18

Find the number of revolutions made by a circular wheel of area $1.54$ m² in rolling a distance of $176$ m.

Solution. The area is given but the circumference is what a rolling wheel lays down, so recover the radius in between.

$$\pi r^2 = 1.54 \implies \frac{22}{7}r^2 = 1.54 \implies r^2 = 1.54 \times \frac{7}{22} = 0.49 \implies r = 0.7 \text{ m}$$

The circumference:

$$2\pi r = 2 \times \frac{22}{7} \times 0.7 = 4.4 \text{ m}$$

The revolutions:

$$\frac{176}{4.4} = 40$$

The middle step is the whole question, and it is the same shape as Question 1: an area is given, a length is wanted, and the radius is the bridge between them. Dividing the distance by the area would give $\frac{176}{1.54} \approx 114$, a number with the wrong units and no meaning.

Question 19

Find the difference of the areas of two segments of a circle formed by a chord of length $5$ cm subtending an angle of $90^\circ$ at the centre.

Solution. A chord splits the circle into a minor segment and a major segment. The difference between them is asked for, and there is a neat way to get it without finding either.

The radius. The chord subtends $90^\circ$, so the triangle formed by the two radii and the chord is right angled at the centre with the chord as hypotenuse:

$$r^2 + r^2 = 5^2 \implies 2r^2 = 25 \implies r^2 = \frac{25}{2}$$

The two segments. Writing $C$ for the circle’s area and $T$ for that triangle’s:

$$\text{minor segment} = \underbrace{\frac{C}{4}}_{\text{quadrant}} – T, \qquad \text{major segment} = C – \text{minor segment} = \frac{3C}{4} + T$$

The difference:

$$\left(\frac{3C}{4} + T\right) – \left(\frac{C}{4} – T\right) = \frac{C}{2} + 2T$$

Now substitute. The circle:

$$C = \pi r^2 = \frac{25\pi}{2}$$

and the triangle, with the two radii as perpendicular legs:

$$T = \frac12 r^2 = \frac12 \times \frac{25}{2} = \frac{25}{4}$$

So

$$\text{difference} = \frac12 \times \frac{25\pi}{2} + 2 \times \frac{25}{4} = \frac{25\pi}{4} + \frac{25}{2} \text{ cm}^2$$

$\approx 19.63 + 12.5 = 32.13$ cm².

Note that the triangle is added, not cancelled. It is subtracted from the minor segment and added to the major one, so it appears twice in the difference. Expecting it to cancel — and answering $\frac{C}{2} = \frac{25\pi}{4}$ — is the error this question is built around.

Question 20

Find the difference of the areas of a sector of angle $120^\circ$ and its corresponding major sector of a circle of radius $21$ cm.

Solution. The two sectors together make the whole circle, so their angles are $120^\circ$ and

$$360^\circ – 120^\circ = 240^\circ$$

The difference in angle is $240^\circ – 120^\circ = 120^\circ$, and since both sectors have the same radius their areas are proportional to their angles. So the difference in area is the area of a $120^\circ$ sector:

$$\frac{120}{360} \times \pi r^2 = \frac13 \times \frac{22}{7} \times 441 = \frac13 \times 1386 = 462 \text{ cm}^2$$

The long way, as a check. The major sector is $\frac{240}{360} \times 1386 = 924$ cm² and the minor is $\frac{120}{360} \times 1386 = 462$ cm², and $924 – 462 = 462$ cm² ✓.

Compare with Question 19. There the two regions were segments, and the triangle between them meant the difference was not simply proportional to the angles — it picked up an extra $2T$. Here the two regions are sectors, both bounded by the same two radii, so nothing extra appears and the subtraction is clean.

That contrast is why the two questions are placed together at the end of the exercise.

Common mistakes

• Questions 1 and 18 — multiplying a rate by the wrong quantity. Fencing is priced per metre, so the circumference is needed; a rolling wheel covers its circumference, not its area. In both the given quantity is an area, and the radius is the bridge.
• Question 5 — the value of $\pi$. The chapter’s convention is $\frac{22}{7}$ unless a question says otherwise, and this one does not. The printed answer uses $3.14$ regardless, so the two differ by about ₹$2.79$. State which you used.
• Question 13 — giving the ratio of the circles rather than the regions. The circles are in the ratio $1 : 4 : 9$; the three regions are $1 : 3 : 5$, because each ring is a difference of consecutive squares.
• Question 19 — expecting the triangle to cancel. It is subtracted from one segment and added to the other, so the difference contains $2T$, not $0$. Write both segments out in full before subtracting.
• Questions 3, 6, 7, 9 and 17 — evaluating sectors one at a time. Whenever the radii are equal, add the angles first. Three corners of a triangle give half a circle; four corners of a quadrilateral give a whole one; four semicircles give two circles.
• Question 11 — counting tiles wrongly, or mixing units. Convert the $50$ cm diameter to $0.5$ m before dividing into the $5$ m and $4$ m sides. The $\frac{\pi}{4} = 78.5\%$ coverage fraction is an independent check that needs no count at all.
• Question 12 — not noticing that the rhombus must be a square. A rhombus inscribed in a circle is cyclic, and a cyclic parallelogram has right angles. Without that the diagonals are not determined and the question cannot be finished.
• Question 10 — taking the radius as a quarter of the area rather than of the side. Four circles in a $2 \times 2$ arrangement fit four radii across the square, so $4r = 28$ gives $r = 7$. The square’s area plays no part beyond supplying the side.

Practise next

• Exercise 12.1 — Surface Areas and Volumes, the next unit, where these same circles become the cross-sections of cylinders, cones and spheres.
• Exercise 11.2 — worth revisiting after Question 16, which constructs the exact counterexample its Question 9 asks you to recognise.
• Exercise 11.3 — its Questions 11 to 15 are the angle-sum shortcut in its simplest form, and are the fastest way to make that habit automatic.