NCERT Class 10 Mathematics — Areas Related to Circles, Exercise 11.1. All 14 questions solved.

This is the only exercise in the chapter, and it asks one question in fourteen disguises: what fraction of a circle is this? Once the angle at the centre is known, three formulae do everything. For a circle of radius $r$ and a sector of angle $\theta$ (in degrees):

$$\text{area of sector} = \frac{\theta}{360} \times \pi r^2 \qquad\qquad \text{length of arc} = \frac{\theta}{360} \times 2\pi r$$

$$\text{area of segment} = \text{area of sector} – \text{area of the triangle formed by the two radii and the chord}$$

The major sector and major segment are whatever is left of the circle once the minor one is taken away.

Key insight. Find the angle at the centre first. A minute hand, an umbrella rib, a brooch wire and a lighthouse beam are all just sectors, and the angle is always $360^\circ$ shared out: $30^\circ$ for five minutes of a clock, $45^\circ$ between eight ribs, $36^\circ$ for ten equal sectors. After that, every answer is $\frac{\theta}{360}$ of a whole circle, less a triangle if a chord is involved.

Unless stated otherwise, use $\pi = \dfrac{22}{7}$.

Question 1

Find the area of a sector of a circle with radius $6$ cm if angle of the sector is $60^\circ$.

Solution. A $60^\circ$ sector is $\frac{60}{360} = \frac{1}{6}$ of the circle:

$$\text{Area} = \frac{60}{360} \times \frac{22}{7} \times 6^2 = \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7}\ \text{cm}^2 \approx 18.86\ \text{cm}^2$$

Question 2

Find the area of a quadrant of a circle whose circumference is $22$ cm.

Solution. The radius is not given directly, so recover it from the circumference:

$$2\pi r = 22 \quad\Longrightarrow\quad 2 \times \frac{22}{7} \times r = 22 \quad\Longrightarrow\quad r = \frac{7}{2}\ \text{cm}$$

A quadrant is a $90^\circ$ sector, one quarter of the circle:

$$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{77}{8}\ \text{cm}^2 = 9.625\ \text{cm}^2$$

Question 3

The length of the minute hand of a clock is $14$ cm. Find the area swept by the minute hand in $5$ minutes.

Solution. The minute hand goes round once, $360^\circ$, in $60$ minutes, so it turns $6^\circ$ a minute and $30^\circ$ in $5$ minutes. It sweeps a sector of radius $14$ cm and angle $30^\circ$:

$$\text{Area} = \frac{30}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{154}{3}\ \text{cm}^2 \approx 51.33\ \text{cm}^2$$

Question 4

A chord of a circle of radius $10$ cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use $\pi = 3.14$)

Solution. Let the chord be $\mathrm{AB}$ with $\angle \mathrm{AOB} = 90^\circ$. The minor sector $\mathrm{OAB}$ is a quadrant:

$$\text{area of minor sector} = \frac{90}{360} \times 3.14 \times 10^2 = 78.5\ \text{cm}^2$$

(i) Triangle $\mathrm{OAB}$ is right-angled at $\mathrm{O}$, so its two radii serve as base and height:

$$\text{ar}(\triangle \mathrm{OAB}) = \tfrac{1}{2} \times 10 \times 10 = 50\ \text{cm}^2$$

The minor segment is the shaded part of the sector outside the triangle:

$$\text{minor segment} = 78.5 – 50 = 28.5\ \text{cm}^2$$

(ii) The major sector is the rest of the circle once the minor sector is removed — no triangle is involved:

$$\text{major sector} = 3.14 \times 10^2 – 78.5 = 314 – 78.5 = 235.5\ \text{cm}^2$$

Question 5

In a circle of radius $21$ cm, an arc subtends an angle of $60^\circ$ at the centre. Find:

(i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord

Solution. $\theta = 60^\circ$ is one sixth of the circle, and $\pi = \frac{22}{7}$ since the question does not say otherwise.

(i) $$\text{arc} = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22\ \text{cm}$$

(ii) $$\text{sector} = \frac{60}{360} \times \frac{22}{7} \times 21^2 = \frac{1}{6} \times 1386 = 231\ \text{cm}^2$$

(iii) Let the chord be $\mathrm{AB}$. In triangle $\mathrm{OAB}$, $\mathrm{OA} = \mathrm{OB}$, so the base angles are equal, and with $\angle \mathrm{AOB} = 60^\circ$ they are $60^\circ$ each. The triangle is equilateral with side $21$ cm:

$$\text{ar}(\triangle \mathrm{OAB}) = \frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4}\ \text{cm}^2$$

$$\text{segment} = \left(231 – \frac{441\sqrt{3}}{4}\right)\text{cm}^2 \approx 40.04\ \text{cm}^2$$

Question 6

A chord of a circle of radius $15$ cm subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Solution. As in question 5(iii), a $60^\circ$ angle at the centre makes triangle $\mathrm{OAB}$ equilateral, here with side $15$ cm.

$$\text{sector} = \frac{60}{360} \times 3.14 \times 15^2 = \frac{706.5}{6} = 117.75\ \text{cm}^2$$

$$\text{ar}(\triangle \mathrm{OAB}) = \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73 \times 225}{4} = 97.3125\ \text{cm}^2$$

$$\text{minor segment} = 117.75 – 97.3125 = 20.4375\ \text{cm}^2$$

The major segment is the whole circle minus the minor segment:

$$\text{major segment} = 3.14 \times 225 – 20.4375 = 706.5 – 20.4375 = 686.0625\ \text{cm}^2$$

Question 7

A chord of a circle of radius $12$ cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Solution. The segment asked for is the minor one, cut off by the chord $\mathrm{AB}$.

$$\text{sector} = \frac{120}{360} \times 3.14 \times 12^2 = \frac{1}{3} \times 452.16 = 150.72\ \text{cm}^2$$

Triangle $\mathrm{OAB}$ is isosceles but not right-angled, so we need its base and height. Draw $\mathrm{OM} \perp \mathrm{AB}$. In an isosceles triangle this perpendicular bisects both the base and the vertex angle, so $\angle \mathrm{AOM} = 60^\circ$ and, in right triangle $\mathrm{OMA}$,

$$\mathrm{OM} = 12\cos 60^\circ = 6\ \text{cm}, \qquad \mathrm{AM} = 12\sin 60^\circ = 6\sqrt{3}\ \text{cm}, \qquad \mathrm{AB} = 12\sqrt{3}\ \text{cm}$$

$$\text{ar}(\triangle \mathrm{OAB}) = \tfrac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3} = 36 \times 1.73 = 62.28\ \text{cm}^2$$

$$\text{segment} = 150.72 – 62.28 = 88.44\ \text{cm}^2$$

Question 8

A horse is tied to a peg at one corner of a square shaped grass field of side $15$ m by means of a $5$ m long rope (see Fig. 11.8). Find

(i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were $10$ m long instead of $5$ m. (Use $\pi = 3.14$)

Solution. Fig. 11.8 shows the horse tethered at a corner $\mathrm{P}$ of the field. The horse can reach every point within $5$ m of the peg, but the two sides of the field meet at the corner at $90^\circ$, so the part of that circle lying inside the field is a quadrant.

(i) $$\text{grazing area} = \frac{90}{360} \times 3.14 \times 5^2 = \frac{78.5}{4} = 19.625\ \text{m}^2$$

(ii) With a $10$ m rope the region is still a quadrant, because $10$ m is less than the $15$ m side, so the arc stays inside the field:

$$\text{new grazing area} = \frac{1}{4} \times 3.14 \times 10^2 = 78.5\ \text{m}^2$$

$$\text{increase} = 78.5 – 19.625 = 58.875\ \text{m}^2$$

A note on the answer key. The textbook prints the answer to part (ii) as $58.875\ \text{cm}^2$. The number is right, but the unit is a misprint: the field and the rope are measured in metres, and the key itself gives part (i) in $\text{m}^2$. The increase is $58.875\ \text{m}^2$.

Question 9

A brooch is made with silver wire in the form of a circle with diameter $35$ mm. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors as shown in Fig. 11.9. Find:

(i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

Solution. The wire has two parts: the circle itself, and the five diameters laid across it.

(i) $$\text{circumference} = \pi d = \frac{22}{7} \times 35 = 110\ \text{mm}$$

$$\text{five diameters} = 5 \times 35 = 175\ \text{mm}$$

$$\text{total wire} = 110 + 175 = 285\ \text{mm}$$

(ii) Five diameters give ten equal sectors of $36^\circ$ each, so each is $\frac{1}{10}$ of the circle, with radius $\frac{35}{2}$ mm:

$$\text{each sector} = \frac{1}{10} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^2 = \frac{1}{10} \times \frac{22}{7} \times \frac{1225}{4} = \frac{385}{4}\ \text{mm}^2 = 96.25\ \text{mm}^2$$

Question 10

An umbrella has $8$ ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius $45$ cm, find the area between the two consecutive ribs of the umbrella.

Solution. Eight equally spaced ribs divide the circle into eight equal sectors of $\frac{360^\circ}{8} = 45^\circ$ each. The area between two consecutive ribs is one of them:

$$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 45^2 = \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22275}{28}\ \text{cm}^2 \approx 795.54\ \text{cm}^2$$

Question 11

A car has two wipers which do not overlap. Each wiper has a blade of length $25$ cm sweeping through an angle of $115^\circ$. Find the total area cleaned at each sweep of the blades.

Solution. Each blade sweeps a sector of radius $25$ cm and angle $115^\circ$. Because the wipers do not overlap, no part of the glass is counted twice, and the total is simply two such sectors:

$$\text{total} = 2 \times \frac{115}{360} \times \frac{22}{7} \times 25^2 = 2 \times \frac{23}{72} \times \frac{22}{7} \times 625 = \frac{158125}{126}\ \text{cm}^2 \approx 1254.96\ \text{cm}^2$$

Question 12

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of $16.5$ km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)

Solution. The lit region is a sector of radius $16.5$ km and angle $80^\circ$:

$$\text{Area} = \frac{80}{360} \times 3.14 \times 16.5^2 = \frac{2}{9} \times 3.14 \times 272.25 = 189.97\ \text{km}^2$$

The division comes out exactly, because $272.25 = \frac{9}{2} \times 60.5$.

Question 13

A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is $28$ cm, find the cost of making the designs at the rate of ₹ $0.35$ per $\text{cm}^2$. (Use $\sqrt{3} = 1.7$)

Solution. Fig. 11.11 shows a regular hexagon drawn inside the circular cover, with its six corners on the circle. The six designs fill the six segments between the sides of the hexagon and the circle (shaded above).

Each side of the hexagon subtends $\frac{360^\circ}{6} = 60^\circ$ at the centre, so, as in question 5, the hexagon is made of six equilateral triangles of side $28$ cm. The area of the designs is the circle less the hexagon:

$$\text{circle} = \frac{22}{7} \times 28^2 = 2464\ \text{cm}^2$$

$$\text{hexagon} = 6 \times \frac{\sqrt{3}}{4} \times 28^2 = 1176\sqrt{3} = 1176 \times 1.7 = 1999.2\ \text{cm}^2$$

$$\text{designs} = 2464 – 1999.2 = 464.8\ \text{cm}^2$$

$$\text{cost} = 464.8 \times 0.35 = ₹\,162.68$$

Question 14

Tick the correct answer in the following:

Area of a sector of angle $p$ (in degrees) of a circle with radius $\mathrm{R}$ is

Solution. The area of a sector of angle $p$ is $\dfrac{p}{360} \times \pi \mathrm{R}^2$. None of the options is written that way, so rewrite it. Multiplying top and bottom by $2$:

$$\frac{p}{360} \times \pi \mathrm{R}^2 = \frac{p}{720} \times 2\pi \mathrm{R}^2$$

which is (D). The others fail for clear reasons: (C) is the length of the arc; (A) is twice the arc length; and (B) is twice the area. (A) and (C) are not even areas, since they contain $\mathrm{R}$ rather than $\mathrm{R}^2$.

Common mistakes

• Question 3, taking the angle as $5^\circ$. Five minutes is $\frac{5}{60}$ of a revolution, which is $30^\circ$, not $5^\circ$. The minute hand turns $6^\circ$ every minute.
• Question 4(ii), subtracting the triangle. A major sector is bounded by two radii and the major arc, so it is the circle minus the minor sector. The triangle only comes in when the boundary is a chord, as in a segment.
• Question 7, using $\tfrac{1}{2}r^2$ for the triangle. That works only when the angle at the centre is $90^\circ$, as in question 4. For $120^\circ$, drop the perpendicular $\mathrm{OM}$ and find the base and height, or the area comes out as $72$ instead of $62.28$.
• Question 8, using a semicircle or the full circle. The peg is at a corner, where the two sides of the field meet at a right angle, so only a quarter of the circle lies in the field. And part (ii) asks for the increase, not the new area.
• Question 9, counting ten diameters. Five diameters cut the circle into ten sectors, but the wire used is still five diameters, $5 \times 35$ mm. Counting ten radii as ten diameters doubles that part of the wire.
• Question 11, answering for one wiper. The car has two blades, and since they do not overlap, the area cleaned is exactly double one sector.
• Using the wrong value of $\pi$. The default here is $\frac{22}{7}$, but questions 4, 6, 7, 8 and 12 say $3.14$. Mixing them changes the last digits, and the key’s exact decimals such as $20.4375$ come out only with the value the question specifies.

Practise next

• Chapter 12, Exercise 12.1 — surface areas of combined solids, where circle areas and circumferences return as the faces of cylinders, cones and hemispheres.
• Chapter 8, Exercise 8.2 — the trigonometric ratios of $30^\circ$, $45^\circ$ and $60^\circ$ that give the triangle’s height and base in question 7.