CBSE · NCERT · Class 10 Maths · Chapter 11

NCERT Solutions: Class 10 Maths Chapter 11 - Areas Related to Circles

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Chapter-wise NCERT intext questions and exercise answers for Areas Related to Circles, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercise 11.1 14
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1Exercise 11.114 questions
Q.1Find the area of a sector of a circle with radius 6 cm if angle of the sector is $60^\circ$.v
Solution

Area of a sector $=\dfrac{\theta}{360^\circ}\pi r^2$. With $r=6$ cm, $\theta=60^\circ$ and $\pi=\dfrac{22}{7}$, area $=\dfrac{60}{360}\times\dfrac{22}{7}\times 6^2=\dfrac{132}{7}$ cm$^2$.

Answer:

The area of the sector is $\dfrac{132}{7}$ cm$^2$ (about $18.86$ cm$^2$).

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Q.2Find the area of a quadrant of a circle whose circumference is 22 cm.v
Solution

The circumference is $2\pi r=22$, so $r=\dfrac{22}{2\times 22/7}=3.5$ cm. Area of a quadrant $=\dfrac14\pi r^2=\dfrac14\times\dfrac{22}{7}\times(3.5)^2=\dfrac{77}{8}$ cm$^2$.

Answer:

The area of the quadrant is $\dfrac{77}{8}$ cm$^2$ or $9.625$ cm$^2$.

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Q.3The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.v
Solution

In 5 minutes, the minute hand sweeps $\dfrac{5}{60}\times 360^\circ=30^\circ$. Area swept $=\dfrac{30}{360}\times\dfrac{22}{7}\times14^2=\dfrac{154}{3}$ cm$^2$.

Answer:

The area swept is $\dfrac{154}{3}$ cm$^2$ (about $51.33$ cm$^2$).

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Q.4A 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 $\pi = 3.14$)v
Solution

For the $90^\circ$ sector, area $=\dfrac{90}{360}\times3.14\times10^2=78.5$ cm$^2$. The triangle formed by the two radii is right-angled, so its area $=\dfrac12\times10\times10=50$ cm$^2$. Minor segment area $=78.5-50=28.5$ cm$^2$. Major sector angle $=270^\circ$, so area $=\dfrac{270}{360}\times3.14\times10^2=235.5$ cm$^2$.

Answer:

(i) Area of the minor segment $=28.5$ cm$^2$.
(ii) Area of the major sector $=235.5$ cm$^2$.

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Q.5In 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 chordv
Solution

Arc length $=\dfrac{60}{360}\times2\times\dfrac{22}{7}\times21=22$ cm. Sector area $=\dfrac{60}{360}\times\dfrac{22}{7}\times21^2=231$ cm$^2$. The chord and two radii form an equilateral triangle of side 21 cm, whose area is $\dfrac{\sqrt3}{4}\times21^2=\dfrac{441\sqrt3}{4}$ cm$^2$. Hence segment area $=231-\dfrac{441\sqrt3}{4}$ cm$^2$.

Answer:

(i) Length of the arc $=22$ cm.
(ii) Area of the sector $=231$ cm$^2$.
(iii) Area of the segment $=\left(231-\dfrac{441\sqrt3}{4}\right)$ cm$^2$ (about $40.04$ cm$^2$).

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Q.6A 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 $\sqrt3 = 1.73$)v
Solution

Sector area $=\dfrac{60}{360}\times3.14\times15^2=117.75$ cm$^2$. The triangle is equilateral with side 15 cm, so its area $=\dfrac{1.73}{4}\times15^2=97.3125$ cm$^2$. Minor segment $=117.75-97.3125=20.4375$ cm$^2$. Circle area $=3.14\times15^2=706.5$ cm$^2$, so major segment $=706.5-20.4375=686.0625$ cm$^2$.

Answer:

Area of the minor segment $=20.44$ cm$^2$ (approximately). Area of the major segment $=686.06$ cm$^2$ (approximately).

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Q.7A 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 $\sqrt3 = 1.73$)v
Solution

Sector area $=\dfrac{120}{360}\times3.14\times12^2=150.72$ cm$^2$. The triangle formed by the two radii has area $\dfrac12\times12\times12\times\sin120^\circ=72\times\dfrac{\sqrt3}{2}=36\sqrt3\approx62.28$ cm$^2$. Therefore, segment area $=150.72-62.28=88.44$ cm$^2$.

Answer:

The area of the corresponding minor segment is $88.44$ cm$^2$ (approximately).

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Q.8A 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 $\pi = 3.14$)v
Solution

The horse can graze a quadrant because it is tied at a corner. For a 5 m rope, area $=\dfrac14\times3.14\times5^2=19.625$ m$^2$. For a 10 m rope, area $=\dfrac14\times3.14\times10^2=78.5$ m$^2$. Increase $=78.5-19.625=58.875$ m$^2$.

Answer:

(i) The grazing area is $19.625$ m$^2$.
(ii) The increase in grazing area is $58.875$ m$^2$.

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Q.9A 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.v
Solution

Circumference of the circle $=\pi d=\dfrac{22}{7}\times35=110$ mm. Five diameters require $5\times35=175$ mm of wire. Total wire $=110+175=285$ mm. The circle is divided into 10 equal sectors, so area of each sector $=\dfrac1{10}\times\dfrac{22}{7}\times(17.5)^2=96.25$ mm$^2$.

Answer:

(i) Total length of silver wire required is 285 mm.
(ii) Area of each sector is $96.25$ mm$^2$.

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Q.10An 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.v
Solution

Eight equally spaced ribs divide the circular umbrella into 8 equal sectors. Area between two consecutive ribs $=\dfrac18\times\dfrac{22}{7}\times45^2=\dfrac{111375}{140}$ cm$^2\approx795.54$ cm$^2$.

Answer:

The area between two consecutive ribs is $\dfrac{111375}{140}$ cm$^2$ or about $795.54$ cm$^2$.

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Q.11A 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.v
Solution

One wiper cleans a sector of radius 25 cm and angle $115^\circ$. Area for two wipers $=2\times\dfrac{115}{360}\times\dfrac{22}{7}\times25^2=1254.96$ cm$^2$ approximately.

Answer:

The total area cleaned is about $1254.96$ cm$^2$.

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Q.12To 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 $\pi = 3.14$)v
Solution

The light covers a sector of radius 16.5 km and angle $80^\circ$. Area $=\dfrac{80}{360}\times3.14\times(16.5)^2=189.97$ km$^2$ approximately.

Answer:

The warned area is $189.97$ km$^2$ (approximately).

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Q.13A 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 cm$^2$. (Use $\sqrt3 = 1.7$)v
Solution

Each design is a minor segment formed by a $60^\circ$ sector of radius 28 cm. Area of one sector $=\dfrac{60}{360}\times\dfrac{22}{7}\times28^2=\dfrac{1232}{3}$ cm$^2$. The triangle in one sector is equilateral, so its area $=\dfrac{1.7}{4}\times28^2=333.2$ cm$^2$. One design area $=\dfrac{1232}{3}-333.2=77.466\ldots$ cm$^2$. Six designs have area $464.8$ cm$^2$, so cost $=464.8\times0.35=₹162.68$.

Answer:

The cost of making the designs is $₹162.68$ (approximately).

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Q.14Tick the correct answer in the following : Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ isv
  1. A. $\dfrac{p}{180}\times2\pi R$
  2. B. $\dfrac{p}{180}\times\pi R^2$
  3. C. $\dfrac{p}{360}\times\pi R^2$
  4. D. $\dfrac{p}{720}\times2\pi R^2$
Solution

A full circle has angle $360^\circ$ and area $\pi R^2$. Therefore a sector of angle $p^\circ$ has area $\dfrac{p}{360}\times\pi R^2$.

Answer:

Correct option: (C) $\dfrac{p}{360}\times\pi R^2$.

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