# Areas Related to Circles : Exercise 12.2 (Mathematics NCERT Class 10th)

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

Q.1       Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60º
Sol.        We know that the area A of a sector of angle $\theta$ in a circle of radius r is given by $A = {\theta \over {360}} \times \pi {r^2}$.
Here r = 6 cm, $\theta = {60^o}$
Therefore, $A = \left[ {{{60} \over {360}} \times {{22} \over 7} \times 36} \right]c{m^2} = \left( {{{132} \over 7}} \right)c{m^2}$

Q.2      Find the area of a quadrant of a circle whose circumference is 22 cm.
Sol.        Let r be the radius of the circle. Then,
Circumference = 22 cm
$\Rightarrow$ $2\pi r = 22$
$\Rightarrow$ $2 \times {{22} \over 7} \times r = 22$
$\Rightarrow$ $r = {7 \over 2}cm$
Area of the quadrant of a circle
$= {1 \over 4}\pi {r^2} = \left( {{1 \over 4} \times {{22} \over 7} \times {{49} \over 4}} \right)c{m^2}$
$= {{539} \over {56}}c{m^2} = {{77} \over 8}c{m^2}$

Q.3      The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Sol.        Clearly, minute hand of a clock describes a circle of radius equal to its length i.e., 14 cm. Since the minutes hand rotates through 6º in one minute, therefore, area swept by the minute hand in one minute is the area of  sector of angle 6º in a circle of radius 14 cm. Hence, the required area i.e., the area swept in 5 minutes.
$= \left( {{\theta \over {360}} \times \pi {r^2} \times 5} \right)c{m^2}$
$= \left[ {{6 \over {360}} \times {{22} \over 7} \times {{\left( {14} \right)}^2} \times 5} \right]c{m^2}$
$= \left( {{1 \over {60}} \times {{22} \over 7} \times 196 \times 5} \right)c{m^2} = {{154} \over 3}c{m^2}$

Q.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)
Sol.        Here, r = 10 cm, $\theta = {90^o}$
(i) Area of the minor segment
$= {r^2}\left[ {{{\pi \theta } \over {360}} - {1 \over 2}\sin \theta } \right]$
$= {\left( {10} \right)^2}\left[ {{{3.14 \times 90} \over {360}} - {1 \over 2}\sin {{90}^o}} \right]c{m^2}$
$= 100\left( {0.785 - 0.5} \right)c{m^2}$
$= \left( {100 \times 0.285} \right)c{m^2} = 28.5\,c{m^2}$

(ii) Area of the major sector
$= {\theta \over {360}} \times \pi {r^2}$, where r = 10 and $\theta = {270^o}$
$= \left( {{{270} \over {360}} \times 3.14 \times {{10}^2}} \right)c{m^2}$
$= \left( {{3 \over 4} \times 3.14 \times 100} \right)c{m^2} = 235.5\,c{m^2}$

Q.5      In a circle of a radius 21 cm, an arc substends an angle of 60º 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.
Sol.        Here r = 21 cm $\theta = {60^o}$
(i) Length of the arc, $\ell = {\theta \over {180}} \times \pi r$
$= \left( {{{60} \over {180}} \times {{22} \over 7} \times 21} \right)cm$
$= 22\,cm$

(ii) Area of the sector, $A = {\theta \over {360}} \times \pi {r^2}$
$= \left( {{{60} \over {360}} \times {{22} \over 7} \times 21 \times 21} \right)c{m^2}$
$= 231\,c{m^2}$

(iii) Area of the segment
$= {r^2}\left[ {{{\pi \theta } \over {360}} - {1 \over 2}\sin \theta } \right]$
$= {\left( {21} \right)^2}\left[ {{{22} \over 7} \times {{60} \over {360}} - {1 \over 2}\sin {{60}^o}} \right]c{m^2}$
$= 441\left( {{{11} \over {21}} - {1 \over 2} \times {{\sqrt 3 } \over 2}} \right)c{m^2}$
$= \left( {21 \times 11 - {{441\sqrt 3 } \over 4}} \right)c{m^2}$
$= \left( {231 - {{441\sqrt 3 } \over 4}} \right)c{m^2}$

Q.6      A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the areas of  the corresponding minor and major segment of the circle.
(Use $\pi = 3.14\,and\,\sqrt 3 = 1.73$)
Sol.        Here r = 15 cm and $\theta = {60^o}$
Therefore, Area of the minor segment
$= {r^2}\left[ {{{\pi \theta } \over {360}} - {1 \over 2}\sin \theta } \right]$
$= {\left( {15} \right)^2}\left[ {{{314 \times 60} \over {360}} - {1 \over 2} \times \sin 60} \right]c{m^2}$
$= 225\left[ {{{314} \over 6} - {1 \over 2} \times {{\sqrt 3 } \over 2}} \right]c{m^2}$
$= 225\left( {{{3.14 \times 2 - 3\sqrt 3 } \over {12}}} \right)c{m^2}$
$= 225\left( {{{6.28 - 3 \times 1.73} \over {12}}} \right)c{m^2}$
$= {{225 \times \left( {6.28 - 5.19} \right)} \over {12}}c{m^2}$
$= {{225 \times 1.09} \over {12}}c{m^2} = {{245.25} \over {12}}c{m^2} = 20.4375\,c{m^2}$
Area of the major segment
= Area of the circle – Area of the minor segment
= (3.14 × 225 – 20.4375) $c{m^2}$
= (706.5 – 20.4375)$c{m^2}$ = 686.0625 $c{m^2}$

Q.7      A chord of a circle of radius 12 cm subtends an angle of 120º at the centre. Find the area of the corresponding segment of the circle.
(Use $\pi = 3.14\,and\,\sqrt 3 = 1.73$)
Sol.        Here r = 12 cm $\theta = {120^o}$
Area of the corresponding segment of the circle
= Area of the minor segment
$= {r^2}\left[ {{{\pi \theta } \over {360}} - {1 \over 2}\sin \theta } \right]$
$= {\left( {12} \right)^2}\left[ {{{3.14 \times 120} \over {360}} - {1 \over 2}\sin \,{{120}^o}} \right]c{m^2}$
$= 144\left( {{{3.14} \over 3} - {1 \over 2} \times {{\sqrt 3 } \over 2}} \right)c{m^2}$
$\left[ {\sin ce\,\,\sin \,{{120}^o} = \sin \left( {{{180}^o} - {{60}^o}} \right) = \sin \,{{60}^o} = {{\sqrt 3 } \over 2}} \right]$
$= 144 \times {{3.14 \times 4 - 3\sqrt 3 } \over {12}}c{m^2}$
$= 12 \times \left( {12.56 - 3 \times 1.73} \right)c{m^2}$
$= 12 \times \left( {12.56 - 5.19} \right)c{m^2}$
$= 12 \times 7.37\,c{m^2} = 88.44\,c{m^2}$

Q.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 figure). 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 $\pi = 3.14$)

Sol.           (i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 m.
Then the area of the quadrant of this circle
$= {{\pi \times {5^2}} \over 4}{m^2} = {{3.14 \times 25} \over 4}{m^2}$
$= {{78.5} \over 4}{m^2} = 19.625\,\,{m^2}$ (ii) In the 2nd case, radius = 10 m.
The area of the quadrant of this circle
$= {{\pi \times {{10}^2}} \over 4}{m^2} = {{3.14 \times 100} \over 4}{m^2}$
$= {{314} \over 4}{m^2} = 78.5\,\,{m^2}$ Therefore, Increase in the grazing area
$= \left( {78.5 - 19.625} \right){m^2} = 58.875\,{m^2}$

Q.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 sector as shown in figure. Find : (i) The total length of the silver wire required.
(ii) The area of each sector of the brooch.
Sol.             (i) Silver wire used to make a brooch
$= 2\pi r,\,where\,r = {{35} \over 2}mm$
$= 2 \times {{22} \over 7} \times {{35} \over 2}mm = 110\,mm$
Wire used in 5 diameters = 5 × 35 mm = 175 mm
Therefore, Total wire used = (110 + 175) mm = 285 mm

(ii) The area of each sector of the brooch
$= {1 \over {10}} \times$ Area of the circle
$= {1 \over {10}} \times \pi {r^2},\,where\,r = {{35} \over 2}mm$
$= \left( {{1 \over {10}} \times {{22} \over 7} \times {{35} \over 2} \times {{35} \over 2}} \right)m{m^2}$
$= {{385} \over 4}m{m^2}$

Q.10      An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Sol.           Area between two consecutive ribs
$= {1 \over 8} \times \pi {r^2},\,where\,r = 45\,cm$
$= \left( {{1 \over 8} \times {{22} \over 7} \times 45 \times 45} \right)\,c{m^2}$
$= {{22275} \over {28}}c{m^2}$

Q.11      A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115º. Find the total area cleaned at each sweep of the blades.
Sol.           Here r = 25 cm, $\theta = {115^o}$
Total area cleaned at each sweep of the blades
= 2 × Area of the sector (having r = 25 and $\theta = {115^o}$
$= 2 \times {\theta \over {360}} \times \pi {r^2}$
$= \left( {2 \times {{115} \over {360}} \times {{22} \over 7} \times 25 \times 25} \right)c{m^2}$
$= \left( {{{23 \times 11 \times 625} \over {18\, \times 7}}} \right)c{m^2} = {{158125} \over {126}}c{m^2}$

Q.12      To warn ships for underwater rocks a lighthouse spreads a red coloured light over a sector of angle 80º to a distance of 16.5 km. Find the area of the sea over  which the ships are warned. (Use $\pi = 3.14$)
Sol.           Area of the sea over which the ships are warned
= Area of the sector (having r = 16.5 km and $\theta = {80^o}$)
$= {\theta \over {360}} \times \pi {r^2}$
$= \left( {{{80} \over {360}} \times 3.14 \times 16.5 \times 16.5} \right)sq.\,km$
$= {{68389.2} \over {360}}sq.km$
$= 189.97\,k{m^2}$

Q.13      A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per $c{m^2}$.
(Use $\sqrt 3 = 1.7$) Sol.      Clearly from the figure,
Area of one desgin
= Area of the sector AOB – Area ($\Delta \,AOB$)
$= \left( {{\theta \over {360}} \times \pi {r^2} - {1 \over 2} \times r \times r \times \sin \,{{60}^o}} \right)c{m^2}$, where r = 28 cm
$= \left( {{{60} \over {360}} \times {{22} \over 7} \times 28 \times 28 - {1 \over 2} \times 28 \times 28 \times {{\sqrt 3 } \over 2}} \right)c{m^2}$ $= \left( {{{1232} \over 3} - 333.2} \right)c{m^2}$
$= \left( {{{1232 - 999.6} \over 3}} \right)c{m^2}$
$= {{232.4} \over 3}c{m^2}$
Area of 6 such design $= \left( {6 \times {{232.4} \over 3}} \right)c{m^2} = 464.8\,c{m^2}$
Cost of making such designs @ Rs. 0.35 per $c{m^2}$
= Rs (0.35 × 464.8)
= Rs . 162.68

Q.14      Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is :
(a) ${p \over {180}} \times 2\pi R$                    (b) ${p \over {180}} \times \pi {R^2}$
(c) ${p \over {360}} \times 2\pi R$                    (d) ${p \over {720}} \times 2\pi {R^2}$
Sol.      We know that area A of a sector of angle $\theta$ in a circle of radius r is given by
$A = {\theta \over {360}} \times \pi {r^2}$
But here , r = R and $\theta = p$
Therefore, $A = {p \over {360}} \times \pi {R^2} = {p \over {720}} \times 2\pi {R^2}$
Therefore, (D) is the correct answer.

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