• Areas Related to Circles : Exercise 12.3 (Mathematics NCERT Class 10th)

Unless stated otherwise, $Use\pi = {{22} \over 7}$
Q.1     Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Sol.      Since ROQ is a diameter, therefore, $\angle RPQ = {90^o}$
In rt $\angle d\,\Delta \,PRQ$ $R{Q^2} = R{P^2} + P{Q^2}$
$\Rightarrow$ $R{Q^2} = {7^2} + {24^2} = 49 + 576 = 625$
$\Rightarrow$ $RQ = \sqrt {625} = 25\,cm$
Therefore, Radius $r = {1 \over 2}RQ = {{25} \over 2}cm$
Area of the semi circle $= {1 \over 2}\pi {r^2} = {1 \over 2} \times {{22} \over 7} \times {{25} \over 2} \times {{25} \over 2}c{m^2}$
$= {{6875} \over {28}}c{m^2}$
and area of $\Delta \,RPQ = {1 \over 2} \times RP \times PQ$
$= \left( {{1 \over 2} \times 7 \times 24} \right)c{m^2} = 84\,c{m^2}$
= Area of the semi circle – Area ($\Delta$ RPQ)
$= \left( {{{6875} \over {28}} - 84} \right)c{m^2} = \left( {{{6875 - 2352} \over {28}}} \right)c{m^2}$
$= {{4523} \over {28}}c{m^2}$

Q.2      Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and $\angle AOC = {40^o}$

Sol.     Area of the shaded region
= Area of sector AOC – Area of sector OBD
$= \left( {{{40} \over {360}} \times \pi \times {{14}^2} - {{40} \over {360}} \times \pi \times {7^2}} \right)c{m^2}$
$= {1 \over 9} \times {{22} \over 7}\left( {196 - 49} \right)c{m^2}$
$= \left( {{{22} \over {63}} \times 147} \right)c{m^2} = {{154} \over 3}c{m^2}$

Q.3      Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Sol.       Area of the square ABCD = ${\left( {14} \right)^2}c{m^2} = 196\,c{m^2}$
Diameter of the semicircles = AD or BC = 14 cm
Therefore, Radius of each semicircle = 7 cm
Area of the semicircular regions $= 2 \times {1 \over 2}\pi {r^2} = \pi {r^2}$
$= \left( {{{22} \over 7} \times 49} \right)c{m^2} = 154\,c{m^2}$
Therefore, Area of the shaded portion
= Area of the square ABCD – Area of the semicircular regions
= $\left( {196 - 154} \right)c{m^2} = 42\,c{m^2}$

Q.4      Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Sol.       Area of the circular portion
= Area of the circle – Area of the sector
$= \pi {r^2} - {{60} \over {360}}\pi {r^2} = \pi {r^2}\left( {1 - {1 \over 6}} \right)$
$= {5 \over 6}\pi {r^2},$ where r = 6
$= \left( {{5 \over 6} \times {{22} \over 7} \times 36} \right)c{m^2} = {{660} \over 7}c{m^2}$
Area of the equilateral $\Delta \,OAB$
$= {{\sqrt 3 } \over 4}{\left( {side} \right)^2} = \left( {{{\sqrt 3 } \over 4} \times 144} \right)c{m^2}$
$= 36\sqrt 3 \,c{m^2}$
Therefore, Area of the shaded region
$= \left( {{{660} \over 7} + 36\sqrt 3 } \right)c{m^2}$

Q.5      From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square.

Sol.           The area of the whole square ABCD $= {4^2}c{m^2} = 16\,c{m^2}$
The sum of the area of the four quadrants at the four corners of the square
= The area of a circle of radius 1 cm
$= {{22} \over 7} \times {1^2}c{m^2} = {{22} \over 7}c{m^2}$
The area of the circle of diameter 2 cm,
i.e., radius $1\,c{m^2}$
$= {{22} \over 7} \times {1^2}c{m^2} = {{22} \over 7}c{m^2}$
Therefore, Area of the remaining portion
= The area of the square ABCD
– The sum of the area of 4 quadrants at the four corners of the square
– The area of the circle of diameter 2 cm
$= \left( {16 - {{22} \over 7} - {{22} \over 7}} \right)c{m^2} = \left( {{{112 - 22 - 22} \over 7}} \right)c{m^2}$
$= {{68} \over 7}c{m^2}$

Q.6      In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).

Sol.         Let ABC be an equilateral triangle and let O be the circumcentre of the circumcircle of radius 32 cm.
Area of the circle $= \pi {r^2}$
$= \left( {{{22} \over 7} \times 32 \times 32} \right)c{m^2}$
$= {{22528} \over 7}c{m^2}$

Area of $\Delta$ ABC = 3 × Area of $\Delta$ BOC
$= 3 \times {1 \over 2} \times OB \times OC \times \sin BOC$
$\left( {{3 \over 2} \times 32 \times 32 \times \sin \,{{120}^o}} \right)$
$= \left( {3 \times 16 \times 32 \times {{\sqrt 3 } \over 2}} \right)$
$= 768\sqrt 3 \,c{m^2}$
Therefore, Area of the design (i.e., shaded region)
= Area of the circle – Area of $\Delta$ ABC
$= \left( {{{22528} \over 7} - 768\sqrt 3 } \right)c{m^2}$

Q.7       In figure , ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Sol.           The area of the whole square $ABCD = {14^2}c{m^2} = 196\,c{m^2}$
The sum of the area of the four quadrants at the four corners of the square
= The area of a circle of radius $7\,c{m^2}$
$= \pi {\left( 7 \right)^2}c{m^2} = \left( {{{22} \over 7} \times 49} \right)c{m^2} = 154\,c{m^2}$
Area of the shaded portion = The area of the square ABCD
– The sum of the area of four quadrants at the four corners of the square
$= \left( {196 - 154} \right)c{m^2} = 42\,c{m^2}$

Q.8       Figure depicts a racking track whose left and right ends are semicircular.

The distance between the two inner parallel line segment is 60 m and they are each 106 m long. If the track is 10 m wide, find
(i) The distance around the track along its inner edge.
(ii) The area of the track.
Sol.           We have, OB = O'C = 30 m
and AB = CD = 10 m
OA = O'D = (30 + 10) m = 40 m

(i) The distance around the track around along its inner edge.
= BC + EH + 2 × circumference of the semicrircle of radius OB = 30 m
$= \left( {106 + 106 + 2 \times {1 \over 2} \times 2\pi \left( {30} \right)} \right)m$
$= \left( {212 + 2 \times {{22} \over 7} \times 30} \right)m$
$= \left( {212 + {{1320} \over 7}} \right)m = \left( {{{1484 + 1320} \over 7}} \right)m = {{2804} \over 7}m$

(ii) Area of the track
= Area of the shaded region
= Area of rectangle ABCD + Area of rectangle EFGH
+ 2 (Area of the semicircle of radius 40 m
– Area of the semicircle with radius 30 m)
$= \left[ {\left( {10 \times 106} \right) + \left( {10 + 106} \right) + 2\left\{ {{1 \over 2} \times {{22} \over 7} \times {{\left( {40} \right)}^2} - {1 \over 2} \times {{22} \over 7} \times {{\left( {30} \right)}^2}} \right\}} \right]{m^2}$

$= \left[ {1060 + 1060 + {{22} \over 7}\left( {{{40}^2} - {{30}^2}} \right)} \right]{m^2}$
$= \left[ {2120 + {{22} \over 7}\left( {40 + 30} \right)\left( {40 - 30} \right)} \right]{m^2}$
$= \left( {2120 + {{22} \over 7} \times 70 \times 10} \right){m^2}$
$= \left( {2120 + 2200} \right){m^2}$
$= 4320\,{m^2}$

Q.9      In figure AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Sol.       Area of the sector $= \left( {{{90} \over {360}} \times {{22} \over 7} \times 7 \times 7} \right)c{m^2}$
$= {{77} \over 2}c{m^2}$
Area of $\Delta$ OCB $= {1 \over 2} \times OC \times OB$
$= {1 \over 2} \times 7 \times 7\,c{m^2}$
$= {{49} \over 2}c{m^2}$
Therefore, The area of the segment BPC
$= \left( {{{77} \over 2} - {{49} \over 2}} \right)c{m^2} = {{28} \over 2}c{m^2} = 14\,c{m^2}$
Similarly the area of the segment AQC = $14\,c{m^2}$
Also, the area of the circle with DO as diameter
$= \left( {{{22} \over 7} \times {7 \over 2} \times {7 \over 2}} \right)c{m^2} = {{77} \over 2}c{m^2}$
Hence, the total area of the shaded region
$= \left( {14 + 14 + {{77} \over 2}} \right)c{m^2}$
$= \left( {{{28 + 28 + 77} \over 2}} \right)c{m^2} = {{133} \over 2}c{m^2}$
$= 66.5\,\,c{m^2}$

Q.10      The area of an equilateral triangle ABC is 17320.5 $\,c{m^2}$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length  of the side of the triangle (see figure). Find the area of the shaded region.
(Use $\pi = 3.14\,and\,\sqrt 3 = 1.73205$)

Sol.       Let each side of the triangle be a cm. Then,
Area = $17320.5\,c{m^2}$
$\Rightarrow$ ${{\sqrt 3 } \over 4}{a^2} = 17320.5$ $\left[ {\sin ce\,Area = {{\sqrt 3 } \over 4}{{\left( {side} \right)}^2}} \right]$
$\Rightarrow$ ${a^2} = {{17320.5 \times 4} \over {\sqrt 3 }} = {{17320.5 \times 4} \over {1.73205}} = 40000$
$\Rightarrow$ a = 200
Thus, radius of each circle is 100 cm.
Now, required area
= Area of $\Delta$ ABC – 3 ×(Area of a sector of angle 60º in a circle of 100 cm)
$= \left[ {17320.5 - 3\left( {{{60} \over {360}} \times 3.14 \times 100 \times 100} \right)} \right]c{m^2}$
$= \left( {17320.5 - 15700} \right)c{m^2} = 1620.5\,c{m^2}$

Q.11      On a square hand kerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Sol.      Side of the square ABCD = AB
= 3 × diameter of circular design
= 3 × (2 × 7) cm = 42 cm
Therefore, Area of the square ABCD
$= \left( {42 \times 42} \right)c{m^2} = 1764\,c{m^2}$
Area of one circular design
$= \pi {r^2} = \left( {{{22} \over 7} \times 7 \times 7} \right)c{m^2} = 154\,c{m^2}$
Therefore, Area of 9 such designs
$= \left( {9 \times 154} \right)c{m^2} = 1386\,c{m^2}$
Therefore, Area of the remaining portion of the handkerchief
= Area of the square ABCD – Area of 9 circular designs
$= \left( {1764 - 1386} \right)c{m^2} = 378\,c{m^2}$

Q.12      In figure OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm , find the area of the

Sol.            (i) Area of quadrant $= {1 \over 4}\pi {r^2}$
$= {1 \over 4} \times {{22} \over 7} \times {\left( {3.5} \right)^2}c{m^2}$
$= {1 \over 4} \times {{22} \over 7} \times {7 \over 2} \times {7 \over 2}c{m^2}$
$= {{77} \over 8}c{m^2}$

(ii) Area of $\Delta \,AOD = {1 \over 2}Base\, \times \,Height$
$= {1 \over 2}\left( {OB \times OD} \right)$
$= {1 \over 2}\left( {3.5 \times 2} \right)c{m^2} = {7 \over 2}c{m^2}$
Hence, area of the shaded region
= Area of quadrant – Area of $\Delta$ AOD
$= \left( {{{77} \over 8} - {7 \over 2}} \right)c{m^2} = \left( {{{77 - 28} \over 8}} \right)c{m^2}$
$= {{49} \over 8}c{m^2}$

Q.13       In figure, a square OABC is inscribed in a quadrant OPBQ . If OA = 20 cm, find the area of the shaded region. (Use $\pi = 3.14$

Sol.            Radius of the quadrant = OB = $\sqrt {O{A^2} + A{B^2}}$
$= \sqrt {{{20}^2} + {{20}^2}} \,cm$
$= 20\sqrt {1 + 1} \,cm$
$= 20\sqrt 2 \,cm$
$= {1 \over 4}\pi {r^2} = {1 \over 4} \times 3.14 \times {\left( {20\sqrt 2 } \right)^2}c{m^2}$
$= \left( {{1 \over 4} \times 3.14 \times 800} \right)c{m^2} = 628\,c{m^2}$
Area of the square OABC
$= {\left( {20} \right)^2}c{m^2} = 400\,c{m^2}$
Hence, area of the shaded region
= Area of quadrant – Area of square OABC
$= \left( {628 - 400} \right)c{m^2} = 228\,c{m^2}$

Q.14      AB and CD are respectively across of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If $\angle AOB = {30^o}$, find the area of the shaded region.

Sol.      Let ${A_1}\,and\,{A_2}$ be the areas of sectors OAB and OCD respectively. Then,
${A_1}$ = Area of a sector of angle 30º in a circle of radius 21 cm.
$= \left( {{{30} \over {360}} \times {{22} \over 7} \times 21 \times 21} \right)c{m^2}\left[ {U\sin g\,A = {\theta \over {360}} \times \pi {r^2}} \right]$
$= \left( {{{231} \over 2}} \right)c{m^2}$
${A_2}$ = Area of a sector of angle 30º in a circle of radius 7 cm
$= \left( {{{30} \over {360}} \times {{22} \over 7} \times 7 \times 7} \right)c{m^2}$
$= {{77} \over 6}c{m^2}$
$= {A_1} - {A_2} = \left( {{{231} \over 2} - {{77} \over 6}} \right)c{m^2}$
$= {{693 - 77} \over 6}c{m^2}$
$= {{616} \over 6}c{m^2} = {{308} \over 3}c{m^2}$

Q.15      In figure , ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Sol.      Let 14 cm be the radius of the quadrant with A as the centre.
Then the area of the quadrant ABMC

$= {1 \over 4}\pi {r^2}$
$= \left( {{1 \over 4} \times {{22} \over 7} \times 196} \right)c{m^2}$
$= 154\,c{m^2}$
Area of $\Delta$ BAC
$= {1 \over 2} \times AC \times AB$
$= \left( {{1 \over 2} \times 14 \times 14} \right)c{m^2} = 98\,c{m^2}$
Therefore, Area of the segment of the circle, BMC
= Area of the quadrant ABMC – Area of $\Delta$ BAC
= (154 – 98) $c{m^2} = 56\,c{m^2}$
Now, since AC = AB = 14 cm and $\angle BAC = {90^o}$
Therefore, By Pyrthagoras Theorem,
$BC = \sqrt {A{C^2} + A{B^2}}$
$= \sqrt {{{14}^2} + {{14}^2}}$
$= 14\sqrt 2$
Therefore, Radius of the semicircle BNC = $7\sqrt 2 \,cm$
Area of the semicircle BNC = ${\pi \over 2}{\left( {7\sqrt 2 } \right)^2}c{m^2}$
$= \left( {{1 \over 2} \times {{22} \over 7} \times 98} \right)c{m^2} = 154\,c{m^2}$
Hence, the area of the region between two arcs BMC and BNC
= The area of the shaded region
= The area of semicircle BNC – The area of the segment of the circle BMC
$= \left( {154 - 56} \right)c{m^2} = 98\,c{m^2}$

Q.16      Calculate the area of the the designed region in figure common between the two quadrants of circle of radius 8 cm each.

Sol.       Here, 8 cm is the radius of the quadrants ABMD and BNDC.
Sum of their areas
$= 2 \times {1 \over 4}\pi {r^2} = {1 \over 2}\pi {r^2}$
$= \left( {{1 \over 2} \times {{22} \over 7} \times 64} \right)c{m^2}$
$= {{704} \over 7}c{m^2}$

Area of the square ABCD
$= \left( {8 \times 8} \right)c{m^2}$
$= 64\,c{m^2}$
Area of the designed region
= Area of the shaded region
= Sum of the area of quadrants
– Area of the square ABCD
$= \left( {{{704} \over 7} - 64} \right)c{m^2} = \left( {{{704 - 448} \over 7}} \right)c{m^2}$
$= {{256} \over 7}c{m^2}$