# Area Related to Circles : Previous Year's Questions

Very short answer type questions : -

Q.1     In the fig., O is the centre of a circle. The area of sector OAPB is ${5 \over {18}}$ of the area of the circle. Find x.

[Delhi 2008]

Sol.

Area of sector OAPB = ${{\pi {r^2} \times x} \over {360^\circ }}$
Area of circle = ${\pi {r^2}}$
Given
Area of sector OAPB = ${5 \over {18}}$ area of circle
$\Rightarrow {{\pi {r^2}} \over {360^\circ }} \times x = {5 \over {18}} \times \pi {r^2}$
$\Rightarrow x = {5 \over {18}} \times 360^\circ$
$\Rightarrow x = 100^\circ$

Q.2      Find the perimeter of the given figure, where arc AED is a semicircle and ABCD is a rectangle.

[AI 2008]

Sol.

AB + BC + CD = 20 + 14 + 20 = 54 cm
Diameter of semicircle = 14 cm
$\Rightarrow$ Radius of semicircle i.e. r = 7 cm
Permimeter of Arc AED = $\pi r$ = $7\pi$
Hence, perimeter of given figure = $\left( {7\pi + 54} \right)$ cm

Q.3      A chord of a circle of radius 14 cm subtends a right angle at the centre. What is the area of the minor sector ? $\left[ {\pi = {{22} \over 7}} \right]$

[Delhi 2008 C]

Sol.

Area of sector = ${{\pi {\rm{r}}^{\rm{2}} \theta}\over {360^\circ }}$ [Where $\theta= 90^\circ$]
$= {{22} \over 7} \times {{14 \times 14} \over {360^\circ }} \times 90^\circ$
$= {{22 \times 2 \times 14} \over 4}$
$= 11 \times 14$
$= 154c{m^2}$
Hence, area of minor sector is $154c{m^2}$.

Q.4      What is the perimeter of a sector of angle 45° of a circle with radius 7 cm ? $\left[ {Use\,\pi = {{22} \over 7}} \right]$

[AI 2008 C]

Sol.

Let O be the centre of the circle and $\theta$ be 45º
Perimeter of a sector OAB = ${{\pi r\theta } \over {180^\circ }} + 2r$

$= {{\pi \times 7 \times 45^\circ } \over {180^\circ }} + 2 \times 7$
$= {{22} \over 7} \times {{7 \times 45^\circ } \over {180^\circ }} + 2 \times 7$
$= {{22}\over 4}+14$
$= 5.5 + 14$
$= 19.5$ cm
Hence , perimeter of circle = 19.5 cm

Q.5      If the diameter of a semicircular protractor is 14 cm, then find its perimeter. $\left[ {\pi = {{22} \over 7}} \right]$

[AI 2009]

Sol.

Diameter of semicircle = 14 cm
radius(r) $= {{14}\over 2}$ = 7 cm
Perimeter of semicircle = $\pi r + 2r$

$= {{22} \over 7} \times 7 + 2 \times 7$
= 22 + 14
= 36 cm
Hence , perimeter of semicircular protractor is 36 cm.

Short answer type questions - I

Q.1      In figure APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14cm each. Find the perimeter of the shaded region. [use ${\pi = {{22} \over 7}}$]

[Delhi 2011]

Sol.

Perimeter of semicircle APB = $\pi r$
$= \pi \times 3.5$
$= 3.5\pi$ cm
Perimeter of semicircle ARC = $\pi r$
$= \pi \times 7$
$= 7\pi$ cm
Perimeter of semicircle CQD = $\pi r$
$= \pi \times 3.5$
$= 3.5\pi$ cm
Perimeter of semicircle BSD = $\pi r$
$= \pi \times 7$
$= 7\pi$
Perimeter of shaded region = Perimeter of semicircle (APB + ARC + CQD + BSD)
$= 3.5\pi + 7\pi + 3.5\pi + 7\pi$
$= 21\pi$
$= 21 \times {{22} \over 7}$
$= 3\times 22$
= 66 cm

Q.2     Find the perimeter of the shaded region in figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [use ${\pi = {{22} \over 7}}$]

[AI 2011]

Sol.

Given : Side of square = 14 cm = diameter of semicircle(APB and CPD)
AB = BC = CD = DA = 14 cm
radius of semicircle(r) $= {{14}\over 2} = 7\,cm$
Perimeter of semicircle DPC = $\pi r$

$= \pi \times 7$
$= 7\pi$ cm
Perimeter of semicircle APB = $\pi r$
$= \pi \times 7$
$= 7\pi$ cm
AD + BC = 14 + 14 = 28 cm
Perimeter of shaded region = AD + BC + Perimeter of semicircle (APB + DPC)
$= 28 + 7\pi + 7\pi$
$= 28 + 14\pi$
$= 28 + 14 \times {{22} \over 7}$
$= 28 + 2 \times 22$
= 28 + 44 = 72 cm
Hence,the perimeter of the shaded region is 72 cm.

Q.3      In given figure, a semicircle is drawn with O as centre and AB as diameter. Semicircles are drawn with AO and OB as diameters. If AB = 28 m, find the perimeter of the shaded region. [use ${\pi = {{22} \over 7}}$]

[AI 2011]

Sol.

Let radius of large semicircle be ${r_1}$ cm and radius of smaller semicircle be ${r_2}$ cm and ${r_3}$ cm.
Therefore ${r_1}$ = 14 cm, ${r_2}$ and ${r_3}$ = 7 cm
Perimeter of shaded region = $\pi {r_1} + \pi {r_2} + \pi {r_3}$ [Perimeter of semicircle$=\pi r$]
$= \pi \times 14 + \pi \times 7 + \pi \times 7$
$= 28\pi$
$= 28 \times {{22} \over 7}$
= 88 cm
Hence,the perimeter of the shaded region is 88 cm.

Q.4      In given figure, ABC is a triangle right-angled at B, with AB = 14 cm and BC = 24 cm. With the vertices A, B and C as centres, arcs are drawn each of radius 7 cm. Find the area of the shaded region. [use ${\pi = {{22} \over 7}}$]

[AI 2011]

Sol.

Radius of each sector (r) = 7 cm
Let $\theta _1$ ,$\theta _2$and $\theta _3$ be the angles of sectors (angle of triangle)
$\Rightarrow$ $\theta _1 + \theta _2 + \theta _3 = 180^\circ$(Angle sum-property)

Area of shaded region = Area of $\Delta AC$ – Area of 3 sectors
$= {1 \over 2} \times AB \times BC - {{\pi {r^2}} \over {360^\circ }}\left[ {{\theta _1} + {\theta _2} + {\theta _3}} \right]$[Area of sector=${{\pi r^2 \theta } \over {360^\circ }}$]

$= {1 \over 2} \times 14 \times 24 - {{\pi \times 7 \times 7} \over {360^\circ }} \times 180^\circ$  [AB = 14 cm, BC = 24 cm(given)]

$= 168 - {{22} \over 7} \times {{7 \times 7} \over 2}$
$= 168 - 11 \times 7$
= 168 – 77
$= 91$ $c{m^2}$

Q.5      In the given figure, the shape of the top of a table is that a sector of a circle with centre O and $\angle AOB = 90^\circ$. If AO = OB = 42 cm, then find the perimeter of the top of the table. [use ${\pi = {{22} \over 7}}$]

[Delhi 2012]

Sol.

Given : AO = OB = 42 cm (radius of circle)
$\angle AOB = 90^\circ$
Perimeter of the top of table = Length of major arc + 2r

$= {{2\pi r} \over {360^\circ }} \times \theta + 2r$
$= {{2 \times 22} \over {7 \times 360^\circ }} \times 270^\circ \times 42 + 2 \times 42$ [$\theta= 360^ \circ - 90^ \circ= 270^ \circ$]
= 198 + 84 = 282 cm

Q.6      Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm × 7 cm. Find the area of the remaining cardboard. [use ${\pi = {{22} \over 7}}$]

[Delhi 2013]

Sol.

Let length of cardboard be l and height be h and radius of semicircle be r cm.
l = 14 cm , h = 7 cm and r = 7/2 cm
Area of remaining cardboard = Area of rectangle – Area of two semicircle

$= l \times h - \left[ {{{\pi {r^2}} \over 2} + {{\pi {r^2}} \over 2}} \right]$
$= 14 \times 7 - \left[ {{{\pi {{\left( {{7 \over 2}} \right)}^2}} \over 2} + {{\pi {{\left( {{7 \over 2}} \right)}^2}} \over 2}} \right]$
$= 98 - \left[ {{{22} \over 7} \times {{7 \times 7} \over {2 \times 2 \times 2}} + {{22} \over 7} \times {{7 \times 7} \over {2 \times 2 \times 2}}} \right]$
$= 98 - \left[ {{{77} \over 4} + {{77} \over 4}} \right]$
$= 98 - {{154} \over 4}$
= 98 – 38.5 = 59.5 $c{m^2}$

Q.7      In the given figure, the area of the shaded region between two concentric circles is 286 $c{m^2}$. If the difference of the radii of the two circles is 7 cm, find the sum of their radii. [use ${\pi = {{22} \over 7}}$]

[Foreign 2013]

Sol.

Given : Area of shaded region = 286 $c{m^2}$
Area of shaded region = Area of large circle – Area of small circle

Let radius of large circle be R
and radius of small circle be r
$\Rightarrow$ R-r = 7 cm    [given]
Therefore area of shaded region = $\pi {R^2} - \pi {r^2}$
$\Rightarrow$ 286 $= \pi \left[ {{R^2} - {r^2}} \right]$
$\Rightarrow$ 286 $= \pi \left[ {R + r} \right]\left[ {R - r} \right]$
$\Rightarrow$ $286 = {{22} \over 7} \times 7\left[ {R + r} \right]$
$\Rightarrow R + r = {{286} \over {22}}$
$\Rightarrow R + r = 13$ cm

Short answer type questions - II

Q.1      In the fig., find the perimeter of shaded region where ADC, AEB and BFC are semicircles on diameter AC, AB and BC respectively.
Or
Find the area of the shaded region in the fig., where ABCD is a square of side 1.4 cm.

[Delhi 2008]

Sol.

Let $r_1 \,,\,r_{2\,} and\,r_3$ be the radius of semicircle ADC,AEB and BFC .
${\rm{r}}_{\rm{1}}= {{2.8 + 1.4}\over 2} = {{4.2} \over 2} = 2.1$ cm
${\rm{r}}_{\rm{2}}= {{2.8} \over 2} = 1.4$ cm
${\rm{r}}_{\rm{3}}= {{1.4} \over 2} = 0.7$ cm

$= (\pi r_1 + \pi r_2 + \pi r_3)$   [Perimeter of semicircle = $\pi r$]
$= (\pi \times 2.1 + \pi \times 1.4 + \pi \times 0.7)$
$= \pi (2.1 + 1.4 + 0.7)$
$= 4.2\pi$
$= 4.2 \times {{22} \over 7}$
= 13.2 cm
Or
AB = BC = CD = DA = 1.4 cm(sides of square)
radius of circle (r) = 7/2 cm
Area of shaded region = Area of square – Area of 4 circles

$= (side)^2- 4(\pi r^2 )$
$= 14 \times 14 - 4 \times \pi \times {\left( {{7 \over 2}} \right)^2}$
$= 196 - 4\pi \times {{7 \times 7} \over {2 \times 2}}$
= 196 – 4 × 38.5
= 196 – 154 = 42 $c{m^2}$.

Q.2      In fig., ABC is a right-angled triangle, right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

[AI 2008]

Sol.

Area of shaded region = Area of semicircle or side AB + Area of semicircle side AC – Area of semicircle on side BC + Area of $\Delta ABC$ .............. (1)
By using Pythagoras theorem in $\Delta ABC$
${\left( {AB} \right)^2} + {\left( {AC} \right)^2} = B{C^2}$   [AB = 3 units ,AC = 4 units]
$\Rightarrow {\left( 3 \right)^2} + {\left( 4 \right)^2} = B{C^2}$
$\Rightarrow 9 + 16 = B{C^2}$
$\Rightarrow BC = 5$ cm
Area of semicircle on side AB = ${{\pi {r^2}} \over 2}$
$= {{22} \over 7} \times {\left( {{3 \over 2}} \right)^2} \times {1 \over 2}$
$= {{22} \over 7} \times {9 \over 4} \times {1 \over 2}$
$= {{99} \over {28}}$ units .............. (2)
Area of semicircle on side AC = ${{\pi {r^2}} \over 2}$
${\rm{ = }}{{{\rm{22}}} \over {\rm{7}}} \times {{(2)^2 } \over 2}$
$= {{22} \over 7} \times {4 \over 2}$
$= {{44} \over 7}$ units ............. (3)
Area of semicircle on side BC = ${{\pi {r^2}} \over 2}$
${\rm{ = }}{{{\rm{22}}} \over {\rm{7}}} \times {{(5/2)^2 } \over 2}$
$= {{22} \over 7} \times {{25} \over {4 \times 2}}$
$= {{275} \over {28}}$ units ............... (4)
Area of $\Delta ABC = {1 \over 2} \times AB \times AC$
$= {1 \over 2} \times 3 \times 4 = 6$ units ............. (5)
Put values from equation (1), (2), (3), (4) and (5), we get
Area of shaded region = ${{99} \over {28}} + {{44} \over 7} - {{275} \over {28}} + 6$ = 6 sq. units

Q.3      In the fig., 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.

[Foreign 2008, Delhi 2008 C]

Sol.

Area of shaded = Area of semicircle of side BC – [Area of quadrant ABC – Area of $\Delta ABC$] .......... (1)
In $\Delta ABC$
$A{B^2} + A{C^2} = {\left( {BC} \right)^2}$
$\Rightarrow {\left( {14} \right)^2} + {\left( {14} \right)^2} = {\left( {BC} \right)^2}$   [AB = AC = 14 cm (given)]

$\Rightarrow BC = 14\sqrt 2$ cm
Area of semicircle on side BC = ${{\pi {r^2}} \over 2}$
$= {\pi \over 2}{\left( {7\sqrt 2 } \right)^2}$               [$r = {{BC} \over 2} = {{14\sqrt 2 } \over 2} = 7\sqrt 2$]
$= {{22} \over 7} \times 7 \times {{7 \times 2} \over 2}$
= 154 cm .............(2)
Area of $\Delta ABC$ = ${1 \over 2} \times AB \times AC$
$= {1 \over 2} \times 14 \times 14 = 98c{m^2}$ ............. (3)
Area of quadrant ABC =${{\pi {r^2}} \over 4}$
$= {{\pi \times 14 \times 14} \over 4}$
$= {{22} \over 7} \times {{14 \times 14} \over 4} = 154$ $c{m^2}$ ................(4)
Put values from equation (2), (3) and (4) in equation (1)
Area of shaded region = 154 – [154 – 98]
= 154 – 154 + 98
= 98 $c{m^2}$

Q.4      Find the area of the shaded region in fig. If PR = 24 cm, PQ = 7 cm and O is the centre of the circle.

[Delhi 2008C, 2009]

Sol.

Given : PR = 24 cm , PQ = 7 cm
$\angle RPQ = 90^\circ$ [Angle in the semicircle]

In $\Delta PQR$
${\left( {PQ} \right)^2} + {\left( {PR} \right)^2} = {\left( {RQ} \right)^2}$
$\Rightarrow {\left( 7 \right)^2} + {\left( {24} \right)^2} = {\left( {RQ} \right)^2}$
$\Rightarrow {\left( {RQ} \right)^2} = 49 + 576$ $\Rightarrow {\left( {RQ} \right)^2} = 625$
$\Rightarrow RQ = 25$ cm
Area of shaded region = Area of circle – [Area of semicircle of side QR] – [Area of $\Delta PQR$] .......(1)
Area of $\Delta PQR$ = ${1 \over 2} \times PQ \times PR$
${1 \over 2} \times 7 \times 24 = 84$ $c{m^2}$ ............ (2)

Area of circle = $\pi {r^2}$
$= {{22} \over 7} \times {\left( {{{25} \over 2}} \right)^2}$
$= {{22 \times 625} \over {7 \times 2}} = {{625 \times 11} \over {14}}$ $c{m^2}$ ...... (3)
Area of semi circle on side QR = ${{\pi {r^2}} \over 2}$

$= {{22} \over 7}{\left( {{{25} \over 2}} \right)^2} \times {1 \over 2}$
$= {{22} \over 7} \times {{625} \over 4} \times {1 \over 2}$
$= {{625 \times 11} \over {28}}$ $c{m^2}$ .............. (4)
Put values from equation (2), (3) and (4) in equation (1) -
Area of shaded region = ${{625 \times 11} \over {14}} - {{625 \times 11} \over {28}} - 84$
$= 625 \times 11\left[ {{1 \over {14}} - {1 \over {28}}} \right] - 84$
$= {{625 \times 11} \over {28}} - 84$
= 161.53 $c{m^2}$

Q.5      Find the area of the segment of a circle of radius 14 cm, if the length of the corresponding arc APB is 22 cm. [use ${\pi = {{22} \over 7}}$]
Or
A square OABC is inscribed in a quadrant OPBQ of a circle as shown in fig. If OB = 14 cm, find the area of the shaded region. [use ${\pi = {{22} \over 7}}$]

[AI 2008 C]

Sol.

1st method : -
$l = {{\pi r\theta } \over {180^\circ }}$
$\Rightarrow 22 = {{22} \over 7} \times {{14 \times \theta } \over {180^\circ }}$
$\Rightarrow \theta = 90^\circ$
Area of segment of a circle = ${{\pi {r^2}\theta } \over {360^\circ }} - {1 \over 2}{r^2}\sin \theta$
$= {{22} \over 7} \times {{14 \times 14 \times 90} \over {360^\circ }} - {1 \over 2} \times 14 \times 14 \times \sin 90$
= 154 – 98
= 56 $c{m^2}$
2nd method : -
Area of sector = ${{lr} \over 2}$
$= {{22 \times 14} \over 2} = 154$ $c{m^2}$
Area of $\Delta AOB = {1 \over 2} \times OA \times OB$
$= {1 \over 2} \times 14 \times 14 = 98$ $c{m^2}$
Area of segment = 154 – 98 = 56 $c{m^2}$
OR

Area of quadrant $OQBP = {{\pi {r^2}} \over 4}$
$= {{\pi \times 14 \times 14} \over 4}$
$= {{22} \over 7} \times {{14 \times 14} \over 4}$
= 154 $c{m^2}$
Let side of a square be x cm
Now, in $\Delta OAB$
${\left( {OA} \right)^2} + {\left( {AB} \right)^2} = {\left( {OB} \right)^2}$
$\Rightarrow {x^2} + {x^2} = {\left( {14} \right)^2}$
$\Rightarrow 2{x^2} = 14 \times 14$
$\Rightarrow {x^2} = 7 \times 14$
$\Rightarrow {x^2} = 7 \times 7 \times 2$
$\Rightarrow x = 7\sqrt 2$ cm
Area of square = ${x^2}$ = 98 $c{m^2}$
Area of shaded region = Area of quadrant OQBP - Area of square
= (154 - 98)$c{m^2}$
= 56 $c{m^2}$

Q.6      The area of an equilateral triangle in $49\sqrt 3$ $c{m^2}$. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. [Take $\sqrt 3 = 1.73$] [AI 2009]
Sol.         Let side of an equilateral $\Delta$ be a

Area of an equilateral $\Delta = {{\sqrt 3 {a^2}} \over 4}$
$\Rightarrow$ $49\sqrt 3 = {{\sqrt 3 } \over 4}{a^2}$
$\Rightarrow {a^2} = 49 \times 4$
$\Rightarrow a = 7 \times 2 = 14$ cm
Radius of circle $= {a \over 2} = {{14} \over 2} = 7$ cm
Area of sector = ${{\pi r^2 \theta } \over {360^\circ}}$
$= {{\pi \times 7 \times 7 \times 60^\circ } \over {360^\circ }}$

$= {{22} \over 7} x {{7 \times 7 \times 60^\circ } \over {360^\circ }}$
$= {{77} \over 3}$ $c{m^2}$
Area of 3 sectors = $3 \times {{77} \over 3} = 77$ $c{m^2}$
Area of $\Delta$ not included in the circle = Area of $\Delta$ – Area of 3 secotrs
$= 49\sqrt 3 - 77$
$= 84.77 - 77$
$= 7.77$ $c{m^2}$

Q.7      In fig., the shape of the top of a table in restaurant is that of a sector of a circle with centre O and $\angle BOD = 90^\circ$, if BO = OD = 60 cm find :

(i) the area of the top of the table
(ii) the perimeter of the table top. [Take $\pi = 3.14$]

[Foreign 2009]

Sol.

Given : Radius(r) =OB = OD = 60 cm
(i) Area of top of the table = Area of sector

$= {{\pi {r^2}\theta } \over {360^\circ }}$
$= {{22} \over 7} \times {{{{\left( {60} \right)}^2} \times 270^\circ } \over {360^\circ }}$
$= {{22} \over 7} \times {{60 \times 60 \times 270^\circ } \over {360^\circ }}$
= 8485 $c{m^2}$
(ii) Perimeter of top of the table = ${{\pi {r^2}\theta } \over {180^\circ }} + 2r$
$= {{\pi \times 60 \times 270^\circ } \over {180^\circ }} + 2 \times 60$
$= {{22 \times 60 \times 270^\circ } \over {7 \times 180^\circ }} + 120$
=282.85 + 120
= 402.6 cm

Q.8      In figure the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. [use ${\pi = {{22} \over 7}}$]

Or
Find the area of the shaded region in figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle.

[Delhi 2010]

Sol.

Radius of largest semicircle = 7 cm
Radius of smallest semicircle = ${{3.5} \over 2}$cm
Radius of middle semicircle = ${{(14 - 7)} \over 2} = {7 \over 2}$ cm
= Area of largest semicircle – Area of 2 small semicircles + Area of middle semicircle ............ (1)

Area of largest semicircle = ${{\pi {r^2}} \over 2}$

$= {{22} \over 7} \times {{{{\left( 7 \right)}^2}} \over 2}$
$= {{22} \over 7} \times {{7 \times 7} \over 2}$
$= 77$ $c{m^2}$ .............. (2)
Area of smallest semicircle = ${{\pi {r^2}} \over 2} + {{\pi {r^2}} \over 2}$
$= \pi {r^2}$
$= {{22} \over 7} \times {\left( {{{3.5} \over 2}} \right)^2}$

$= {{22} \over 7} \times {{35 \times 35} \over {4 \times 10 \times 10}}$
$= {{22} \over 7} \times {{35 \times 35} \over {4 \times 100}}$
$= 9.625$ $c{m^2}$ ............ (3)
Area of middle semicircle = ${{\pi {r^2}} \over 2}$
$= {{22} \over 7} \times {1 \over 2} \times {\left( {3.5} \right)^2}$
$= {{22} \over 7} \times {1 \over 2} \times {{35} \over {10}} \times {{35} \over {10}} = 19.25$ $c{m^2}$ .......... (4)
Put values from equation (2), (3) and (4) in equation (1), we get
Area of shaded region = 77 – 9.625 + 19.25
= 86.625 $c{m^2}$
= 86.63 $c{m^2}$
OR
Given :  AC = 24 cm , BC = 10 cm
In $\Delta ABC$

${\left( {AC} \right)^2} + {\left( {BC} \right)^2} = {\left( {AB} \right)^2}$
$\Rightarrow {\left( {24} \right)^2} + {\left( {10} \right)^2} = {\left( {AB} \right)^2}$
$\Rightarrow {\left( {AB} \right)^2} = 576 + 100$
$\Rightarrow {\left( {AB} \right)^2} = 676$
$\Rightarrow AB = 26$ cm
Area of shaded region = Area of semicircle or side AB – Area of $\Delta ABC$
$= {{\pi {r^2}} \over 2} - {1 \over 2} \times AC \times BC$
$= {{22} \over 7} \times {{{{\left( {13} \right)}^2}} \over 2} - {1 \over 2} \times 24 \times 10$[Where $r = {{AB} \over 2}= {{26} \over 2} = 13\,$ cm]

= 265.33 – 120
= 145.33 $c{m^2}$

Q.9      Find the area of the shaded region in figure, where a circular arc of radius 7 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm, as centre. [Foreign 2010]
Sol.

Radius of circle = 7 cm
Area of shaded region = Area of circle – Area of sector + Area of an equilateral $\Delta$ ........... (1)

Area of circle = $\pi {r^2}$
$= {{22} \over 7} \times {\left( 7 \right)^2}$
$= {{22} \over 7} \times 7 \times 7$
$= 154$ $c{m^2}$ ............ (2)
Area of sector = ${{\pi {r^2}\theta } \over {360^\circ }}$
$= {{22} \over 7} \times {{{{\left( 7 \right)}^2} \times 60^\circ } \over {360^\circ }}$
$= {{22} \over 7} \times {{7 \times 7} \over 6}$
$= {{77} \over 3}$ $c{m^2}$ ........... (3)
Area of an equilateral $\Delta$ $= {{\sqrt 3 {a^2}} \over 4}$ where a is side of an equilateral $\Delta$ ABC   [Where a = 12 cm]
$= {{\sqrt 3 } \over 4} \times 12 \times 12$
$= 36\sqrt 3$
$= 68.28$ $c{m^2}$ ............ (4)
Put values from equation (2), (3) and (4) in equation (1), we get
Area of shaded region = $154 - {{77} \over 3} + 62.28$
$= 190.68$ $c{m^2}$

Q.10     Find the area of the major segment APB, in figure of a circle of radius 35 cm and $\angle AOB = 90^\circ$. [use ${\pi = {{22} \over 7}}$]

[Delhi 2011]

Sol.

Given :  Radius of circle (r) = 35 cm
Area of circle = $\pi {r^2}$

$= {{22} \over 7} \times {\left( {35} \right)^2}$
$= {{22} \over 7} \times 35 \times 35$
$=3850$ $c{m^2}$
Area of minor segement of circle = ${{\pi {r^2}\theta } \over {360^\circ }} - {1 \over 2}{r^2}\sin \theta$
$= {{{\rm{22}} \times {\rm{(35)}}^{\rm{2}}\times 90^\circ} \over {{\rm{7}} \times {\rm{360}}^ \circ}} - {1 \over2}(35)^2 \,Sin\,90^\circ$  [Since $\theta= 90^\circ$]
$= {{22} \over 7} \times {{35 \times 35 \times 1} \over 4} - {1 \over 2} \times 35 \times 35 \times 1$
$= 962.5 - 612.5 = 350$ $c{m^2}$
Area of major segmenet APB = Area of circle – Area of minor segment
$= 3850 - 350$
$= 3500$ $c{m^2}$

Q.11     In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and $\angle BOD = 90^\circ$. Find the area of the shaded region. [use $\pi = 3.14$]

[AI 2012]

Sol.

$\angle BOD = 90^\circ$ [Given]
$\angle CAB = 90^\circ$ [Angle is semicircle]
In $\Delta CAB$
${\left( {AC} \right)^2} + {\left( {AB} \right)^2} = {\left( {BC} \right)^2}$  [By Pythagoras Theorem]
$\Rightarrow {\left( {24} \right)^2} + {\left( 7 \right)^2} = {\left( {BC} \right)^2}$
$\Rightarrow {\left( {BC} \right)^2} = 576 + 49$
$\Rightarrow {\left( {BC} \right)^2} = 625$
$\Rightarrow BC = 25$ cm
Area of shaded region = Area of circle – Area of $\Delta CAB$ – Area of sector COD ................ (1)
Area of circle = $\pi {r^2}$
$= {{22} \over 7}{\left( {{{25} \over 2}} \right)^2}$   [ Since r= ${{BC} \over 2} = {{25} \over 2}$ cm]
$= {{6875} \over {14}}$ $c{m^2}$ .............. (2)
Area of $\Delta CAB = {1 \over 2} \times AC \times AB$

$= {1 \over 2} \times 24 \times 7 = 84$ $c{m^2}$ ............... (3)
Area of sector $COD = {{\pi {r^2}\theta } \over {360^\circ }}$
$= {{22} \over 7} \times {\left( {{{25} \over 2}} \right)^2} \times {{90^\circ } \over {360^\circ }}$

$= {{22} \over 7} \times {{625} \over 4} \times {1 \over 4}$
$= {{6875} \over {56}}$ $c{m^2}$ ................ (4)
Put values from equation (2), (3) and (4) in equation (1)
Area of shaded region $= {{6875} \over {14}} - 84 - {{6875} \over {56}}$
$= 283.968$ $c{m^2}$

Q.12      In the given figure, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region. [use ${\pi = {{22} \over 7}}$]

[Delhi 2013]

Sol.

Area of shaded region = Area of circle – Area of $\Delta ACD$ – Area of semicircle on side CD + Area of smaller circle ............... (1)
Area of circle = $\pi {r^2}$
$= {{22} \over 7}{\left( 7 \right)^2} = 154$ $c{m^2}$ ............ (2)
Area of smaller circle = $\pi {r^2}$
$= {{22} \over 7}{\left( {{7 \over 2}} \right)^2} = {{22} \over 7} \times {{7 \times 7} \over {2 \times 2}}$

$= {77 \over 2}$ $c{m^2}$ .............. (3)
Area of $\Delta CAD$ = Area of $\Delta AOC$ + Area of $\Delta AOD$
$= {1 \over 2} \times 7 \times 7 + {1 \over 2} \times 7 \times 7$
$= 2 \times {1 \over 2} \times 7 \times 7$
$= 49$ $c{m^2}$ .................. (4)
Area of semicircle on side CD = ${{\pi {r^2}} \over 2}$
$= {{22} \over 7} \times {{{{\left( 7 \right)}^2}} \over 2}$
$= {{22} \over 7} \times {{7 \times 7} \over 2}$
$= 77$ $c{m^2}$ ............ (5)
Put values from equation (2), (3), (4), (5) in equation (1), we get
Area of shaded region $= 154 - 49 - 77 + {{77} \over 2}$
$= 154 - 126 + {{77} \over 2}$
$= {{133} \over 2}$ $c{m^2}$
$= 66.5$ $c{m^2}$

Q.13      In the given figure, 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. Find the area of the shaded region. [use $\pi = 3.14$]

[Foreign 2013]

Sol.

Area of one quadrant $= {{\pi {r^2}} \over 4}$
$= {{\pi \times {{\left( 1 \right)}^2}} \over 4}$
$= {\pi \over 4}$ $c{m^2}$
Therefore area of 4 quadrants = $4 \times {\pi \over 4}$
$= \pi$ $c{m^2}$
$= 3.14$ $c{m^2}$
Area of circle = $\pi {r^2}$
$= \pi {\left( 1 \right)^2}$
$= \pi c{m^2}$
$= 3.14$ $c{m^2}$
Area of square  $=(Side)^2$
$=(4)^2$
= 4 ×4

= 16 $c{m^2}$
Area of shaded region = Area of square – Area of circle – Area of 4 quadrants
= (16 – 3.14 – 3.14 ) $c{m^2}$
= 16 - 6.28 $c{m^2}$
$= 9.72$ $c{m^2}$

Long answer type questions : -

Q.1      ABC is a right triangle, right angled at A. Find the area of shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of $\Delta ABC$. [Take $\pi = 3.14$]

[Delhi 2009]

Sol.

Given : AB = 6 cm , BC = 10 cm
Area of shaded region = Area of right angle $\Delta$ ABC – Area of circle ............. (1)

Area of right angle $\Delta$ ABC $= {1 \over 2} \times AB \times AC$
$= {1 \over 2} \times 6 \times 8$
$=24$ $c{m^2}$ ............. (2)
Therefore in $\Delta ABC$

${\left( {AB} \right)^2} + {\left( {AC} \right)^2} = {\left( {BC} \right)^2}$
$\Rightarrow {\left( 6 \right)^2} + {\left( {AC} \right)^2} = {\left( {10} \right)^2}$
$\Rightarrow 36 + {\left( {AC} \right)^2} = 100$
$\Rightarrow {\left( {AC} \right)^2} = 64$
$\Rightarrow AC = 8$ cm
$OQ \bot AB$ and $OR \bot AC$ [Because a tangent to the circle is perpendicular to the radius through the point of contact]
Therefore OQAR is a square
Let OR = x
Because OQAR is a square
Therefore OR = OQ = AQ = AB = x cm
BP = BQ [Because length of tangents drawn from a external point to a circle are equal]
and CR = CP  [Because length of tangents drawn from a external point to a circle are equal]
Therefore BP = BQ = 6 – x
and CP = CR = 8 – x
and BC = CP + BP
$\Rightarrow$ 10 = 8 – x + 6 – x
$\Rightarrow$ 10 = 14 – 2x
$\Rightarrow$ 2x = 4
$\Rightarrow$ x = 2 cm
Therefore radius of circle = 2 cm
Area of circle = $\pi {r^2}$
$= {{22} \over 7} \times {\left( 2 \right)^2} = 3.14 \times 4 = 12.56$ $c{m^2}$ ............ (3)
Put values from equation (2) and (3) in equation (1), we get
Area of shaded region = 24 – 12.56 = 11.44 $c{m^2}$

Q.2      In the given figure, 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 of shaded region enclosed between these three circles. [use ${\pi = {{22} \over 7}}$]

[AI 2011]

Sol.

As radius of each circle is 3.5 cm
Therefore AC = 3.5 + 3.5 = 7 cm
BC = 3.5 + 3.5 = 7 cm
AB = 3.5 + 3.5 = 7 cm
$\Rightarrow$ ABC is an equilateral $\Delta$ with side 7 cm.
Area of an equilateral $\Delta$ $= {{\sqrt 3 } \over 4}{\left( 7 \right)^2}$
$= {{\sqrt 3 } \over 4} \times 7 \times 7$ $c{m^2}$
$= {{49\sqrt 3 }\over 4}$
Area of one sector = ${{\pi {r^2}} \over {360^\circ }} \times \theta$
$= {{\pi {r^2}} \over {360^\circ }} \times 60^\circ$ [$\theta$ ia an angle of equilateral $\Delta$]

$= {{22} \over 7} \times {\left( {3.5} \right)^2} \times {{60^\circ } \over {360^\circ }}$
$= {{22} \over 7} \times {{35} \over {10}} \times {{35} \over {10}} \times {1 \over 6}$
$= {{77} \over {12}}$ $c{m^2}$
Therefore area of 3 sectors = $3 \times {{77} \over {12}} = {{77} \over 4}$ $c{m^2}$
Area of shaded region = Area of an equilateral – Area of 3 sectors
$= {{49\sqrt 3 } \over 4} - {{77} \over 4}$
$= {1 \over 4}\left[ {49\sqrt 3 - 77} \right]$
$= {{84.77 - 77} \over 4}$
$= {{7.77} \over 4} = 1.9425$ $c{m^2}$

Q.3     In given figure, an equilateral triangle has been inscribed in a circle of radius 6 cm. Find the area of the shaded region. [Use $\pi = 3.14$]

[AI 2011]

Sol.          Draw $OM \bot BC$

As ABC is an equilateral $\Delta$
Therefore $\angle BOC = 2\angle BAC$ [Because angle subitendend at centre is double the angle at alternate segment]
$\Rightarrow \angle BOC = 2 \times 60^\circ$
$\Rightarrow \angle BOC = 120^\circ$
Now, in $\Delta BOC$
OB = OC [Radii of circle]
$\Rightarrow$ $\angle OBC = \angle OCB$ [Because angle opposite to equal sides are equal]
Now in $\Delta BOC$
$\angle BOC + \angle OBC + \angle OCB = 180^\circ$
$\Rightarrow 120^\circ + \angle OBC + \angle OBC = 180^\circ$
$\Rightarrow 2\angle OBC = 60^\circ$
$\Rightarrow \angle OBC = 30^\circ$
Now, in $\Delta BOM$
cos30°=${{BM} \over {OB}}$
$\Rightarrow {{\sqrt 3 } \over 2} = {{BM} \over 6}$ [ OB is radius of circle]
$\Rightarrow BM = 3\sqrt 3$ cm
As $OM \bot BC$
Therefore BM = CM [Because perpendicular from the centre to the chord bisects the chord]
$\Rightarrow BC = 2BM$
$\Rightarrow BC = 2 \times 3\sqrt 3$
$\Rightarrow BC = 6\sqrt 3$ cm
Area of an equilateral $\Delta$ = ${{\sqrt 3 } \over 4}{\left( {6\sqrt 3 } \right)^2}$
$= {{\sqrt 3 } \over 4} \times 36 \times 3$
$= 27\sqrt 3$ $c{m^2}$
$= 46.71$ $c{m^2}$
Area of circle = $\pi {r^2}$
$= \pi {\left( 6 \right)^2}$
$= {{22} \over 7} \times 36$
$= 3.14 \times 36$
$= 113.04$ $c{m^2}$
Area of shaded region = Area of circle – Area of an equilateral $\Delta$
$= 113.04 - 46.71$
$= 66.33$ $c{m^2}$

Q.4      From a thin metallic piece, in the shape of a trapezium ABCD in which AB || CD and $\angle BCD = 90^\circ$, a quarter circle BFEC is removed (See figure). Given AB = BC = 3.5cm and DE = 2 cm, calculate the area of the remaining (shaded) part of the metal sheet. [use ${\pi = {{22} \over 7}}$]

[Foreign 2011]

Sol.

AB = BC = 3.5 cm [Given]
AE = BC = 3.5 cm
Area of shaded region = Area of trapezium – Area of quarter circle
$= {1 \over 2}\left( {AB + CD} \right) \times AE - {{\pi {r^2}} \over 4}$
$= {1 \over 2}\left( {3.5 + 5.5} \right) \times 3.5 - {{22} \over 7} \times {\left( {{{\left( {3.5}^2 \right)} \over 4}} \right)^2}$