# Areas Related to Circles

## Class 10 NCERT

### NCERT

1   The radii of two circles are $19 cm$ and $9 cm$ respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

##### Solution :

Radius $\left(r_1\right)$ of 1st circle = $19 cm$$\\ Radius \left(r_2\right) or 2nd circle = 9 cm$$\\$ Let the radius of 3rd circle be $r$$\\ Circumference of 1st circle = 2 \pi r_1 = 2\pi \left(19\right) = 38\pi$$\\$ Circumference of 2nd circle = $2 \pi r_2 = 2\pi \left(9\right) = 18\pi$ $\\$ Circumference of 3rd circle = $2\pi r$ $\\$ Given that, $\\$ Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle$\\$ $2 \pi r = 38 \pi + 18 \pi = 56 \pi \\ r= \dfrac{56\pi}{2\pi} = 28$ $\\$ Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is $28 cm.$

2   The radii of two circles are $8 cm$ and $6 cm$ respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

##### Solution :

Radius of $\left( r_1\right)$ 1st circle = $8 cm $$\\ Radius of \left(r_2 \right) 2nd circle = 6 cm$$\\$ Let the radius of 3rd circle be $r$$\\ Area of 1st circle = \pi r_1^2 = \pi \left(8^2\right) = 64\pi$$\\$ Area of 2nd circle = $\pi r_2^2 = \pi \left(6^2\right) = 36\pi$ $\\$ Given that,$\\$ Area of 3rd circle = Area of 1st circle + Area of 2nd circle $\\$ $\pi r^2 = \pi r_1^2 + \pi r_2^2 \\ \pi r^2 = 64\pi + 36\pi \\ \pi r^2 = 100\pi \\ r^2 =100 \\ r= \pm 10$ $\\$ However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is $10 cm.$

3   Given figure depicts an archery target marked with its five scoring areas from the center outwards as $Gold, Red, Blue, Black$ and $White.$ The diameter of the region representing Gold score is $21 cm$ and each of the other bands is $10.5 cm$ wide. Find the area of each of the five scoring regions.$\left[Use \pi = \dfrac {22}{7}\right]$

##### Solution :

Radius$\left( r_1\right)$ of gold region (i.e., 1st circle) =$21^2 = 10.5 cm$ $\\$ Given that each circle is $10.5 cm$ wider than the previous circle.$\\$ Therefore, radius$\left( r_2\right)$of 2nd circle =$10.5 + 10.5 = 21 cm$$\\ Radius \left( r_3\right) of 3rd circle = 21 + 10.5 = 31.5 cm$$\\$ Radius $\left( r_4\right)$ of 4th circle = $31.5 + 10.5 = 42 cm $$\\ Radius \left( r_5\right) of 5th circle = 42 + 10.5 = 52.5 cm$$\\$ Area of gold region = Area of 1st circle = $\pi r_1^2 = \pi \left(10.5\right)^2 = 346.5 cm^2$ $\\$ Area of red region = Area of $2^{nd}$ circle - Area of $1^{st }$circle $\\$ $=\pi r_2^2 - \pi r_1^2 \\ =\pi \left(21\right)^2 - \left(10.5\right)^2 \\ =441\pi - 110.25\pi = 330.75\pi \\ = 1039.5 cm^2$$\\ Area of blue region = Area of 3^{rd} circle - Area of 2^{nd} circle \\ = \pi r_3^2 - \pi r_1^2 \\ = \pi \left(31.5\right)^2 - \pi \left(21 \right)^2 \\ = 992.25\pi - 441\pi = 551.25\pi \\ = 1732.5 cm^2$$\\$ Area of black region = Area of $4^{th}$ circle -Area of $3^{rd}$ circle $\\$ $\pi r_4^2 - \pi r_3^2 \\ \pi \left(42\right)^2 -\pi \left(31.5\right)^2 \\ = 1764\pi - 992.25\pi \\ = 771.75\pi = 2425.5 cm^2 \\$$\\ Area of white region = Area of 5^{th} circle -Area of 4^{th} circle \\ = \pi r_5^2 - \pi r_4^2 \\ \pi \left(52.5\right )^2 - \pi \left(42\right)^2 \\ = 2756.25\pi - 1764\pi \\ =992.25\pi = 3118.5 cm^2$$\\$ Therefore, areas of gold, red, blue, black, and white regions are $346.5 cm^2, 1039.5 cm^2 , 1732.5 cm^2 , 2425.5 cm^2$ and $3118.5 cm^2$ respectively.

4   The wheels of a car are of diameter $80 cm$ each. How many complete revolutions does each wheel make in $10$ minutes when the car is traveling at a speed of $66 km$per hour?

##### Solution :

Distance travelled by the car in $10 minutes$$\\ = 110000 × 10 = 1100000 cm$$\\$ Let the number of revolutions of the wheel of the car be $n$.$\\$ $n$ × Distance travelled in $1$ revolution (i.e., circumference) $\\$ = Distance travelled in $10 minutes $$\\ n * 80\pi = 1100000 \\ n= \dfrac{1100000 *7}{80*22} = 4375$$\\$ Therefore, each wheel of the car will make $4375$ revolutions.

Diameter of the wheel of the car =$80 cm $$\\ Radius \left(r\right) of the wheel of the car = 40 cm$$\\$ Circumference of wheel = $2\pi r$ $\\$ $= 2\pi \left(40\right) = 80\pi cm $$\\ Speed of car = 66 km /hour$$\\$ $=\dfrac{66*100000}{60} cm /min \\ = 110000 cm/min $$\\ Distance travelled by the car in 10 minutes$$\\$ $= 110000 × 10 = 1100000 cm $$\\ Let the number of revolutions of the wheel of the car be n.\\ n × Distance travelled in 1 revolution (i.e., circumference) \\ = Distance travelled in 10 minutes$$\\$ $n * 80\pi = 1100000 \\ n= \dfrac{1100000 *7}{80*22} = 4375$$\\ Therefore, each wheel of the car will make 4375 revolutions. 5 Find the area of a quadrant of a circle whose circumference is 22 cm. [Use \pi = \dfrac{22}{7}] ##### Solution : Let the radius of the circle be r.\\ Circumference = 22 cm$$\\$ $2\pi r = 22 \\ r = \dfrac{22}{2\pi} = \dfrac{11}{\pi}$ $\\$ Quadrant of circle will subtend $90^o$ angle at the center of the circle.$\\$ Area of such quadrant of the circle $\\$ $= \dfrac{90^o}{360^o}*\pi*r^2 \\ \dfrac{1}{4}*\pi *\left(\dfrac{11}{\pi}\right)^2 \\ = \dfrac{121}{4\pi} \\ = \dfrac{121*7}{4*22}\\ = \dfrac{77}{8} cm^2$

6   The length of the minute hand of a clock is $14 cm$. Find the area swept by the minute hand in $5$ minutes. [Use $\pi = \dfrac{22}{7}$]

##### Solution :

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates $360^o$.$\\$ In $5$ minutes, minute hand will rotate = $\dfrac{360^o}{60^o} * 5 = 30^o$ $\\$ Therefore, the area swept by the minute hand in $5$ minutes will be the area of a sector of $30$ in a circle of $14 cm$ radius.

Area of sector of angle $\theta = \dfrac{\theta}{360^o}* \pi r^2$ $\\$ Area of sector of $30^o = \dfrac{30^o}{360^o}*\dfrac{27}{7}*14*14$ $\\$ $= \dfrac{22}{12}*2814 \\ \dfrac{11*14}{3}\\ \dfrac{154}{3}cm^2$ $\\$ Therefore, the area swept by the minute hand in $5$ minutes is $\dfrac{154}{3} cm^2$

7   A chord of a circle of radius $10 cm$ subtends a right angle at the center. Find the area of the corresponding:

Area of $\Delta OAB =\dfrac{\sqrt{3}}{4}*\left(side^2\right)^2$ $\\$ $=\dfrac{\sqrt{3}}{4}*\left(21\right)^2 = \dfrac{441\sqrt{3}}{4} cm^2 $$\\ Area of segment ACB = Area of sector OACB - Area of \Delta OAB \\ = \left(231 - \dfrac{441\sqrt{3}}{4}\right) cm^2 9 A chord of a circle of radius 15 cm subtends an angle of 60^o at the center. Find the areas of the corresponding minor and major segments of the circle. [Use \pi= 3.14 and \sqrt{ 3} = 1.73 ] ##### Solution : Radius \left(r\right) of circle = 15 cm$$\\$ Area of sector $OPRQ = \dfrac{60^o}{360^o} *\pi r^2 $$\\ = \dfrac{1}{6}*3.14\left(15\right)^2 \\ = 117.75 cm^2 \\ In \Delta OPQ, \\ \angle OPQ =\angle OQP(As radius OP=OQ)\\ \angle OPQ +\angle OQP +\angle POQ = 180^o \\ 2\angle OPQ = 120^o \\ \angle OPQ = 60^o$$\\$ $\Delta OPQ,$ is an equilateral triangle. $\\$ Area of $\Delta OPQ = \dfrac{\sqrt{3}}{4}*\left(side\right)^2 $$\\ \dfrac{\sqrt{3}}{4}*\left(15\right)^2 = \dfrac{225\sqrt{3}}{4} cm^2 \\ 97.3125 cm^2$$\\$ Area of minor segment $PRQP$ =Area of sector $OPRQ$ -Area of $\Delta OPQ$ =$117.75-97.3125 = 20.4375 cm^2$ $\\$ Area of major segment $PSQP$ = Area of circle-Area of minor segment $PRQP$$\\ \pi \left(15\right)^2 - 20.4375 \\ = 3.14 *225 - 20.4375 \\ = 686.0625 cm^2 10 Tick the correct answer in the following and justify your choice\colon If the perimeter and the area of a circle are numerically equal, then the radius of the circle is ##### Solution : Let the radius of the circle be r.$$\\$ Circumference of circle =$2\pi r$$\\ Area of circle = \pi r^2 \\ Given that, the circumference of the circle and the area of the circle are equal. \\ This implies 2\pi r = \pi r^2$$\\$$2 = r$$\\$ Therefore, the radius of the circle is $2 units$. Hence, the correct answer is $A$.

11   Find the area of a sector of a circle with radius $6 cm$ if angle of the sector is $60^o$ [use $\pi = \dfrac{22}{7}$]

##### Solution :

Let $OACB$ be a sector of the circle making $60^o$ angle at centre $O$ of the circle.$\\$ Area of sector of angle $\theta = 360\theta × \pi r^2$

Area of sector $OACB =\dfrac{60^o}{360^o}*\dfrac{22}{7}*\left(6^2\right)$$\\ = \dfrac{1}{6}*\dfrac{22}{7}*6*6 = \dfrac{132}{7} cm^2 \\ Therefore, the area of the sector of the circle making 60^o at the centre of the circle is \dfrac{132}{7} cm^2 12 A chord of a circle of radius 12 cm subtends an angle of 120^o at the center. Find the area of the corresponding segment of the circle. [Use \pi = 3.14 and \sqrt{3} = 1.73 ] ##### Solution : Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.\\$$ SV = VT \\ In \Delta OVS, OV OS = \cos60^o \\ \dfrac{OV}{ 12} =\dfrac{1}{ 2}\\ OV = 6cm$ $\\$ $\dfrac{SV}{SO} = \sin60^o = \dfrac{\sqrt{3}{2}} \\ \dfrac{SV}{12} = \dfrac{\sqrt{3}}{2}\\ =ST = 2SV = 2*6\sqrt{3} = 12\sqrt{3} cm$ $\\$ Area of $\Delta OST = \dfrac{1}{2}× ST × OV = \dfrac{1}{ 2}× 12\sqrt{3} × 6 = 36 \sqrt{3} = 36 × 1.73 = 62.28 cm^2 $$\\ Area of sector OSUT = \dfrac{120}{ 360} × \pi \left( 12 \right)^2 = \dfrac{1}{2}* 3.14 *144 = 150.72 cm^2 \\ Area of segment SUTS = Area of sector OSUT - Area of \Delta OST = 150.72 - 62.28 = 88.44 cm^2$$\\$ Area of segment $SUTS$ = Area of sector $OSUT$ - Area of $\Delta OST = 150.72 - 62.28 = 88.44 cm^2$

13   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 the given 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 $\pi = 3.14$]

##### Solution :

From the figure, it can be observed that the horse can graze a sector of $90^o$ in a circle of $5 m$ radius. $\\$ Area that can be grazed by horse = Area of sector $OACB$$\\ = \dfrac{90^o}{360^o}\pi r^2 \\ =\dfrac{1}{4}*3.14*\left(5\right)^2\\ = 19.625 m^2$$\\$ Area that can be grazed by the horse when length of rope is $10 m$ long $= \dfrac{90^o}{360^o}* \pi * \left( 10 \right)^2 \\ = \dfrac{1}{ 4}* 3.14 * 100 = 78.5 m^2$ Increase in grazing area = $\left(78.5 -19.625\right) m^2 = 58.875 m^2$

14   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 the figure. Find. $\\$ (i) The total length of the silver wire required. $\\$ (ii) The area of each sector of the brooch $\left [use \quad \pi = \frac{22}{7} \right]$ $\\$

##### Solution :

Total length of wire required will be the length of $5$ diameters and the circumference of the brooch.$\\$ Radius of circle = $\dfrac{35}{2} mm$$\\ Circumference of brooch = 2 \pi r$$\\$ $=2*\dfrac{22}{7}*\left(\dfrac{35}{2}\right)\\ 110 mm $$\\ Length of wire required=110+5*35 =285 mm$$\\$ It can be observed from the figure that each of 10 sectors of the circle is subtending $36^o$ (i.e. $\dfrac{360^o}{10}=36^o$) at the center of the circle.$\\$

Therefore, area of each sector $= \frac{36^o}{360^o}*\pi r^2$ $\\$ $= \dfrac{1}{10}*\dfrac{22}{7}*\left(\dfrac{35}{2}\right)*\left(\dfrac{35}{2}\right) \\ =\dfrac{385}{4} mm^2$

15   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.

##### Solution :

There are$8$ ribs in an umbrella. The area between two consecutive ribs is subtending=$\dfrac{ 360^o}{8} = 45$at the centre of the assumed flat circle.$\\$

We know that area of sector of angle $\theta =\dfrac{\theta}{360^o}*\pi R^2$$\\ Area of sector of angle P =\dfrac{p}{360^o}\left(\pi R^2\right)$$\\$ $=\left(\dfrac{p}{720^o}\right)\left(2 \pi R^2\right)$$\\ Hence, (D) is the correct answer. 20 Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the center of the circle. [Use \pi =\dfrac{22}{7}] ##### Solution : 21 Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the center of the circle. [Use \pi =\dfrac{22}{7}] ##### Solution : It can be observed that RQ is the diameter of the circle.\\ Therefore, \angle RPQ will be 90^o$$\\$ By applying Pythagoras theorem in $\Delta PQR,$$\\ RP^2 + PQ^2 = RQ^2 \\ \left(7\right)^2 + \left(24\right)^2 = RQ^2 \\ RQ = \sqrt{625} = 25$$\\$ Radius of circle, $OR = \dfrac{RQ}{ 2} = \dfrac{25}{ 2} $$\\ Since RQ is the diameter of the circle, it divides the circle in two equal parts.\\ Area of semicircle RPQOR =\dfrac{1}{2}\pi r^2$$\\$ $= \dfrac{1}{2}\pi \left(\dfrac{25}{2}\right)^2 \\ =\dfrac{1}{2}*\dfrac{22}{7}*\dfrac{625}{4}\\ =\dfrac{6875}{28} cm^2 $$\\ Area of \Delta PQR = \dfrac{1}{ 2} * PQ * PR$$\\$$= \dfrac{1}{2}* 24 * 7\\ = 84 cm^2$$\\$ Area of shaded region = Area of semi -circle $RPQOR$ -Area of $\Delta PQR $$\\ \dfrac{6875}{28}-84 \\ =\dfrac{6874-2352}{28} \\ = \dfrac{4523}{28} cm^2 22 Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and \angle AOC = 40^o.[Use \pi =\dfrac{22}{7}] ##### Solution : Radius of inner circle = 7 cm Radius of outer circle = 14 cm Area of shaded region = Area of sector OAFC - Area of sector OBED$$\\$ $= \dfrac{40^o}{360^o}*\pi \left(14\right)^2 = \dfrac{40^o}{360^o}*\pi \left(7\right)^2 \\ = \dfrac{1}{9} *\dfrac{22}{7}*14814 -\dfrac{1}{9}*\dfrac{22}{7}*7*7 \\ = \dfrac{616}{9}-\dfrac{154}{9} \\= \dfrac{154}{3} cm^2$

23   Find the area of the shaded region in the given figure, if $ABCD$ is a square of side $14 cm$ and $APD$ and $BPC$ are semicircles.[Use $\pi =\dfrac{22}{7}$]

It can be observed from the figure that the radius of each semi-circle is $7 cm.$$\\ Area of each semi-circle =\dfrac {1}{2}\pi r^2$$\\$ $= \dfrac{1}{2}*\dfrac{22}{7}*7*7 \\ = 77 cm^2 $$\\ Area of square ABCD = (Side)^2 = \left(14\right)^2 = 196 cm^2$$\\$ Area of the shaded region = Area of square $ABCD$- Area of semi-circle $APD$-Area of semi-circle$BPC$$\\ = 196 -77 - 77 = 196 - 154 = 42 cm^2 24 Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as a center. \left [ Use \ \pi = \dfrac{22}{7} \right ] \\ ##### Solution : We know that each interior angle of an equilateral triangle is of measure60^o. \\ Area of sector OCDE = \frac{60^o}{360^o} \pi r^2 \\ = \frac{1}{6}*\frac{22}{7}*6*6 \\ = \frac{132}{7} cm^2 \\ Area of \Delta{OAB} = \frac{\sqrt{3}}{4}(12)^2 = \frac{\sqrt{3}*12*12}{4} = 36\sqrt{3} cm^2 \\ Area of circle = \pi r^2 = \frac{22}{7} * 6*6 = \frac{792}{7} cm^2 \\ Area of shaded region = Area of \Delta{OAB} + Area of circle - Area of sector OCDE \\ = 36 \sqrt{3} + \frac{792}{7} - \frac{132}{7} \\ = \left( 3 \sqrt{3} + \frac{660}{7} \right) cm^2 \\ 25 From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining portion of the square. [use \pi = \dfrac{22}{7}] ##### Solution : Each quadrant is a sector of 90^o in a circle of 1 cm radius.\\ Area of each quadrant =\dfrac{90^o}{360^o}\pi r^2$$\\$ $\\ \dfrac{1}{4}*\dfrac{22}{7}*\left(1\right)^2 = \dfrac{22}{28} cm^2$$\\ Area of square = (Side)^2 = \left(4\right)^2 = 16 cm 2\\ Area of circle = \pi r^2 = \pi \left(1\right)^2$$\\$ $=\dfrac{22}{7} cm^2$ $\\$ Area of the shaded region = Area of square -Area of circle - $4$ * Area of quadrant$\\$ $= 16-\dfrac{22}{7}-4*\dfrac{22}{28}\\ = 16-\dfrac{22}{7}-\dfrac{22}{7}\\ =\dfrac{112-44}{7} \\ =\dfrac{68}{7} cm^2$

26   In a circular table cover of radius $32 cm$, a design is formed leaving an equilateral triangle $ABC$ in the middle as shown in the given figure. Find the area of the design (Shaded region). [use $\pi =\dfrac{22}{7}$]

Radius $(r)$ of circle = $32 cm$$\\ AD is the median of ABC.\\ OA = \dfrac{2}{3} AD\\ AD = 48 cm\\ In \Delta ABD,\\ AB^2 = AD^2 + BD^2\\ AB62 =\left(48\right)^2 + \left(\dfrac{48}{2}\right)^2 \\ \dfrac{3AB^2}{4} =(48)^2 \\ AB= \dfrac{48*2}{\sqrt{3}} = 32\sqrt{3} cm \\ Area of equilateral triangle,\\ = \dfrac{\sqrt{3}}{4}*32*32*3 =96 *8*\sqrt{3} \\ 768\sqrt{3}$$\\$ Area of circle = $\pi r^2$$\\ =\dfrac{22}{7}*(32)^2\\ =\dfrac{22528}{7} cm^2$$\\$ Area of design = Area of circle - Area of $\Delta ABC$$\\ = \left(\dfrac{22528}{7}-768\sqrt{3}\right) cm^2 27 In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.[Use \pi= \dfrac{22}{7}] ##### Solution : Area of each of the 4 sectors is equal to each other and is a sector of 90^o in a circle of 7 cm radius.\\ Area of each sector = \dfrac{90^o}{360^o}\pi (7)^2$$\\$ $= \dfrac{1}{4}*\dfrac{22}{7}*7*7\\ =\dfrac{77}{2} cm^2$ $\\$ Area of square $ABCD = (Side)^2 = (14)^2 = 196 cm^2$$\\ Area of shaded portion = Area of square ABCD -4 * Area of each sector\\ =196 -4* \dfrac{77}{ 2}= 196 -154=42 cm^2$$\\$

28   The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is $60 m$ and they are each $106 m$ long. If the track is $10 m$ wide, find $\colon$

##### Solution :

Area of the track = (Area of $GHIJ$ - Area of $ABCD$) + (Area of semi-circle $HKI$ - Area of semi-circle $BEC$) + (Area of semi-circle $GLJ$ - Area of semi- circle $AFD$)$\\$ $= 106*80-106*60+\dfrac{1}{2}*\dfrac{22}{7}*(40)^2-\dfrac{1}{2}*\dfrac{22}{7}*(30)^2+\dfrac{1}{2}*\dfrac{22}{7}*(40)^2 - \dfrac{1}{2}*\dfrac{22}{7}*(30)^2\\ =106(80-60)+ \dfrac{22}{7}*(40)^2-\dfrac{22}{7}*(30)^2\\ =2120+\dfrac{22}{7}(40-30)(40+30)\\ =2120+\left(\dfrac{22}{7}(10)(70)\right) \\ =4320 m^2 $$\\ Therefore, the area of the track is 4320 m^2 . Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA \\ = 106 +\dfrac{1}{2 } * 2 \pi r + 106 +\dfrac{1}{2} * 2 \pi r \\ = 212 + \dfrac{1}{2} * 2 * \dfrac{22}{7} * 30 *\dfrac{1}{2} * 2 * \dfrac{22}{7} * 30 \\ = 212 + 2 * \dfrac{22}{7 } * 30 \\ = \dfrac{1484+1320}{7}\\ =\dfrac{2804}{7}m \\ Area of the track = (Area of GHIJ - Area of ABCD) + (Area of semi-circle HKI - Area of semi-circle BEC) + (Area of semi-circle GLJ - Area of semi- circle AFD)\\ = 106*80-106*60+\dfrac{1}{2}*\dfrac{22}{7}*(40)^2-\dfrac{1}{2}*\dfrac{22}{7}*(30)^2+\dfrac{1}{2}*\dfrac{22}{7}*(40)^2 - \dfrac{1}{2}*\dfrac{22}{7}*(30)^2\\ =106(80-60)+ \dfrac{22}{7}*(40)^2-\dfrac{22}{7}*(30)^2\\ =2120+\dfrac{22}{7}(40-30)(40+30)\\ =2120+\left(\dfrac{22}{7}(10)(70)\right) \\ =4320 m^2$$\\$ Therefore, the area of the track is $4320 m^2$ .

29   In the given 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

##### Solution :

Radius ($r^1$) of larger circle = $7 cm$$\\ Radius (r^2 ) of smaller circle =\dfrac{7}{2} cm Area of smaller circle = \pi r_1^2$$\\$ $= \dfrac{22}{7}*\dfrac{7}{2}*\dfrac{7}{2}\\ \dfrac{77}{2} cm^2 $$\\ Area of semi-circle AECFB of larger circle= \dfrac{1}{2}\pi r+1^2 \\ =\dfrac{1}{2}*\dfrac{22}{7}*(7)^2\\ = \dfrac{77}{2} cm^2 \\ Area of \Delta ABC = \dfrac{1}{2}*AB*OC$$\\$ $= \dfrac{1}{2}*14*7=49 cm^2 $$\\ Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB - Area of \Delta ABC$$\\$ $= \dfrac{77}{2} + 77 -49 \\ = 28+38.5 =66.5 cm^2$

30   The area of an equilateral triangle $ABC$ is $17320.5 cm^2$ . 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 the given figure). Find the area of shaded region. [Use $\pi = 3.14$ and $3 = 1.73205$ ]

Let the side of the equilateral triangle be a. Area of equilateral triangle = $17320.5 cm^2$$\\ \dfrac{\sqrt{3}}{4}(a)^2 = 17320.5 \\= \dfrac{1.73205}{4}a^2 = 17320.5 \\ =a^2 = 4*10000\\ = a=200 cm$$\\$ Each sector is of measure $60^o$.$\\$ Area of sector $ADEF=\dfrac{60^o}{360^o}*\pi*r^2 \\ =\dfrac{1}{6}*\pi *(100)^2 \\ =\dfrac{15700}{3}cm^2$$\\ Area of shaded region = Area of equilateral triangle -3 * Area of each sector\\ = 17320.5 - 3* \dfrac{15700}{3}\\ =1620.5 cm^2 31 On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief.[Use \pi = \dfrac{22}{7}] ##### Solution : From the figure, it can be observed that the side of the square is 42 cm. Area of square = (Side)^2 = (42)^2 = 1764 cm$$\\$ Area of square = $(Side)^2 = (42)^2 = 1764 cm^2 $$\\ Area of each circle = \pi r^2 = \dfrac{22}{7} *(7)^2 = 154 cm^2 \\ Area of 9 circles = 9 × 154 = 1386 cm 2$$\\$ Area of the remaining portion of the handkerchief = $1764 - 1386 = 378 cm^2$

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

(i) Since $OACB$ is a quadrant, it will subtend $90^o$ angle at $O$.$\\$ Area of quadrant $OACB = \dfrac{90^o}{360^o}*\pi r^2 $$\\ =\dfrac{1}{4}*\dfrac{22}{7}*(3.5)^2 = \dfrac{1}{4} *\dfrac{22}{7}*\left(\dfrac{7}{2}\right)^2 \\ =\dfrac{77}{8} cm^2$$\\$ (ii) Area of $\Delta OBD = \dfrac{1}{2}*OB *OD $$\\ = \dfrac{1}{2}* \dfrac{7}{2}*2\\ =\dfrac{7}{2 } cm^2 \\ Area of the shaded region = Area of quadrant OACB -Area of GOBD \\ = \dfrac{77}{8}- \dfrac{7}{2} = \dfrac{49}{8 } cm^2 33 In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [Use \pi = 3.14] ##### Solution : In \Delta OAB, \\ OB^2 = OA^2 + AB^2 \\ = (20)^2 +(20)^2\\ OB=20\sqrt{2} \\ Radius (r) of circle 20\sqrt{2} cm$$\\$ Area of quadrant $OPBQ =\dfrac{90^o}{360^o}*3.14*\left(20\sqrt{2}\right)^2 $$\\ =\dfrac{1}{4}*3.14*800\\ 628 cm^2$$\\$ Area of $OABC = (Side)^2 = (20)^2 = 400 cm^2$$\\ Area of shaded region = Area of quadrant OPBQ - Area of OABC$$\\$ $= (628 - 400) cm^2 = 228 cm^2$

34   $AB$ and $CD$ are respectively arcs of two concentric circles of radii$21 cm$ and $7 cm$ and centre $O$ (see the given figure). If $\Delta AOB = 30^o$, find the area of the shaded region.[Use $\pi = \dfrac{22}{7}$]

Area of the shaded region = Area of sector $OAEB$- Area of sector $OCFD$$\\ =\dfrac{30^o}{360^o}*(21)^2 - \dfrac{30^o}{360^o} *\pi *(7)^2 \\ = \dfrac{1}{12}\pi\left[\left(21-7\right)\left(21+7\right)\right]\\ =\dfrac{22*14*28}{7*12} \\ =\dfrac{308}{3} cm^2 35 In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. [Use \pi=\dfrac{22}{7}] ##### Solution : As ABC is a quadrant of the circle, \angle BAC will be of measure 90^o. In \Delta ABC , \\ BC^2 = AC^2 + AB^2 \\ =(14)^2+(14)^2 \\ BC= 14\sqrt{2}$$\\$ Radius ($r_1$ ) of semi-circle drawn on $BC = \dfrac{14\sqrt{2}}{2} = 7\sqrt{2} cm $$\\ Area of sector ABCD = \dfrac{90^o}{360^o} * \pi r^2$$\\$ $= \dfrac{1}{4}* \dfrac{22}{7} *14*14 \\ = 154 cm^2$ Area of semi-circle drawn on $BC =\dfrac{1}{2}*\pi r_1^2 =\dfrac{1}{2}*\dfrac{22}{7}* \left(7\sqrt{2}\right)^2 \\ = 154 cm^2$ $\\$ Area of shaded region = Area of semi-circle on $BC$ - (Area of sector$ABDC$ - Area of $\Delta ABC$)$\\$ $= 154 - (154 - 98) = 98 cm^2$

36   Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius $8 cm$ each. [Use $\pi = \dfrac{22}{7}$]

##### Solution :

The designed area is the common region between two sectors $BAEC$ and $DAFC$.$\\$ Area of sector $BAEC =\dfrac{90^o}{360^o} *\dfrac{22}{7} *(8)^2 $$\\ =\dfrac{1}{4} *\dfrac{22}{7}* 64 \\ \dfrac{352 }{7}cm^2$$\\$ Area of $\Delta BAC = \dfrac{1}{2}*Ba*BC $$\\ = \dfrac{1}{2} * 8*8 =32 cm^2$$\\$ Area of the designed portion = $2$* (Area of segment $AEC$) = $2$* (Area of sector$BAEC$- Area of $\Delta BAC$)$\\$ $= 2 * \left(\dfrac{352}{7}-32\right ) =\dfrac{2*128}{7}\\ =\dfrac{256}{7} cm^2$