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:

Solution :

Let AB be the chord of the circle subtending $90^o$ angle at center $O$ of the circle.$\\$ Area of minor sector $OACB = \left(\dfrac{360^o - 90^o}{360^o}\right) * \pi r^2 = \left(\dfrac{270^o}{360^o}\right)\pi r^2 $$\\$ $= \dfrac{3}{4}*3.14810*10 \\ 235.5 cm^2 $ $\\$ Area of minor segment $ACBA = \dfrac{90^o}{360^o}*\pi r^2$$\\$ $=\dfrac{1}{4}*3.14*10*10\\ =78.5 cm^2 $ $\\$ Area of $\Delta OAB = \dfrac{1}{2}*Oa*OB = \dfrac{1}{2}*10*10 \\ =50 cm^2 $$\\$ Area of minor segment $ACBA$ = Area of minor sector $OACB$ -Area of $\Delta OAB= 78.5-50 = 28.5 cm^2$

8   In a circle of radius $21 cm$, an arc subtends an angle of $60^o$ at the centre.Find:

Solution :

Radius $\left(r\right)$ of circle =$ 21 cm$$\\$ Angle subtended by the given arc =$ 60^o$$\\$ Length of an arc of a sector of angle $\theta = \dfrac{\theta}{360^o}*2\pi r$$\\$ Length of arc $ACB = \dfrac{60^o}{360^o}*2*\dfrac{22}{7}*21$$\\$ $= \dfrac{1}{6}*2*22*3 =22 cm $$\\$ $In \Delta OAB, \\ \angle OAB = \angle OBA $(As radius OA=OB)$\\$ $\angle OAB +\angle AOB +\angle OBA =180^o \\ 2\angle OAB +60^o =180^o \\ \angle OAB =60^o$ $\\$ Threfore , $\Delta OAB$ is an equilateral triangle

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 $at the centre of the assumed flat circle.$\\$

Area between two consecutive ribs of circle =$\dfrac{45^o}{360^o}*\pi r^2 $$\\$ $= \dfrac{1}{8}*\dfrac{22}{7}*\left(45^2\right)^o \\ =\dfrac{22275}{28} cm^2$

16   A car has two wipers which do not overlap. Each wiper has blade of length $25 cm$ sweeping through an angle of $115^o$. Find the total area cleaned at each sweep of the blades.[Use $\pi =\dfrac{22}{7}$]

Solution :

It can be observed from the figure that each blade of wiper will sweep an area of a sector of $115^o$ in a circle of $25 cm$ radius.$\\$ Area of such sector = $\dfrac{115^o}{360^o}*\pi*\left(25\right)^2$$\\$ $ = \dfrac{23}{72}*\dfrac{22}{7}*25*25 = \dfrac{158125}{252}cm^2 $$\\$ Area swept by $2$ blades = $2*\dfrac{158125}{252}=\dfrac{158125}{126} cm^2 $

17   To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^o$ to a distance of $16.5 km$. Find the area of the sea over which the ships warned. [Use $\pi= 3.14$]

Solution :

It can be observed from the figure that the lighthouse spreads light across a sector of $80^o$ in a circle of $16.5 km$ radius.$\\$ Area of sector $OACB =\dfrac{80^o}{360^o}*\pi r^2$ $\\$ $ = \dfrac{2}{9}*3.14*16.5*16.5\\ =189.97 km^2 $

18   A round table cover has six equal designs as shown in the figure. If the radius of the cover is $28 cm$, find the cost of making the designs at the rate of $Rs.0.35$ /$ cm^2$. $[Use \sqrt {3} = 1.7 ] $

Solution :

It can be observed that these designs are segments of the circle. Consider segment $APB$. Chord $AB$ is a side of the hexagon. Each chord will substitute at $\dfrac{360^o}{ 6} = 60^o$ at the centre of the circle. $\\$ $ In \Delta OAB, \angle OAB = \angle OBA ( As OA = OB ) \\ \angle OAB + \angle OBA +\angle AOB = 180^2 \\ 2 \angle OAB = 180^2 -60^o =120^o \\ \angle OAB =60^o $$\\$ Therefore, $\Delta OAB$ is an equilateral triangle.$\\$ Area of $\Delta OAB = \dfrac{\sqrt{3}{4}}*\left(side\right)^2$$\\$ $= \dfrac{\sqrt3}{4}*\left(28\right)^2 = 196\sqrt{3}=333.2 cm^2 $$\\$ Area of sector $OAPB =\dfrac{60^o}{360^o}*\pi r^2 $$\\$ $\dfrac{1}{6}*\dfrac{22}{7}*28*28 \\ \dfrac{1232}{3}cm^2 $$\\$ Area of segment $APBA$ = Area of sector $OAPB$ - Area of $\Delta OAB$$\\$ $= \left(\dfrac{1232}{3}-333.2\right) cm^2 $$\\$ Therefore, area of designs =$6*\left(\dfrac{1232}{3}-333.2\right) cm^2 $$\\$ $= 464.8 cm^2 $$\\$ Cost of making $1 cm^2$ designs = $Rs 0.35$$\\$ Cost of making $464.76 cm^2 $ designs =$ 464.8 * 0.35 $= Rs$ 162.68 $ Therefore, the cost of making such designs is Rs$ 162.68. $

19   Tick the correct answer in the following$\colon$$\\$ Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is

Solution :

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 $\pi =\dfrac{22}{7}$]

Solution :

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 measure$60^o$. $\\$

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 $\pi =\dfrac{22}{7}$]

Solution :

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 $3 = 1.73205$ ]

Solution :

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

Solution :

(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 $\pi = \dfrac{22}{7}$]

Solution :

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 $