NCERT Solutions Class 10 Math Chapter 12 Areas Related To Circle

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## Topics Covered in Class 10 Math Chapter 12

12.1 Introduction

12.2 Perimeter and Area of Circle

12.3: Area of Sector and Segment of Circle

12.4: Area of Combination of Plane Figure

In this chapter, students will learn the following 5 points:

The formula to calculate the circumference of circle is 2πr.

The formula to calculate the area of circle is πr2

Length of an arc of a sector of a circle with radius r and angle with degree measure θ is given as $\frac{0}{360}$ 2πr.

Area of a sector of a circle with radius r and with degree measure θ is given as $\frac{0}{360}$ πr2.

Area of segment of circle is given as Area of Corresponding Sector - Area of Triangle

Exercise 12.1

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.

Ans: Given that,

Radius of 1st circle = \[{r_1}\] = 19 cm

Radius of 2nd circle = \[{r_2}\] = 9 cm

Circumference of 3rd circle = Circumference of 1st circle + Circumference of

2nd circle

Let the radius of the 3rd circle be \[r\].

Now,

Circumference of 1st circle = \[2\pi {r_1}\]

= \[2\pi (19)\]

= \[38\pi \]

Circumference of 2nd circle = \[2\pi {r_2}\]

= \[2\pi (9)\]

= \[18\pi \]

Circumference of 3rd circle = \[2\pi {r_{}}\]

Using given condition,

\[2\pi r = 38\pi + 18\pi \]

\[ = 56\pi \]

\[ \Rightarrow r = \frac{{56\pi }}{{2\pi }}\]

\[ = 28\].

Therefore, the radius of the circle which having 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.

Ans:

Given that,

Radius of 1st circle = \[{r_1}\] = 8 cm

Radius of 2nd circle = \[{r_2}\] = 6 cm

Area of 3rd circle = Area of 1st circle + Area of 2nd circle

Let the radius of 3rd circle be \[r\].

Area of 1st circle = \[\pi {r_1}^2\]

\[ = \pi {(8)^2}\]

\[ = 64\pi \]

Area of 2nd circle = \[\pi {r_2}\]

=\[\pi {r_2}^2\]

=\[36\pi \]

Using given condition,

\[\pi {r^2} = \pi {r_1}^2 + \pi {r_2}^2\]

\[ = 64\pi + 36\pi \]

\[ = 100\pi \]

\[ \Rightarrow {r^2} = 100\]

\[r = \pm 10\]

We know that, the radius cannot be negative. Therefore, the radius of the circle

having area equal to the sum of the areas of the other 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. [

\[\mathbf{Use}\text{ }\pi \text{ }\!\!~\!\!\text{ }=\frac{22}{7}\]

]

(Image Will Be Updated Soon)

Solution 3:

(Image Will Be Updated Soon)

Radius of gold region (i.e., 1st circle) = \[{r_1}\]

\[ = \frac{{21}}{2}\]

\[ = 10.5\].

Given that each circle is 10.5 cm wider than the previous circle.

Thus, radius of 2nd circle = \[{r_2}\]

= \[10.5 + 10.5\]

= 21 cm

Radius of 3rd circle = \[{r_3}\]

= \[21{\rm{ }} + {\rm{ }}10.5\]

= 31.5 cm

Radius of 4th circle = \[{r_4}\]

= \[31.5{\rm{ }} + {\rm{ }}10.5\]

= 42 cm

Radius of 5th circle = \[{r_5}\]

= \[42{\rm{ }} + {\rm{ }}10.5\]

= 52.5 cm

According to given condition,

Area of gold region = Area of 1st circle

\[ \Rightarrow \pi {r_1}^2 = \pi {(10.5)^2}\]

\[ = 346.5c{m^2}\]

Area of red region = Area of 2nd circle − Area of 1st circle

\[ = \pi {r_2}^2 - \pi {r_1}^2\]

\[ = \pi {(21)^2} - \pi {(10.5)^2}\]

\[ = 441\pi - 110.25\pi \]

\[ = 330.75\pi \]

\[ = 1039.5c{m^2}\]

Area of blue region = Area of 3rd circle − Area of 2nd circle

\[ = \pi {r_3}^2 - \pi {r_2}^2\]

\[ = \pi {(31.5)^2} - \pi {(21)^2}\]

\[ = 992.25\pi - 441\pi \]

\[ = 551.25\pi \]

\[ = 1732.5c{m^2}\]

Area of black region = Area of 4th circle − Area of 3rd circle

\[ = \pi {r_4}^2 - \pi {r_3}^2\]

\[ = \pi {(42)^2} - \pi {(31.5)^2}\]

\[ = 1764\pi - 992.25\pi \]

\[ = 771.75\pi \]

\[ = 2425.5c{m^2}\]

Area of white region = Area of 5th circle − Area of 4th circle

\[ = \pi {r_5}^2 - \pi {r_4}^2\]

\[ = \pi {(52.5)^2} - \pi {(42)^2}\]

\[ = 2756.25\pi - 1764\pi \]

\[ = 992.25\pi \]

\[ = 3118.5c{m^2}\]

Therefore, areas of gold, red, blue, black, and white regions are \[346.5c{m^2}\], \[1039.5c{m^2}\], \[1732.5c{m^2}\], \[2425.5c{m^2}\] and \[3118.5c{m^2}\] 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?

Ans:

Given that,

Diameter of the wheel of the car = 80 cm

Radius of the wheel of the car = \[r\] =40 cm

Speed of car = 66 km/hour

We know that,

Circumference of wheel = \[2\pi r\]

= \[2\pi (40)\]

= \[80\pi cm\]

Speed of car \[ = \frac{{66 \times 100000}}{{60}}cm/\min \]

\[ = 1,10,000cm/\min \]

Now, distance travelled by the car in 10 minutes \[ = 110000 \times 10\]

\[ = 11,00,000cm\]

Let the number of revolutions of the wheel of the car be n.

We know that,

Distance travelled in 10 minutes = n × Distance travelled in 1 revolution (i.e., circumference)

\[ \Rightarrow 1100000 = n \times 80\pi \]

\[ = \frac{{35000}}{8}\]

\[ = 4375\]

Therefore, each wheel of the car will make 4375 revolutions.

5. Tick the correct answer in the following and justify your choice: If the

perimeter and the area of a circle are numerically equal, then the radius of the

circle is

2 units (B) π units (C) 4 units (D) 7 units

Ans:

Given that,

the circumference and the area of the circle are equal.

Let the radius (to be find) of the circle be \[r\]

Thus,

Circumference of circle \[ = 2\pi r\] and

Area of circle \[ = \pi {r^2}\]

According to given condition,

\[2\pi r = \pi {r^2}\]

\[ \Rightarrow 2 = r\]

Therefore, the radius of the circle is 2 units.

Hence, the correct answer is A.

Exercise (12.2)

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

Ans:

(Image Will Be Updated Soon)

\[\frac{{132}}{7}c{m^2}\]

Given that,

Radius of the circle = \[r = 6cm\]

Angle made by the sector with the center, \[\theta = {60^ \circ }\]

Let OACB be a sector of the circle making \[{60^ \circ }\] angle at center O of the circle.

We know that area of sector of angle, \[ = \frac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}\]

Thus, Area of sector OACB \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \frac{{22}}{7} \times {(6)^2}\]

\[ = \frac{1}{6} \times \frac{{22}}{7} \times 6 \times 6\]

\[ = \frac{{132}}{7}c{m^2}\]

Therefore, the area of the sector of the circle making 60° at the center of the circle is \[\frac{{132}}{7}c{m^2}\].

2. Find the area of a quadrant of a circle whose circumference is 22 cm. $pi =\dfrac{22}{7}$

Ans:

(Image Will Be Updated Soon)

Given that,

Circumference = 22 cm

Let the radius of the circle be \[r\].

According to the given condition,

\[2\pi r = 22\]

\[ \Rightarrow r = \frac{{22}}{{2\pi }}\]

\[ = \frac{{11}}{\pi }\]

We know that, quadrant of circle subtends \[{90^ \circ }\] angle at the center of the circle.

Area of such quadrant of the circle \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {r^2}\]

\[ = \frac{1}{4} \times \pi \times {\left( {\frac{{11}}{\pi }} \right)^2}\]

\[ = \frac{{121}}{{4\pi }}\]

\[ = \frac{{77}}{8}c{m^2}\]

Hence, the area of a quadrant of a circle whose circumference is 22 cm is \[ = \frac{{77}}{8}c{m^2}\].

3.The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

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

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of clock or circle = \[r\] = 14 cm.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates \[{360^ \circ }\].

Thus, in 5 minutes, minute hand will rotate \[ = \frac{{{{360}^ \circ }}}{{{{60}^ \circ }}} \times 5\]

\[ = {30^ \circ }\]

Now,

the area swept by the minute hand in 5 minutes = the area of a sector of \[{30^ \circ }\] in a circle of 14 cm radius.

Area of sector of angle \[\theta = \frac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}\]

Thus, Area of sector of \[{30^ \circ } = \frac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \frac{{22}}{7} \times 14 \times 14\]

\[ = \frac{{11 \times 14}}{3}\]

\[ = \frac{{154}}{3}c{m^2}\]

Therefore, the area swept by the minute hand in 5 minutes is \[\frac{{154}}{3}c{m^2}\].

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

$[\text{Use }\pi =3.14]$

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of the circle \[ = r = 10cm\]

Angle subtended by the cord = angle for minor sector\[ = {90^ \circ }\]

Angle for minor sector \[ = {360^ \circ } - {90^ \circ } = {270^ \circ }\]

i) Minor segment

Ans: It is evident from the figure that,

Area of minor segment ACBA = Area of minor sector OACB − Area of ΔOAB

Thus,

Area of minor sector OACB \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\] \[ = \frac{1}{4} \times 3.14 \times {(10)^2}\] \[ = 78.5c{m^2}\]

Area of ΔOAB \[ = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times {(10)^2} = 50c{m^2}\]

Area of minor segment ACBA \[ = 78.5 - 50 = 28.5c{m^2}\]

Hence, area of minor segment is \[28.5c{m^2}\]

ii) Major sector

Ans: It is evident from the figure that,

Area of major sector OADB \[ = \frac{{{{270}^ \circ }}}{{{{360}^ \circ }}} = \frac{3}{4} \times 3.14 \times {(10)^2} = 235.5c{m^2}\].

Hence, area of major sector is \[235.5c{m^2}\].

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.

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

Find:

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of circle = \[r = \] 21 cm

Angle subtended by the given arc = \[\theta = {60^ \circ }\]

i) The length of the arc

Ans: We know that, Length of an arc of a sector of angle \[\theta = \frac{\theta }{{{{360}^ \circ }}} \times 2\pi r\]

Thus, Length of arc ACB \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \frac{{22}}{7} \times 21\]

\[ = 22cm\]

Hence, length of the arc of given circle is \[22cm\].

ii) Area of the sector formed by the arc

Ans: We know that, Area of sector OACB \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = 231c{m^2}\]

Hence, area of the sector formed by the arc of the given circle is \[231c{m^2}\].

iii) Area of the segment formed by the corresponding chord

Ans: In \[OAB\],

As radius \[OA = OB\]

\[ \Rightarrow \angle OAB = \angle OBA\]

\[\angle OAB + \angle AOB + \angle OBA = {180^ \circ }\]

\[2\angle OAB + {60^ \circ } = {180^ \circ }\]

\[\angle OAB = {60^ \circ }\]

Therefore, \[OAB\] is an equilateral triangle.

Now, area of \[OAB\] \[ = \frac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}\]

\[ = \frac{{\sqrt 3 }}{4} \times {\left( r \right)^2}\]

\[ = \frac{{\sqrt 3 }}{4} \times {\left( {21} \right)^2}\]

\[ = \frac{{441\sqrt 3 }}{4}c{m^2}\]

We know that, Area of segment ACB = Area of sector OACB − Area of

\[ = \left( {231 - \frac{{441\sqrt 3 }}{4}} \right)c{m^2}\].

Hence, Area of the segment formed by the corresponding chord in circle is

\[\left( {231 - \frac{{441\sqrt 3 }}{4}} \right)c{m^2}\].

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments of the circle.

[\text{Use}\text{ }\pi =\dfrac{22}{7}\text{ }\mathbf{and}\text{ }\sqrt{3}=1.73\text{ }]

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of circle = \[r = \] 15 cm

Angle subtended by chord \[ = \theta = {60^ \circ }\]

Area of circle \[ = \pi {r^2}\] \[ = 3.14{\left( {15} \right)^2}\]

\[ = 706.5c{m^2}\]

Area of sector OPRQ \[ = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \dfrac{1}{6} \times 3.14{(15)^2} = 117.75c{m^2}\]

Now, for the area of major and minor segments,

In \[OPQ\],

Since, \[OP = OQ\]

\[ \Rightarrow \angle OPQ = \angle OQP\]

\[\angle OPQ = {60^ \circ }\]

Thus, \[OPQ\] is an equilateral triangle.

Area of \[OPQ\] \[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}\]

\[ = \dfrac{{\sqrt 3 }}{4} \times {(r)^2}\]

\[ = \dfrac{{225\sqrt 3 }}{4}\] \[ = 97.3125c{m^2}\].

Now,

Area of minor segment PRQP = Area of sector OPRQ − Area of \[OPQ\]

\[ = 117.75 - 97.3125\]

\[ = 20.4375c{m^2}\]

Area of major segment PSQP = Area of circle − Area of minor segment PRQP

\[ = 706.5 - 20.4375\]

\[ = 686.0625c{m^2}\]

Therefore, the areas of the corresponding minor and major segments of the circle are \[20.4375c{m^2}\] and \[686.0625c{m^2}\] respectively.

7. A chord of a circle of radius 12 cm subtend an angle of 120° at the center. Find the area of the corresponding segment of the circle. [\text{ }\mathbf{Use}\text{ }\pi =\frac{22}{7}\text{ }\mathbf{and}\text{ }\sqrt{3}=1.73\text{ }]

Ans:

(Image Will Be Updated Soon)

Draw a perpendicular OV on chord ST bisecting the chord ST such that SV = VT

Now, values of OV and ST are to be found.

Therefore,

In \[OVS\],

\[\cos {60^ \circ } = \frac{{OV}}{{OS}}\]

\[ \Rightarrow \frac{{OV}}{{12}} = \frac{1}{2}\]

\[ \Rightarrow OV = 6cm\]

Also, \[\frac{{SV}}{{SO}} = \sin {60^ \circ }\]

\[ \Rightarrow \frac{{SV}}{{12}} = \frac{{\sqrt 3 }}{2}\]

\[ \Rightarrow SV = 6\sqrt 3 \]

Now, \[ST = 2SV\]

\[ = 2 \times 6\sqrt 3 = 12\sqrt 3 cm\]

Area of \[OST = \dfrac{1}{2} \times ST \times OV\]

\[ = \dfrac{1}{2} \times 12\sqrt 3 \times 6\]

\[ = 62.28c{m^2}\]

Area of sector OSUT \[ = \dfrac{{{{120}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(12)^2}\]

\[ = 150.42c{m^2}\]

Area of segment SUTS = Area of sector OSUT − Area of \[OVS\]

\[ = 150.72 - 62.28\]

\[ = 88.44c{m^2}\]

Hence, the area of the corresponding segment of the circle is \[88.44c{m^2}\].

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see the given figure). [\mathbf{Use}\text{ }\pi =3.14]

(Image Will Be Updated Soon)

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.

Ans:

(Image Will Be Updated Soon)

From the above figure, it is clear that the horse can graze a sector of \[{90^ \circ }\] in a circle of 5 m radius.

Hence,

\[\theta \text{ }\!\!~\!\!\text{ }={{90}^{{}^\circ }}\]

\[r=5m\]

(i) The area of that part of the field in which the horse can graze.

Ans: It is evident from the figure,

Area that can be grazed by horse = Area of sector OACB

\[ = \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}\]

\[ = \dfrac{1}{4} \times 3.14 \times {(5)^2}\] \[ = 19.625{m^2}\]

(ii) The increase in the grazing area if the rope were 10 m long instead of 5 m.

Ans: It is evident from the figure,

Area that can be grazed by the horse when length of rope is 10 m long

\[ = \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {(10)^2}\]

\[ = 78.5{m^2}\]

Therefore, the increase in grazing area for horse \[ = (78.5 - 19.625){m^2} = 58.875{m^2}\].

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Find:

The total length of the silver wire required.

The area of each sector of the brooch.

Ans:

Given that,

Radius of the circle \[ = r\] \[ = \frac{{diameter}}{2}\]\[ = \frac{{35}}{2}mm\]

(Image Will Be Updated Soon)

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° (i.e., 360°/10=36°) at the center of the circle.

(i) The total length of the silver wire required.

Ans:

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Circumference of brooch \[ = 2\pi r\]

\[ = 2 \times \frac{{22}}{7} \times \left( {\frac{{35}}{2}} \right)\]

\[ = 110mm\]

Length of wire required \[ = 110 + \left( {5 \times 35} \right)\]

\[ = 285mm\]

Therefore, The total length of the silver wire required is \[285mm\].

(ii) The area of each sector of the brooch.

Ans:

Area of each sector \[ = \frac{{{{36}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{{10}} \times \frac{{22}}{7} \times {\left( {\frac{{35}}{2}} \right)^2}\]

\[ = \frac{{385}}{4}m{m^2}\]

Hence, The area of each sector of the brooch is \[\frac{{385}}{4}m{m^2}\].

10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

(Image Will Be Updated Soon)

Ans:

Given that,

Radius of the umbrella \[ = r\]\[ = 45cm\]

There are 8 ribs in an umbrella.

The angle between two consecutive ribs is subtending \[\frac{{{{360}^ \circ }}}{8} = {45^ \circ }\] at the center of the assumed flat circle.

Area between two consecutive ribs of the assumed circle \[ = \frac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{8} \times \frac{{22}}{7} \times {\left( {45} \right)^2}\]

\[ = \frac{{22275}}{{28}}c{m^2}\]

Hence, the area between the two consecutive ribs of the umbrella is \[\frac{{22275}}{{28}}c{m^2}\].

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

(Image Will Be Updated Soon)

(Image Will Be Updated Soon)

Ans:

Given that,

Each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of sector \[ = \frac{{{{115}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {\left( {25} \right)^2}\]

\[ = \frac{{158125}}{{252}}c{m^2}\]

Area swept by 2 blades \[ = 2 \times \frac{{158125}}{{252}}\]

\[ = \frac{{158125}}{{126}}c{m^2}\].

Therefore, the total area cleaned at each sweep of the blades is \[\frac{{158125}}{{126}}c{m^2}\].

12. To warn ships for underwater rocks, a lighthouse spreads a red colored light over a sector of angle \[{80^ \circ }\] to a distance of 16.5 km. Find the area of the sea over which the ships warned. [\mathbf{Use}\text{ }\pi =3.14]

Ans:

(Image Will Be Updated Soon)

Given that,

The lighthouse spreads light across a sector (represented by shaded part in the figure) of \[{80^ \circ }\] in a circle of 16.5 km radius.

Area of sector OACB \[ = \frac{{{{80}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{2}{9} \times 3.14 \times {(16.5)^2}\]

\[ = 189.97k{m^2}\]

Hence, the area of the sea over which the ships are warned is \[189.97k{m^2}\].

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

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given in the figure,

The designs are segments of the circle.

Radius of circle is 28cm.

Consider segment APB and chord AB is a side of the hexagon.

Each chord will substitute at \[\frac{{{{360}^ \circ }}}{6} = {60^ \circ }\] at the center of the circle.

In \[OAB\],

Since, \[OA = OB\]

\[ \Rightarrow \angle OAB + \angle OBA + \angle AOB = {180^ \circ }\]

\[2\angle OAB = {180^ \circ } - {60^ \circ } = {120^ \circ }\]

\[\angle OAB = {60^ \circ }\]

Therefore, \[OAB\] is an equilateral triangle.

Area of \[OAB\] \[ = \frac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}\]

\[ = \frac{{\sqrt 3 }}{4} \times {\left( {28} \right)^2}\]

\[ = 333.2c{m^2}\]

Area of sector OAPB \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{6} \times \frac{{22}}{7} \times {(28)^2}\]

\[ = \frac{{1232}}{3}c{m^2}\]

Area of segment APBA = Area of sector OAPB − Area of ∆OAB

\[ = \left( {\frac{{1232}}{3} - 333.2} \right)c{m^2}\]

Therefore, area of designs \[ = 6 \times \left( {\frac{{1232}}{3} - 333.2} \right)c{m^2}\]

\[ = 464.8c{m^2}\]

Now, given that Cost of making 1 \[c{m^2}\] designs = Rs 0.35

Cost of making 464.76 \[c{m^2}\] designs \[ = 464.8 \times 0.35 = \]162.68

Therefore, the cost of making such designs is Rs 162.68.

14. Tick the correct answer in the following: Area of a sector of angle \[p\] (in degrees) of a circle with radius \[R\] is

\[(A)\frac{P}{{180}} \times 2\pi R\] \[(B)\frac{P}{{180}} \times 2\pi {R^2}\] \[(C)\frac{P}{{180}} \times \pi R\] \[(D)\frac{P}{{720}} \times 2\pi {R^2}\]

Ans:

We know that area of sector of angle \[\theta = \frac{\theta }{{{{360}^ \circ }}} \times \pi {R^2}\]

So, Area of sector of angle \[P = \frac{P}{{{{360}^ \circ }}}\left( {\pi {R^2}} \right)\]

\[ = \left( {\frac{P}{{{{720}^ \circ }}}} \right)\left( {2\pi {R^2}} \right)\]

Hence, (D) is the correct answer.

Exercise 12.3

1. 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. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

From the given figure,

RQ is the diameter of the circle which implies that \[\angle RPQ = {90^ \circ }\]

Thus,

By applying Pythagoras theorem in \[PQR\],

\[ \Rightarrow R{P^2} + P{Q^2} = R{Q^2}\]

\[ \Rightarrow {(7)^2} + {(24)^2} = R{Q^2}\]

\[ \Rightarrow RQ = 25\]

Thus, Radius of circle, \[OR = \frac{{RQ}}{2} = \frac{{25}}{2}\]

We know that, RQ is the diameter of the circle, it divides the circle in two equal parts.

So, for area of shaded region,

Area of semicircle \[ = \frac{1}{2}\pi {r^2}\]

\[ = \frac{1}{2}\pi {\left( {\frac{{25}}{2}} \right)^2}\]

\[ = \frac{{6875}}{{28}}c{m^2}\]

Area of \[PQR\] \[ = \frac{1}{2} \times PQ \times PR\]

\[ = \frac{1}{2} \times 24 \times 7\]

\[ = 84c{m^2}\]

Area of shaded region = Area of semi - circle RPQOR − Area of \[PQR\]

\[ = \frac{{6875}}{{28}} - 84 = \frac{{4532}}{{28}}c{m^2}\]

Therefore, the area of the shaded region in the given figure is \[161.85 c{m^2}\].

2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with center O are 7 cm and 14 cm respectively and \[\angle AOC = {40^ \circ }\]. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of inner circle = 7 cm = \[{r_1}\]

Radius of outer circle = 14 cm = \[{r_2}\]

Angle subtended is \[{40^ \circ }\] = \[{{\theta _1}}\]

Hence, \[{{\theta _2}}\] = \[{40^0}\]

Now,

Area of shaded region = Area of sector OAFC − Area of sector OBED

\[ = \frac{{{\theta _1}}}{{{{360}^0}}} \times \pi {r_1}^2 - \frac{{{\theta _2}}}{{{{360}^0}}} \times \pi {r_2}^2\]

\[ = \frac{{{{40}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(14)^2} - \frac{{{{40}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(7)^2}\]

\[ = \frac{{616}}{9} - \frac{{159}}{9}\]

\[ = \frac{{154}}{3}c{m^2}\]

Therefore, the area of the shaded region in the given figure is \[ = 51.33 c{m^2}\].

3. 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. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

From the above figure it is evident that the radius of each semi-circle is 7 cm.

For area of shaded region,

So, Area of each semi-circle \[ = \frac{1}{2}\pi {r^2}\]

\[ = \frac{1}{2} \times \frac{{22}}{7} \times {(7)^2}\]

\[ = 77c{m^2}\]

Area of square ABCD \[ = {(side)^2} = {(14)^2} = 196c{m^2}\]

Area of the shaded region = Area of square ABCD − Area of semi-circle APD − Area of semi-circle BPC

\[ = 196 - 77 - 77 = 42c{m^2}\]

Therefore, the area of the shaded region in the given figure is \[42c{m^2}\].

4. 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 center. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of the circle is 6cm

We know that each interior angle of an equilateral triangle is of measure \[{60^ \circ }\] = \[\theta \].

For area of shaded region,

Area os sector OCDE = \[\frac{\theta }{{{{360}^0}}} \times \pi {r^2}\]

Area of sector OCDE \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}\]

\[ = \frac{1}{6} \times \frac{{22}}{7} \times {(6)^2}\]

\[ = \frac{{132}}{7}c{m^2}\]

Area of \[OAB\] \[ = \frac{{\sqrt 3 }}{4}{(12)^2}\]

\[ = 36\sqrt 3 c{m^2}\]

Area of circle \[ = \pi {r^2}\]

\[ = \frac{{22}}{7} \times {(6)^2}\]

\[ = \frac{{792}}{7}c{m^2}\]

Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE

\[ = 36\sqrt 3 + \frac{{792}}{7} - \frac{{132}}{7}\]

\[ = \left( {36\sqrt 3 + \frac{{660}}{7}} \right)c{m^2}\]

Hence, the area of the shaded region in the given figure \[\left( {36\sqrt 3 + \frac{{660}}{7}} \right)c{m^2}\].

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 the given figure. Find the area of the remaining portion of the square. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

It is evident from the above figure that each quadrant is a sector of \[{90^ \circ }\] in a circle of 1 cm radius.

For area of shaded region,

Area of quadrant \[ = \frac{\theta }{{{{360}^0}}} \times \pi {r^2}\]

Area of each quadrant \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {(1)^2}\]

\[ = \frac{{22}}{{28}}c{m^2}\]

Area of square \[ = {(side)^2}\]

\[ = {(4)^2}\]

\[ = 16c{m^2}\]

Area of circle \[ = \pi {r^2}\]

\[ = \pi {(1)^2}\] \[ = \frac{{22}}{7}c{m^2}\]

Area of the shaded region = Area of square − Area of circle – (4 × Area of quadrant)

\[ = 16 - \frac{{22}}{7} - \left( {4 \times \frac{{22}}{{28}}} \right)\]

\[ = 16 - \frac{{44}}{7}\]

\[ = \frac{{68}}{7}c{m^2}\]

Therefore, the area of the remaining portion of the square is \[\frac{{68}}{7}c{m^2}\].

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 the given figure. Find the area of the design (Shaded region). $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of circle = \[r\] = 32 cm

AD is the median of \[ABC\]

\[OA = \frac{2}{3}AD\]

\[AD = 48cm\]

In \[ABD\],

Using Pythagoras Theorem,

\[A{B^2} = A{D^2} + B{D^2}\]

\[A{B^2} = {(48)^2} + {\left( {\frac{{BC}}{2}} \right)^2}\]

\[A{B^2} = {(48)^2} + {\left( {\frac{{AB}}{2}} \right)^2}\]

\[ \Rightarrow \frac{{3A{B^2}}}{4} = {(48)^2}\]

\[ \Rightarrow AB = \frac{{48 \times 2}}{{\sqrt 3 }} = \frac{{96}}{{\sqrt 3 }}\]

\[ = 32\sqrt 3 \] cm

For area of design,

Area of equilateral triangle \[ = \frac{{\sqrt 3 }}{4} \times {(32)^2} \times 3\]

\[ = 96 \times 8 \times \sqrt 3 \]

\[ = 768\sqrt 3 c{m^2}\]

Area of circle \[ = \pi {r^2}\]

\[ = \frac{{22}}{7} \times {(32)^2}\]

\[ = \frac{{22528}}{7}c{m^2}\]

Area of design = Area of circle − Area of \[ABC\]

\[ = \left( {\frac{{22528}}{7} - 768\sqrt 3 } \right)c{m^2}\]

Hence, the area of the design (Shaded region) is \[\left( {\frac{{22528}}{7} - 768\sqrt 3 } \right)c{m^2}\].

7. In the given figure, ABCD is a square of side 14 cm. With centers 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. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

It is evident that,

Area of each of the 4 sectors is equal to each other

Sector of \[{90^ \circ }\] in a circle of 7 cm radius.

For the area of shaded region,

Area of each sector \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {(7)^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {\left( 7 \right)^2}\]

\[ = \frac{{77}}{2}c{m^2}\]

Area of square ABCD \[ = {(side)^2}\]

\[ = {(14)^2} = 196c{m^2}\]

Area of shaded portion = Area of square ABCD – (4 × Area of each sector)

\[ = 196 - \left( {4 \times \frac{{77}}{2}} \right)\]

\[ = 42c{m^2}\]

Therefore, the area of shaded portion is \[42c{m^2}\].

8. The given figure depicts a racing track whose left and right ends are semicircular.

(Image Will Be Updated Soon)

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

Find

Ans:

(Image Will Be Updated Soon)

The distance around the track along its inner edge

Ans: For inner edge,

Radius = \[r\]= 30cm

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA

\[ = 106 + \left( {\frac{1}{2} \times 2\pi r} \right) + 106 + \left( {\frac{1}{2} \times 2\pi r} \right)\]

\[ = 212 + \left( {\frac{1}{2} \times 2 \times \frac{{22}}{7} \times 30} \right) + \left( {\frac{1}{2} \times 2 \times \frac{{22}}{7} \times 30} \right)\]

\[ = 212 + \left( {2 \times \frac{{22}}{7} \times 30} \right)\]

\[ = \frac{{2804}}{7}m\]

Hence, distance around the track along its inner edge is \[\frac{{2804}}{7}m\].

The area of the track

Ans: Radius of inner edge = 30cm

Radius of outer edge = 40cm

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 semicircle AFD)

\[ = \left( {106 \times 80} \right) - \left( {106 \times 60} \right) + \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(40)}^2}} \right) - \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(30)}^2}} \right) + \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(40)}^2}} \right) - \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(30)}^2}} \right)\]

\[ = 106\left( {80 - 60} \right) + \left( {\frac{{22}}{7} \times {{\left( {40} \right)}^2}} \right) - \left( {\frac{{22}}{7} \times {{\left( {30} \right)}^2}} \right)\]

\[ = 2120 + \frac{{22}}{7}\left[ {{{\left( {40} \right)}^2} - {{\left( {30} \right)}^2}} \right]\]

\[ = 2120 + 2200\]

\[ = 4320{m^2}\]

Therefore, the area of the track is \[4320{m^2}\].

9. In the given figure, AB and CD are two diameters of a circle (with center 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.

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

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of larger circle \[ = {r_1}\] = 7 cm

Radius of smaller circle \[ = {r_2}\] = \[\frac{7}{2}\] cm

For area of shaded region,

Area of smaller circle \[ = \pi {r_2}^2\]

\[ = \frac{{22}}{7} \times {\left( {\frac{7}{2}} \right)^2}\]

\[ = \frac{{77}}{2}c{m^2}\]

Area of semi-circle AECFB of larger circle \[ = \frac{1}{2}\pi {r_2}^2\]

\[ = \frac{1}{2} \times \frac{{22}}{7} \times {\left( 7 \right)^2}\]

\[ = \frac{{77}}{2}c{m^2}\]

Area of \[ABC\]\[ = \frac{1}{2} \times AB \times OC\]

\[ = \frac{1}{2} \times 14 \times 7\]

\[ = 49c{m^2}\]

Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of \[ABC\]

\[ = \frac{{77}}{2} + 77 - 49\]

\[ = 28 + 38.5 = 66.5c{m^2}\]

Therefore, the area of shaded region is \[66.5c{m^2}\].

10. The area of an equilateral triangle ABC is \[17320.5c{m^2}\]. With each vertex of the triangle as center, 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. [\mathbf{Use}\text{ }\pi =3.14 \text{ }\mathbf{and}\text{ }\sqrt{3}=1.73205]

(Image Will Be Updated Soon)

Ans:

Let the side of the equilateral triangle be a.

Given that,

Area of equilateral triangle \[ = 17320.5c{m^2}\]

\[ \Rightarrow \frac{{\sqrt 3 }}{4}{(a)^2} = 17320.5\]

\[ \Rightarrow \frac{{1.73205}}{4}{a^2} = 17320.5\]

\[ \Rightarrow {a^2} = 4 \times 10000\]

\[ \Rightarrow a = 200cm\]

(Image Will Be Updated Soon)

It is evident from the figure that, each sector is of measure \[{60^ \circ }\]

Area of sector ADEF \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {r^2}\]

\[ = \frac{1}{6} \times \pi \times {\left( {100} \right)^2}\]

\[ = \frac{{3.14 \times 10000}}{6}\]

\[ = \frac{{15700}}{3}c{m^2}\]

Area of shaded region = Area of equilateral triangle – (3 × Area of each sector)

\[ = 17320.5 - 3 \times \frac{{15700}}{3}\]

\[ = 17320.5 - 15700\]

\[ = 1620.5c{m^2}\]

Therefore, area of given shaded region is \[1620.5c{m^2}\].

11. 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. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

It is evident from the above figure, that the side of the square is 42 cm.

So, for area of the remaining portion of handkerchief,

Area of square \[ = {(side)^2} = {(42)^2}\]

\[ = 1764c{m^2}\]

Area of each circle \[ = \pi {r^2}\]\[ = \frac{{22}}{7} \times {(7)^2}\]

\[ = 154c{m^2}\]

Area of 9 circles \[ = 9 \times 154\]

\[ = 1386c{m^2}\]

Area of the remaining portion of the handkerchief \[ = 1764 - 1386\]

\[ = 378c{m^2}\]

Therefore, the area of the remaining portion of the handkerchief is \[ = 378c{m^2}\].

12. In the given figure, OACB is a quadrant of circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that radius is 3.5cm.

Quadrant OACB

Ans:

Since OACB is a quadrant, the angle at O is \[{90^ \circ }\].

Area of quadrant OACB \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {\left( {3.5} \right)^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {\left( {\frac{7}{2}} \right)^2}\]

\[ = \frac{{77}}{8}c{m^2}\]

Therefore, the area of quadrant OACB is \[\frac{{77}}{8}c{m^2}\].

Shaded region

Ans:

For the area of the shaded region,

Area of \[OBD\] \[ = \frac{1}{2} \times OB \times OD\]

\[ = \frac{1}{2} \times 3.5 \times 2\]

\[ = \frac{7}{2}c{m^2}\]

Area of the shaded region = Area of quadrant OACB − Area of \[OBD\]

\[ = \frac{{77}}{8} - \frac{7}{2}\]

\[ = \frac{{49}}{8}c{m^2}\].

Hence, the area of the shaded region in the given figure is \[\frac{{49}}{8}c{m^2}\].

13. In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [\mathbf{Use}\text{ }\pi =3.14]

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

For radius

In \[OAB\],

Using Pythagoras Theorem,

\[O{B^2} = O{A^2} + O{B^2}\]

\[ = {(20)^2} + {(20)^2}\]

\[ \Rightarrow OB = 20\sqrt 2 \]

Radius of circle \[ = r = \]\[20\sqrt 2 \] cm

For the area of shaded region,

Area of quadrant OPBQ \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times 3.14 \times {\left( {20\sqrt 2 } \right)^2}\]

\[ = \frac{1}{4} \times 3.14 \times 800\]

\[ = 628c{m^2}\]

Area of square OABC \[ = {\left( {side} \right)^2}\]

\[ = {(20)^2}\]

\[ = 400c{m^2}\]

Area of shaded region = Area of quadrant OPBQ − Area of square OABC

\[ = \left( {628 - 400} \right)c{m^2}\]

\[ = 228c{m^2}\]

Therefore, the area of the shaded region is \[228c{m^2}\].

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

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that,

Radius for sector OAEB = 21cm

Radius for sector OCFD = 7cm

Angle subtended is \[{30^ \circ }\]

Area of the shaded region = Area of sector OAEB − Area of sector OCFD

\[ = \left[ {\frac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {{\left( {21} \right)}^2}} \right] - \left[ {\frac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {{\left( 7 \right)}^2}} \right]\]

\[ = \frac{1}{{12}}\pi \left[ {\left( {21 - 7} \right)\left( {21 + 7} \right)} \right]\]

\[ = \frac{{22 \times 14 \times 28}}{{12 \times 7}} = \frac{{22 \times 14 \times 28}}{{12 \times 7}}\]

\[ = \frac{{308}}{3}c{m^2}\]

Therefore, the area of shaded region is \[\frac{{308}}{3}c{m^2}\].

15. 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. $pi =\dfrac{22}{7}$

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that,

Radius of circle = \[r = \] 14cm

As ABC is a quadrant of the circle, \[\angle BAC\] will be of measure \[{90^ \circ }\].

In \[ABC\],

Using Pythagoras Theorem,

\[ \Rightarrow B{C^2} = A{C^2} + A{B^2}\]

\[ = {(14)^2} + {(14)^2}\]

\[ \Rightarrow BC = 14\sqrt 2 \]

Radius of semi-circle drawn on BC = \[{r_1}\] \[ = \frac{{14\sqrt 2 }}{2} = 7\sqrt 2 cm\]

Now, for area of the shaded region,

Area of \[ABC\] \[ = \frac{1}{2} \times AB \times AC\]

\[ = \frac{1}{2} \times {(14)^2}\]

\[ = 98c{m^2}\]

Area of sector ABDC \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {(14)^2}\]

\[ = 154c{m^2}\]

Area of semi-circle drawn on BC \[ = \frac{1}{2} \times \pi \times {r_1}^2\]

\[ = \frac{1}{2} \times \frac{{22}}{7} \times {\left( {7\sqrt 2 } \right)^2}\]

\[ = 154c{m^2}\]

Area of shaded region = Area of semi-circle on BC – (Area of sector ABDC – Area of ∆ABC)

\[ = 154 - \left( {154 - 98} \right)\]

\[ = 98c{m^2}\]

Hence, the area of the shaded region of the given figure is \[98c{m^2}\].

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

(Image Will Be Updated Soon)

Ans:

(Image Will Be Updated Soon)

Given that ,

Radius of each circle is 8cm

The designed area is the common region between two sectors BAEC and DAFC.

For the area of the designed region,

Area of sector \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \frac{{22}}{7} \times {\left( 8 \right)^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times 64\]

\[ = \frac{{22 \times 16}}{7}\]

\[ = \frac{{352}}{7}c{m^2}\]

Area of \[BAC\] \[ = \frac{1}{2} \times BA \times BC\]

\[ = \frac{1}{2} \times {(8)^2}\]

\[ = 32c{m^2}\]

Area of the designed portion = 2 × (Area of segment AEC) = 2 × (Area of sector BAEC − Area of \[BAC\])

\[ = 2 \times \left( {\frac{{352}}{7} - 32} \right)\]

\[ = \frac{{2 \times 128}}{7}\]

\[ = \frac{{256}}{7}c{m^2}\]

Hence, the area of the designed region is \[\frac{{256}}{7}c{m^2}\].

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles - PDF Download

To make the study process easier and seamless, we provide the students with an option of PDF download. This provides the liberty to learn and revise from the Class 10 Maths Revision Notes Chapter 12, anytime, anywhere, the students wish to. The students can also get the hardcopy printed, which comes in handy when the students are studying. This also eliminates the need to open the web every time while studying.

You can opt for Chapter 12 - Areas Related to Circles NCERT Solutions for Class 10 Maths PDF for upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

### NCERT Solutions for Class 10 Maths

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Pair of Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometry

Chapter 8 - Introduction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 13 - Surface Areas and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

### NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles

The expert mentors compiled the important techniques in NCERT Solutions for Class 10 Maths Chapter 12. The solutions for all types of problems in NCERT Chapter Areas Related to Circles are precisely explained in Ch 12 Class 10 Questions with Solutions. After covering the concepts of area and perimeter of various polygons, areas related to circles chapter deals with the areas and perimeter of circles, segments, and sectors have been explained.

Areas Related to Circles Class 10 Ncert Solutions provide conceptual clarity on various topics. The chapter is divided into multiple sections. The first section covers the areas and perimeters of circles. Secondly, it covers the area and perimeter of Segments and Sectors. Lastly, the problems related to the area and perimeters of combinational figures are dealt with.

### Area and Perimeter of Circle

NCERT Class 10 Maths Chapter 12 highlights various applicational problems of the concept ‘Area and Perimeter of Circle’. The usage of the formula of area and perimeter of the circle and the same analytical skill is vital for this part of the chapter. The types of problems covered in Chapter 12 Maths Class 10 are finding radii of a circle that covers the area or perimeter of the sum area or perimeter covered by two other circles, finding the area of rings, and also the calculation of distance covered in the given the number of revolutions.

### Area of Sector and Segment of a Circle

The formula for finding the area of sector and segment has been used in Class 10 Areas Related to Circle in the form of application-based sums.

The part of the circular region enclosed by the two radii and the corresponding arc is called the sector of the circle. The part of the circular region enclosed between a chord and the corresponding arc is called the segment of the circle. The segment is classified as a minor segment and a major segment. The major segment covers a larger area corresponding to the minor segment. Similarly, the sector is classified into minor and major where the major sector covers a larger area.

The area and perimeter significantly depend on the angle subtended by the arc at the centre of the circle. When applicational problems are to be solved, the ability to pull out the angle subtended is crucial. For example, the angle subtended by a minute and hour hand at a certain time in the wall clock is repetitive. Also, a lot of questions are based on how the circle is divided equally, thus the technique to find the angle of each equally divided arc is to be known by the student for earning better marks. These are asked in the forms of segments by umbrella or sectors when a regular polygon is inscribed in a circle.

Although it may seem easy, Chapter 12 Class 10 Maths can easily cause students to make silly mistakes. So, practice is very significant for avoiding losing marks in a section where the possibility of scoring a hundred per cent is pretty high. Areas related to Circle Class 10 Solutions provide problems for practice as well.

### We Cover All The Exercises In The Chapter Given Below:-

Chapter 12 Areas Related to Circles All Exercises in PDF Format | |

Exercise 12.1 | 5 Questions & Solutions |

Exercise 12.2 | 14 Questions & Solutions |

Exercise 12.3 | 16 Questions & Solutions |

### Areas of Combinations of Plane Figures

Areas of Combinations of Plane Figures involves a lot of analytical and critical thinking as the type of question asked in the CBSE Boards from this section can not be predicted. This section mostly involves calculating the area of various designs and patterns using many plane figures and polygons. Mastering the formulas for finding the area of fundamental polygons is also to be concentrated by the student during the exam. NCERT Maths Solution Class 10 Chapter 12 helps students draft preparation and includes most of the formulas required with detailed explanations.

## Download Class 10 Maths Areas Related to Circles NCERT Solutions PDF

Get the free PDF version of the solutions for all the chapters. Once downloaded, you can access them anytime according to your study schedules. Focus on how to proceed with the preparation of Class 10 Maths Areas Related to Circles chapter and develop more confidence to solve its problems.

## FAQs

### What is the formula of ch 12 area related to circle class 10? ›

Area of a circle is **πr ^{2}**, where π=22/7 or ≈ 3.14 (can be used interchangeably for problem-solving purposes) and r is the radius of the circle. π is the ratio of the circumference of a circle to its diameter.

**How many exercises are there in areas related to circles Class 10? ›**

There are **3 exercises** in class 10 math chapter 12 (Areas Related to Circles). In first exercise (Ex 12.1), there are only 5 questions. In second exercise (Ex 12.2), there are in all 14 questions. In third exercise (Ex 12.3), there are in all 16 questions.

**How is area related to circle grade 10? ›**

You know that area of a circle (in fact of a circular region or disc) is **πr2.** **360 r θ × π .** **θ πr** , where r is the radius of the circle and θ the angle of the sector in degrees.

**What is the area of the circle solutions? ›**

The area of a circle is pi times the radius squared (**A = π r²**).

**Why are of circle is πr2? ›**

The usual definition of pi is the ratio of the circumference of a circle to its diameter, so that **the circumference of a circle is pi times the diameter, or 2 pitimes the radius**. ... This give a geometric justification that the area of a circle really is "pi r squared".

**What are all the formulas for area? ›**

Figures | Area Formula | Variables |
---|---|---|

Area of Rectangle | Area = l × w | l = length w = width |

Area of Square | Area = a^{2} | a = sides of the square |

Area of a Triangle | Area = 1/2 b×h | b = base h = height |

Area of a Circle | Area = πr^{2} | r = radius of the circle |

**Which exercises are deleted for class 10 maths? ›**

**Deleted Topics — Euclid's division lemma, Decimal representation of rational numbers as terminating or non-terminating recurring decimals.**

- 2.) Chapter - 2. Polynomials.
- 3.) Chapter - 3.
- Pair of linear equations in two variables.
- 4.) Chapter - 4. Quadratic equations. ...
- 5.) Chapter - 5. Arithmetic Progressions. ...
- 6.) Chapter - 6.

**How many circles are there in 2 points? ›**

**Infinitely many** circles can be drawn passing through two points, as shown in the figure. Q.

**How many circles can be made with 3 points? ›**

It is observed that **only a unique circle** will pass through all three points.

**What is area easy answer? ›**

Area is defined as **the total space taken up by a flat (2-D) surface or shape of an object**. The space enclosed by the boundary of a plane figure is called its area. The area of a figure is the number of unit squares that cover the surface of a closed figure. Area is measured in square units like cm² and m².

### What grade level is area of a circle? ›

Area of a circle - Common Core: **7th Grade** Math.

**What are the 2 formulas for area of a circle? ›**

Area of a circle can be calculated by using the formulas: **Area = π × r ^{2}, where 'r' is the radius.**

**Area = (π/4) × d**.

^{2}, where 'd' is the diameter**How do you solve a circle equation step by step? ›**

To find the equation of a circle when you know the radius and centre, use the formula **( x − a ) 2 + ( y − b ) 2 = r 2** , where represents the centre of the circle, and is the radius. This equation is the same as the general equation of a circle, it's just written in a different form.

**What is the formula for solving a circle? ›**

Explanation: The formula for the equation of a circle is **(x – h) ^{2}+ (y – k)^{2} = r^{2}**, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.

**What is 2πr of a circle? ›**

The Circumference (or) perimeter of circle = 2πR

Pi (π) is a special mathematical constant; it is **the ratio of circumference to diameter of any circle**.

**How does 3.14 relate to a circle? ›**

Pi Day will be celebrated on March 14—3.14. Pi (Greek letter “π”) is the symbol used in mathematics to represent a constant—**the ratio of the circumference of a circle to its diameter**—which is approximately 3.14159.

**What is the value of 2πr? ›**

Circumference of a circle = 2πr, where, 'r' is the radius of the circle and π(pi) is a special mathematical constant with a value approximated to 3.14159 or π = 22/7.

**What is the easiest way to find area? ›**

Area is calculated by **multiplying the length of a shape by its width**. In this case, we could work out the area of this rectangle even if it wasn't on squared paper, just by working out 5cm x 5cm = 25cm² (the shape is not drawn to scale).

**What is the formula for area answer? ›**

The area is measurement of the surface of a shape. To find the area of a rectangle or a square you need to multiply the length and the width of a rectangle or a square. Area, A, is **x times y**.

**How can I get 95% in class 10? ›**

**Solve each paper as if it were your final**. You will become familiar with solving the paper within three hours. You will be able to score 95% if you practice and learn time management skills by solving previous year question papers.

### How can I get 100 marks in maths in 10th? ›

**How to Score 100 in Class 10 Maths**

- Step 1: Make Notes for Formulae, Theorems and Methods using NCERT.
- Step 2: Try To Find Solutions of NCERT Exemplar Yourself.
- Step 3: Practicing From The Sample And Previous Year Papers.
- Step 4: Set the Target to Know Your Weak Points.
- Step 5: Cover All Your Syllabus and Be Confident.

**What if I fail in maths in 10th boards? ›**

A candidate who has failed In the examination in the first attempt shall be required, to re-appear in all the subjects at the subsequent annual examination of the Board.

**What are 3 parts of a circle called? ›**

**Radius, diameter, center, and circumference**--all are parts of a circle.

**What are the 7 parts of a circle? ›**

What are the parts of a circle? The parts of a circle are the **radius, diameter, circumference, arc, chord, secant, tangent, sector and segment**.

**Do 3 points always make a circle? ›**

**Three points uniquely define a circle**. If you circumscribe a circle around a triangle, the circumcenter of that triangle will also be the center of that circle.

**Can you split a circle into 3? ›**

Take your ruler and draw a perpendicular line from this point to the outside diameter of the circle. This will give you two points. Connect these two points to the center point and you have the circle divided into three.

**How do you find the equation of a circle passing through 3 points? ›**

Let the three given points be A,B,C respectively. The general form of the equation of the circle is **x2+y2+2gx+2fy+c=0** where (−g,−f) is the centre of the circle. We need to substitute the given points in the place of (x,y) in the equation of a circle.

**How many circles can be drawn from 4 points? ›**

So, at most **one circle** can be drawn through a given set of four distinct points. Q. The least number of non-collinear points required to draw a unique circle passing through them is: Q.

**What is area answer Brainly? ›**

Explanation: Area is **the amount of space occupied by a two-dimensional figure**. In other words, it is the quantity that measures the number of unit squares that cover the surface of a closed figure.

**How many types of area are there? ›**

...

Area of 2D Shapes Formula.

Shape | Area | Terms |
---|---|---|

Square | a^{2} | a = length of side |

Rectangle | l × w | l = length w = width |

Parallelogram | b × h | b=base h=vertical height |

Trapezium | ½(a+b) × h | a and b are the length of parallel sides h = height |

### What grade is 60% in a level maths? ›

B 70% 79% C **60%-69%** D 50%-59% E 40%-49% (this is considered a pass)

**What percentage is a grade A in A level maths? ›**

**What is S1 and S2 in circle? ›**

**S1 is tangent to x-axis and S2 is tangent to y axis** and the straight line y = mx touches both circle at their common point.

**How many theorems are there in circles Class 10? ›**

Thus, the **two** important theorems in Class 10 Maths Chapter 10 Circles are: Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

**How many formulas are there in areas related to circles? ›**

Circle formulae:

Circumference of a circle = πd or 2πr, where d is diameter and r is a radius of the circle. Perimeter of a semicircle = πr + 2r. **Area of a circle = πr²** Area of a semicircle = 1/2 πr²

**How do you solve the circle theorem 2? ›**

When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference. So **angle AOB = 2 × angle ACB**.

**What is the area of 4 circle? ›**

The area of a circle with a diameter of 4 inches is 4π square inches, or approximately **12.566 square inches**.

**What is value of pi? ›**

Regardless of the circle's size, this ratio will always equal pi. In decimal form, the value of pi is **approximately 3.14**. But pi is an irrational number, meaning that its decimal form neither ends (like 1/4 = 0.25) nor becomes repetitive (like 1/6 = 0.166666...). (To only 18 decimal places, pi is 3.141592653589793238.)

**What is the equation of a circle PDF? ›**

**x2 + y2 = r2** , and this is the equation of a circle of radius r whose centre is the origin O(0, 0). The equation of a circle of radius r and centre the origin is x2 + y2 = r2 .

**What is C in circle equation? ›**

Let C(h, k) be **the centre of the circle** and P(x, y) be any point on the circle. Therefore, the radius of a circle is CP.

### What is H and K in a circle? ›

(h, k) is the center. r is the radius. (x, y) is any point on the circle. All points (x, y) on the circle are a fixed distance (radius) away from the center (h, k). The h value of your center is the first value of your ordered pair and the k value of your center is the second value of your ordered pair.

**How do you find the area of a 12 circle? ›**

First you need to find the radius, the key term to find the radius is diameter=2⋅radius ; therefore the radius r is 122=6 inches. To find the area of a circle, you need to use the formula **A=π⋅r2** ; therefore the area A is π⋅62≈113 square inches.

**What is the formula for circles Class 10? ›**

**Circumference of a circle = 2 π r**. Area of a circle = π r. Arc length of sector of circle with radius r and angle θ is ( θ/360) x 2 π r. The area of sector of a circle with radius 'r' and θ angle = ( θ/360) x π r.

**What is the equation of circle class 12? ›**

General form of Equation of a Circle

Therefore, **x ^{2} + y^{2} + 2gx + 2fy + c = 0**, represents the circle with centre (−g,−f) and radius equal to a

^{2}= g

^{2}+ f

^{2}− c.

**How do you find the area of a 12? ›**

The area is measurement of the surface of a shape. To find the area of a rectangle or a square you need to **multiply the length and the width of a rectangle or a square**. Area, A, is x times y.

**What is the area of 4 diameter circle? ›**

Answer and Explanation: The area of a circle with a diameter of 4 inches is **4π square inches**, or approximately 12.566 square inches. In general, if a circle has a diameter of length d units, then the area, A, of the circle can be found using the following formula: A=14πd2.

**How much area is a 16 ft diameter circle? ›**

Multiply π (pi) by the diameter square: π(16 ft)² = **804.25 ft²**. Divide that result by four: 804.25 ft²/4 = 201.06 ft². That's the square footage of your circle.

**What is the area of 12 by 8? ›**

Summary: The area of a rectangle that is 8 inches wide and 12 inches long is **96 square inches**.

**What is the formula for full circle? ›**

We know that the general equation for a circle is **( x - h )^2 + ( y - k )^2 = r^2**, where ( h, k ) is the center and r is the radius.

**How do you solve a circle equation? ›**

To find the equation of a circle when you know the radius and centre, use the formula **( x − a ) 2 + ( y − b ) 2 = r 2** , where represents the centre of the circle, and is the radius.