NCERT Solutions for Class 10 Maths Chapter 12 - Areas Related To Circles

Exercise 12.1: This exercise involves questions that are based on finding the perimeter and area of circles, semicircles, and quarter circles. The exercise contains a total of six questions that test the student's understanding of the formulas and concepts related to circles and their properties. The solutions to each question provide step-by-step instructions on how to find the perimeter and area of the given figures.

Exercise 12.2: This exercise involves questions based on finding the area of shaded regions, which consist of circles, triangles, and rectangles. The exercise contains a total of five questions that test the student's understanding of the concepts related to the area of circles and their related figures. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

Exercise 12.3: This exercise focuses on questions based on finding the area and circumference of concentric circles, length of the chord of a circle, and finding the radius of the circle. The exercise contains a total of six questions that test the student's understanding of the formulas and concepts related to circles and their properties. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

Exercise 12.4: This exercise involves questions based on finding the areas of sectors and segments of a circle. The exercise contains a total of seven questions that test the student's understanding of the formulas and concepts related to circles and their properties. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

Overall, the exercises in NCERT Solutions for Class 10 Maths Chapter 12 "Areas Related to Circles" are designed to help students develop their understanding of the concepts and formulas related to circles and their properties, and to enhance their problem-solving skills.

Access NCERT Solutions for Class 10 Mathematics Chapter 12 – Areas related to circles

NCERT Solutions Class 10 Math Chapter 12 Areas Related To Circle

NCERT Solutions for Class 10 Math Chapter 12 Areas Related to Circles is an important study material required for the students to study inClass 10. These NCERT Solutions for Class 10 Math help the students in understanding the topics covered in the chapter in a better way. The main purpose of delivering NCERT solutions for Class 10 Maths Chapter 12 Area related to circles is to help students in preparing for their academic and competitive examination.

The NCERT Solutions for Class 10 Math, available at Vedantu, will help students ace their Class 10 Board examination by preparing for them well in advance. Get free Math NCERT Solutions for Class 10 Chapter 12 Areas Related To Circle and clear your doubts for all the complex topics covered in the chapter. . To assist the students of Class 10 for preparing the examination, the  subject experts at Vedantu have formulated these NCERT Solutions based on the latest syllabus issued by the CBSE board

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:

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

2. The formula to calculate the area of circle is πr2

3. 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.

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

5. 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

1. 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:

1. The total length of the silver wire required.

2. 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 cm². [\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)

1. 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 \[\frac{{2804}}{7}m\].

2. 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.

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

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:-

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. 

Benefits of Following the Exercise Solutions for Class 10 Maths Areas Related to Circles

• Exercise problems solved in a simpler way for easy learning and self-sufficiency in problem-solving.

• Solutions developed to conveniently prepare for the chapter and resolve doubts instantly.

• Access expert-framed solutions to clarify exercise questions and enhance understanding.

• Utilize solutions to improve time-efficient problem-solving skills and manage exam time effectively.

• Important chapter introducing concepts of sector and segment, integral in various mathematical areas.

• Circular region concepts used in trigonometry and calculus applications.

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• Practice exercises in the textbook or on Vedantu’s platform for reinforcement and skill development.

Conclusion

Vedantu's NCERT Solutions for Class 10 Maths Chapter 6 - Triangles provide a comprehensive and accessible resource for students to grasp the intricacies of triangle geometry. With a diverse range of well-structured exercises and step-by-step explanations, these solutions promote a deeper understanding of key concepts. By incorporating real-life applications, students can appreciate the relevance of triangles in everyday scenarios. Vedantu's expertly crafted solutions foster self-confidence in solving complex problems, bolstering students' problem-solving abilities. The user-friendly platform encourages interactive learning, making the study process engaging and enjoyable. As a reliable aid, Vedantu's NCERT Solutions for Class 10 Maths Chapter 6 empower students to excel in their academics and develop a strong foundation in geometry.