NCERT Solutions for Class 10 Maths Chapter 11 - Areas Related to Circles

Exercise 11.1: 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.

Access NCERT Solutions for Class 10 Maths Chapter 11 – Areas Related to Circles

Exercise 11.1

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 coloured 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.

NCERT Solutions for Class10 Maths Chapter 11 PDF

In Class 10 Maths, the chapter "Areas Related to Circles" helps students understand how to calculate the area and perimeter (circumference) of circles and their parts. This includes learning to find the area of a whole circle, as well as parts of circles like sectors and segments. Understanding these concepts is useful not only for solving math problems but also for real-life applications, such as determining the size of circular objects like wheels, gardens, or pizzas. The chapter breaks down complex shapes into simpler parts to make calculations easier and more intuitive.

Area and Perimeter of Circle

NCERT Class 10 Maths Chapter 11 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 11 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 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.

Overview of Deleted Syllabus for CBSE Class 10 Maths Areas Related to Circles

Class 10 Maths Chapter 11: Exercises Breakdown

Conclusion

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles by Vedantu provides students with clear and detailed explanations to tackle problems involving the areas of circles, sectors, and segments. Understanding these concepts is crucial as they form the basis for various practical applications and higher-level geometry. Key points to focus on include mastering the formulas for the area of a circle, the area of a sector, and the area of a segment.

Additionally, practising the problems regularly will help solidify these concepts. In previous years' exams, around 3–4 questions have been asked from this chapter, often focusing on real-life applications of calculating areas. Ensuring a strong grasp of these topics will help students perform well in their exams.

Other Study Material for CBSE Class 10 Maths Chapter 11

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.