NCERT Solutions for Class 10 Maths Chapter 12 Area Related to Circles

Ans : Radius \( \left(r_{1}\right)\) of the 1st circle = 19cm
Radius \( \left(r_{2}\right)\) of the 1st circle = 9 cm
Let the radius of 3rd circle be r.
Circumference of 1st circle = \( 2 \pi r_{1}=2 \pi(19)\) = \( 38 \pi\)
Circumference of 2nd= \( 2 \pi r_{1}=2 \pi(9)\) = \( 18 \pi\)
Circumference of 3rd circle =\( 2r\pi\)
That given ,
Circumference of 3rdcircle = Circumference of 2nd circle  = Circumference of 1st circle
\( 2 \pi r=38\pi+18\pi=56 \pi\)
\( r=\frac{56 \pi}{2 \pi}=28\)
Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm. 

Ans : Radius \( \left(r_{1}\right)\) of the 1st circle = 8 cm
Radius \( \left(r_{2}\right)\) of the 1st circle = 6 cm
Let the radius of 3rd circle be r.
Area of 1ts circle = \( \pi r_{1}^{2}=\pi(8)^{2}=64 \pi\)
Area of 1ts circle =  \( \pi r_{1}^{2}=\pi(6)^{2}=36 \pi\)
That given,
Area of 3rd circle =Area of 2nd circle  =Area of 1st circle
\( \pi r^{2}=\pi r_{1}^{2}+\pi r_{2}^{2}\)
\( \pi r^{2}=64 \pi+36 \pi\)
\( \pi r^{2}=100 \pi\)
\( r=\pm 10\) 
However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm. 

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

Ans : Diameter of the wheel of the car = 80 cm 
Radius (r) of the wheel of the car = 40 cm 
Circumference of wheel =\( 2 \pi r\)
\( =2 \pi(40)=80 \pi \mathrm{cm}\)
Speed of car = 66 km/hr
\( =\frac{66 \times 100000}{60} \mathrm{cm} / \mathrm{min}\)
\( =\frac{35000}{8}=4375\)
Therefore , each wheel of the car will make 4375 revolutions. 

Ans : Let the radius of the circle be r. 
Circumference of circle = \( 2 \pi r\)
Area of circle =\( \mathrm\pi r^{2}\)
Given that, the circumference of the circle and the area of the circle equal. 
This implies  \( 2 \pi r\) = \( \mathrm\pi r^{2}\)
r=2
Therefore, the radius of the circle is 2 units. 
Hence, the correct answer is A.