Question

The sum of diameter of two circle is 35 cm .and difference of their circumference is 22cm.what is are of smaller circle?

Answer 1

Let the diameter of the smaller circle be $d_{1}$ and the diameter of the larger circle be $d_{2}$.

       According to the question : -

Sum of both the diameters = 35 cm

Diameter of smaller circle + Diameter of larger circle = 35 cm

$d_{1}+ d_{2}$ = 35 cm ---: ( 1 )

Difference of their circumference = 22 cm

Circumference of larger circle - Circumference of smaller circle = 22 cm

$\pie d_{2} - \pie d_{1}$ = 22cm

$\pie ( d_{2} - d_{1} )$ = 22 cm

$\dfrac{22}{7} d_{2}-d_{1}$ = 22 cm

$d_{2}-d_{1} = 22 \: cm \times \dfrac{7}{22}$

$d_{2}-d_{1}$ = 7 cm  ---: ( 2 )

      Subtract ( 2 ) from ( 1 ) : -

$d_{1}+d_{2}$ = 35 cm

$d_{2} - d_{1}$ = 7cm

-           +    = -  

________________

$2\: d_{1}$= 28 cm

________________

$d_{1} =\dfrac{28\:cm}{2}$

$d_{1} = 14 cm$

Therefore the diameter of the smaller circle is 14 cm.

diameter with reference to smaller circle = 14 cm

2 Radius with reference to smaller circle = 14 cm

Radius with reference to smaller circle = 14 / 2 cm

Radius with reference to smaller circle = 7 cm

Now,

Area of smaller circle = πr^2

          = 22 / 7 × ( 21 / 2 cm )^2

         = 22 / 7 × 7 × 7 cm^2

         = 22 × 7 cm^2

         = 154 cm^2

Therefore, area of the smaller circle is 154 cm^2

Answer 2

$if \: d1 = x \\ then \: d2 = 35 - x \\ r1 = x \div 2 \\ r2 =( 35 - x) \div 2 \\ cr1 = 2\pi \: r 1= \pi \: x \\ cr2 =2\pi \: r2 =2 (35 - x) \div 2 = 35\pi - x\pi \\ cr2 - cr1 = 22 \\ 35\pi - \pi \: x - \pi \: x = 22 \\ 35 - 2x\pi = 22 \\ 35 \times \frac{22}{7} - 2 \times \frac{22}{7} x = 22 \\ 110 - \frac{44}{7} x = 22 \\ 110 - 22 = \frac{44}{7} x \\ 88 \times \frac{7}{44} = x \\ x = 14 \: cm \\ therefore \: d1 = 14 \: and \: d2 = 21\\ hence \: the \: smaller \: area = \pi {r}^{2} \\ = \frac{22}{7} \times 7 \times 7 \\ = 154 {cm}^{2}$