Question

The radius of a circle is greater than the radius of other circle by 3m.The sum of their area is 89 pie m square.Find the radius of each circle. ​

Answer 1

★Given :

• The radius of a circle is greater than the radius of other circle by 3m.
• The sum of their area is 89πm².

★To find :

• Radius of each circle.

★Solution :

Let 'r' be the radius of smaller circle.

Then, radius of large circle = r+3

We know,

✦Area of circle = πr²

Given that sum of areas is 89πm².

That is,

→πr² + π(r +3)² =  89π m^2

→π[r² + (r +3)²] =  89π

→r² + (r +3)² = 89

→r²  + r² +  6r + 9 = 89

→2r² +  6r + 9 - 89 = 0

→2r² +  6r - 80 = 0

→r² +  3r  - 40 = 0

Solving this quadratic equation,

→r² +  3r  - 40 = 0

→r² +8r -5r - 40 = 0

→r(r+8) - 5(r+8) = 0

→(r-5)(r+8) = 0

→r - 5 = 0  (or) r+8 = 0

→r = 5 and r = -8

We need to consider only +ve values.

Hence r = 5

Therefore,radius of smaller circle = 5m

And radius of larger circle :

→r + 3

→5 + 3

→8cm

Therefore,

• Radius of smaller circle is 5m
• Radius of larger circle is 8m.

 ______________

Answer 2

Given Question :-

• The radius of a circle is greater than the radius of other circle by 3m. The sum of their area is 89 pie m square. Find the radius of each circle.

ANSWER

GIVEN :-

• The radius of a circle is greater than the radius of other circle by 3m.

• The sum of their area is 89 pie m square.

TO FIND :-

• The radius of each circle.

FORMULA USED :-

$\bigstar \: \: \: { \boxed{ \pink{ \bf \: Area_{(circle)} \: = \: \pi \: {r}^{2} }}}$

CALCULATION :-

• Let radius of smaller circle = 'r' m

So,

• Radius of larger circle = (r + 3) m.

According to statement,

$\rm :\implies\:Area_{(larger \: circle)} + Area_{(smaller \: circle)} = 89\pi$

$\rm :\implies\:\pi \: {(r + 3)}^{2} + \pi \: {r}^{2} = 89\pi$

$\rm :\implies\: {(r + 3)}^{2} + {r}^{2} = 89$

$\rm :\implies\: {r}^{2} + 9 + 6r + {r}^{2} = 89$

$\rm :\implies\: {2r}^{2} + 6r - 80 = 0$

$\rm :\implies\: {r}^{2} + 3r - 40 = 0$

$\rm :\implies\: {r}^{2} + 8r - 5r - 40 = 0$

$\rm :\implies\:r(r + 8) - 5(r + 8) = 0$

$\rm :\implies\:(r + 8)(r - 5) = 0$

$\rm :\implies\:r = 5 \: m$

$\begin{gathered}\begin{gathered}\bf \: Hence- \begin{cases} &\sf{radius_{(smaller \: circle)} = \: 5 \: m} \\ &\sf{radius_{(larger \: circle)} \: = \: 8 \: m} \end{cases}\end{gathered}\end{gathered}$