Question

The area enclosed between two concentric circles is 770$c{{m}^{2}}$. If the radius of the outer circle is 21 cm, then calculate the radius of the inner circle.

Answer 1

Hint: We start solving the problem by recalling the definition of concentric circles. We then assume the radius of the inner circle and draw the figure representing all the information. We then find the area enclosed between the circles by subtracting the area of the inner circle from the area of the outer circle. We then equate the obtained area with 770$c{{m}^{2}}$ and make necessary calculations to get the radius of the inner circle.

Complete step-by-step solution
According to the problem, we are given that the area enclosed between two concentric circles is 770$c{{m}^{2}}$. We need to find the radius of the inner circle if the radius of the outer circle is 21 cm.
Let us recall the definition of concentric circles. We know that if two or more circles are said to be concentric if they all have the same center.
Let us assume the radius of the inner circle is ‘x’ cm. Now, we draw the figure representing the given information.

We know that the area of the circle with radius ‘r’ is defined as $\pi {{r}^{2}}$.
So, we get the area of the inner circle as $\pi {{x}^{2}}c{{m}^{2}}$ and the area of the outer circle as $\pi {{\left( 21 \right)}^{2}}c{{m}^{2}}$.
Now, we are given that the area between these two circles is 770$c{{m}^{2}}$.
We have area enclosed between the circles as $\left( \pi {{\left( 21 \right)}^{2}}-\pi {{x}^{2}} \right)c{{m}^{2}}$.
Now, we get $\pi {{\left( 21 \right)}^{2}}-\pi {{x}^{2}}=770$.
Let us take the value of $\pi $ as $\dfrac{22}{7}$.
$\Rightarrow \dfrac{22}{7}\times \left( 441-{{x}^{2}} \right)=770$.
$\Rightarrow 22\times \left( 441-{{x}^{2}} \right)=5390$.
$\Rightarrow 441-{{x}^{2}}=\dfrac{5390}{22}$.
$\Rightarrow 441-{{x}^{2}}=245$.
$\Rightarrow {{x}^{2}}=196$.
$\therefore x=14$cm.
So, we have found the radius of the inner circles as 14 cm.

Note: We should not make calculation mistakes while solving this problem. We can solve this problem as follows:
Let us assume the radius of the inner circle be $\left( 21-x \right)cm$.
We get $\pi {{\left( 21 \right)}^{2}}-\pi {{\left( 21-x \right)}^{2}}=770$.
Let us take the value of $\pi $ as $\dfrac{22}{7}$.
$\Rightarrow \dfrac{22}{7}\times \left( {{21}^{2}}-{{\left( 21-x \right)}^{2}} \right)=770$.
$\Rightarrow 22\times \left( 441-\left( 441-42x+{{x}^{2}} \right) \right)=5390$.
$\Rightarrow 42x-{{x}^{2}}=245$.
$\Rightarrow {{x}^{2}}-42x+245=0$.
$\Rightarrow {{x}^{2}}-35x-7x+245=0$.
$\Rightarrow x\left( x-35 \right)-7\left( x-35 \right)=0$.
$\Rightarrow \left( x-7 \right)\left( x-35 \right)=0$.
$\Rightarrow x-7=0$ or $x-35=0$.
$\Rightarrow x=7$ or $x=35$.
$\Rightarrow 21-x=14cm$ or $21-x=-14$.
Since we cannot take negative values for the radius of the circle, we get the radius of the inner circle as 14cm.