Question

If the radius of a circle is diminished by $10\%$, then its area is diminished by?

Answer 1

Step 1: State the given data and calculate the new radius

It is given that the radius of the circle is diminished by $10\%$.

Let, the original radius be $r$.

Then, the area of the original circle $\left(A\right)=\mathrm{\pi }{r}^2$

Now, the new radius $\left(r_n\right)$ can be calculated as,

$\Rightarrow r_n=r–\left(10\%\mathrm{of}r\right)$

$\Rightarrow r_n=r–\frac{10}{100}\times r$ $\left[\because P\%=\frac{P}{100}\right]$

$\Rightarrow r_n=\frac{100r-10r}{100}$

$\Rightarrow r_n=\frac{9r}{10}$

Step 2: Calculate the new area of the circle

Now, the new area $\left(A_n\right)$ of the circle can be calculated as,

$A_n=\mathrm{\pi }{\left({\mathrm{r}}_{\mathrm{n}}\right)}^2$ $\left[\because A=\pi r^2\right]$

$\Rightarrow A_n=\mathrm{\pi }{\left(\frac{9\mathrm{r}}{10}\right)}^2$

$\Rightarrow A_n=\mathrm{\pi }\left(\frac{81{\mathrm{r}}^2}{100}\right)$

$\Rightarrow A_n=\frac{81}{100}\mathrm{\pi }{r}^2$

Step 3: Calculate the change in the area

Now, change in the area $\left(∆A\right)$ can be calculated as,

$∆A=A-A_n$ $[\because A>A_n]$

$\Rightarrow ∆A=\mathrm{\pi }{r}^2-\frac{81}{100}{\mathrm{\pi r}}^2$

$\Rightarrow ∆A=\frac{100\mathrm{\pi }{r}^2-81{\mathrm{\pi r}}^2}{100}$

$\Rightarrow ∆A=\frac{19}{100}{\mathrm{\pi r}}^2$

$\Rightarrow ∆A=\frac{19}{100}\times A$ $\left[\because A=\pi r^2\right]$

$\Rightarrow ∆A=19\%\mathrm{of}A$ $[\because \frac{P}{100}=P\%]$

i.e., the change in the area $\left(∆A\right)$ is $19\%$ of the area of the original circle $\left(A\right)$.

Hence, the area of the circle is diminished by $19\%$.