Mensuration Practice Test 8

Area of circle = $\mathit{\pi }{\mathit{r}}^2$
$\mathit{\pi }{\mathit{r}}^2$ = x .... (i)
Increase in area and radius of circle
$\mathit{\pi }{\left(\mathit{r}+1\right)}^2$ = x + 22 ... (ii)
Subtracting (ii) from (i),
$\mathit{\pi }{\left(\mathit{r}+1\right)}^2-\mathit{\pi }{\mathit{r}}^2$ = x + 22 - x
$\mathit{\pi }\left({\mathit{r}}^2+1+2\mathit{r}\right)-\mathit{\pi }{\mathit{r}}^2$ = 22
π + 2πr = 22
π (2r + 1) = 22
2r + 1 = $22\times \frac{7}{22}$
2r = 7 - 1
r = 3 cm.