Mensuration Practice Test 10

Area of circle = $\mathit{\pi }{\mathit{r}}^2$
⇒ $\mathit{\pi }{\mathit{r}}^2$ = 154
${\mathit{r}}^2$ = $\frac{154\times 7}{22}$
${\mathit{r}}^2$ = 7 × 7
r = 7 cm
Medium of the equilateral triangle is divided in the ratio of 2 : 1.
⇒ AD = 21 cm

Altitude of the equilateral triangle = $\frac{\sqrt{3}\mathit{a}}{4}$
= 21 = $\frac{\sqrt{3}}{2}\mathit{a}$
= $\frac{21\times 2}{\sqrt{3}}$ = a
= $\frac{42}{\sqrt{3}}$ = a
Perimeter of the equilateral triangle = 3 × $\frac{42}{\sqrt{3}}$
= 42 $\sqrt{3}$ cm