Mensuration Practice Test 6

Let the perimeter of the square = Perimeter of the equilateral triangle = 48 cm.
Each side of the square = 12 cm
Diagonal = 4a = 2d $\sqrt{2}$
$\frac{4\times 12}{2}$ = $\mathit{d}\sqrt{2}$
24 = $\mathit{d}\sqrt{2}$
$12\sqrt{2}$ = d
Area of circle = $\mathit{\pi }{\left(6\sqrt{2}\right)}^2$ = 72π
Each side of the equilateral triangle = 16 cm
Area of the triangle = $\frac{\sqrt{3}}{4}\times 16\times 16$ = $\frac{64\sqrt{3}}{3}$
$\frac{1}{2}\times 16\times \mathit{h}$ = $\frac{64\sqrt{3}}{3}$
h = $\frac{64}{3}\times \sqrt{3}\times \frac{2}{16}$
h = $\frac{8}{\sqrt{3}}$
Now by Pythagoras theorem,
${\mathit{h}}^2+8^2$ = ${\mathit{r}}^2$
$\frac{64}{3}+64$ = ${\mathit{r}}^2$
r = $\sqrt{\frac{64+192}{3}}$
r = $\sqrt{\frac{256}{3}}$ = $\frac{16}{\sqrt{3}}$
Now area of the circle = π × $\frac{16}{3}\times 16:72\mathit{\pi }$
= $\frac{2}{3}\times 16:9$
= 32 : 27
= 27 : 32