SSC CPO SI and ASI Model Paper 9

(4) Length of wire = Perimeter oftriangle = Perimeter of square= Perimeter of hexagon= Circumference of circle
Area of equilateral triangle
$=\frac{\sqrt{3}}{4}{\left(\mathit{s}\mathit{i}\mathit{d}\mathit{e}\right)}^2=\frac{\sqrt{3}}{4}\times \frac{102}{3}\times \frac{102}{3}$
$=289\times \sqrt{3}=500.56\mathit{c}{\mathit{m}}^2$
Area of square = side × side$=\frac{102}{4}\times \frac{102}{4}=650.25\mathit{c}{\mathit{m}}^2$
Area of regular hexagon
$=\frac{3\sqrt{3}}{2}\times {\left(\mathit{s}\mathit{i}\mathit{d}\mathit{e}\right)}^2$
$=\frac{3\sqrt{3}}{2}\times \frac{102}{6}\times \frac{102}{6}$
$=750.84\mathit{c}{\mathit{m}}^2$
Area of circle $=\mathit{\pi }{\mathit{r}}^2$
$=\mathit{\pi }\times \frac{102}{2\mathit{\pi }}\times \frac{102}{2\mathit{\pi }}$
$=\frac{2601\times 7}{22}=827.59\mathit{c}{\mathit{m}}^2$
$\left[\therefore 102=2\mathit{\pi }\mathit{r}\Rightarrow \mathit{r}\frac{102}{2}\mathit{\pi }\right]$Hence, circle covers the largestsurface area