Question

The area of the equilateral triangle described on the side of a square is $\frac{1}{k}$ the area of the equilateral triangle described on its diagonal. Find k.

Answer 1

In $\square ABCD$, Equilateral triangles $\Delta BCEand\Delta ACF$ have been described on side BC and diagonal AC respectively
Since $\Delta BCE$ and $\Delta ACF$ are equilateral. Therefore, they are equiangular ( each angle being equal to ${60}^o$) and hence $\Delta BCE\sim \Delta ACF$
$\Rightarrow \frac{Area\left(\Delta BCE\right)}{Area\left(\Delta ACF\right)}=\frac{{BC}^2}{{AC}^2}$
$\Rightarrow \frac{Area\left(\Delta BCE\right)}{Area\left(\Delta ACF\right)}=\frac{{BC}^2}{{\left(\sqrt{2BC}\right)}^2}=\frac{1}{2}$
$\Rightarrow \frac{Area\left(\Delta BCE\right)}{Area\left(\Delta ACF\right)}=\frac{1}{2}$.
Hence proved.