Question

A man of height $1.8m$ is standing near a Pyramid. If the shadow of the man is of length $2.7m$ and the shadow of the Pyramid is $210m$ long at that instant, find the height of the Pyramid

Answer 1

Let $AB$ and $DE$ be the heights of the Pyramid and the man respectively.
Let $BC$ and $EF$ be the lengths of the shadows of the Pyramid and the man respectively.
In $△ABC$ and $△DEF$, we have
$\angle ABC=\angle DEF={90}^{\circ }$
$\angle BCA=\angle EFD$ (angular elevation is same at the same instant)
$\therefore △ABC\sim △DEF$ ($AA$ similarity criterion)
Thus, $\frac{AB}{DE}=\frac{BC}{EF}$
$\Rightarrow \frac{AB}{1.8}=\frac{210}{2.7}\Rightarrow =\frac{210}{2.7}\times 1.8=140$.
Hence, the height of the Pyramid is $140m$.