From geometry of figure, shadow length is $C D(=H)$.
From similar triangles $\triangle B G F$ and $\triangle D E F$,
we have $\quad \frac{D E}{B G}=\frac{F E}{G F}$
$\frac{h^{\prime}}{h}=\left(\frac{d-x}{x}\right)$
$\Rightarrow \quad \frac{d}{x}=\frac{h^{\prime}+h}{h} \quad \dots(i)$
Now, from similar triangles $\triangle A B G$ and $\triangle A C E$, we have
$\frac{C E}{A E}=\frac{B G}{A G}$
$\therefore \quad \triangle A B G \cong \triangle F B G$
and $\quad A G=G F=x$
$\Rightarrow \frac{H+h^{\prime}}{d+x}=\frac{h}{x}$
$\Rightarrow \frac{d}{x}=\frac{H+h^{\prime}-h}{h} \quad \dots(ii)$
Equating Eqs. $(i)$ and $(ii)$, we get
$\frac{h^{\prime}+h}{h}=\frac{H+h^{\prime}-h}{h}$
$\Rightarrow \quad 2 h=H$
Hence height of shadow on wall is $2h$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
