Heights and Distances Practice Test 1

Let AB = 5m and CD = 2m be the heights of a lamp post and the men respectively. At any time t BD = x m and shadow of man ED = y m.
Then, $\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=6\mathrm{m}∕\min$
Now, right triangles ABE and CDE are similar; then
$\frac{\mathit{A}\mathit{B}}{\mathit{C}\mathit{D}}=\frac{\mathit{B}\mathit{E}}{\mathit{D}\mathit{E}}\Rightarrow \frac{5}{2}=\frac{\mathit{x}+\mathit{y}}{\mathit{y}}$
$\Rightarrow 5\mathit{y}-2\mathit{y}=2\mathit{x}$
$\Rightarrow 3\mathit{y}=2\mathit{x}$
$\Rightarrow 3\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=2\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}$
$\Rightarrow 3\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=2\times 6$
$\Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=4$
Hence, length of his shadow increases at the rate of 4m/min.