Question

A $5m60cm$ high vertical pole casts a shadow $3m20cm$ long. Find at the same time (i) the length of the shadow cast by another pole $10m50cm$ high (ii) the height of a pole which casts a shadow $5m$ long.

Answer 1

Let $x$ m be the length of shadow of the pole whose height is $10m50cm$.

So, $AB=5m60cm=5.60m,BC=3m20cm=2.20m$
Since, the above two triangles are similar as we are measuring at the same time.
So, the corresponding sides are in proportion.
Clearly, it is a case of direct variation.
$\Rightarrow \frac{AB}{BC}=\frac{PQ}{QR}$
$\Rightarrow \frac{5.60}{3.20}=\frac{10.50}{x}$

$\Rightarrow x=\frac{10.50\times 3.20}{5.60}$
$\therefore x=6$

Hence, the length of the shadow is $6m$.
Similar way,
(ii) Let $ym$ be the length of the pole whose shadow is $5m$ long.

Again the triangle are similar, so the corresponding sides are in proportion.
Thus, $\frac{AB}{BC}=\frac{PQ}{QR}$

$\Rightarrow \frac{5.60}{3.20}=\frac{y}{5}$

$\Rightarrow y=\frac{5.60\times 5}{3.20}$
$\therefore y=8.75$

Hence, the length of the pole is $8.75$ meters.