Question

A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?
(a) 2.4 m
(b) 1.35 m
(c) 1.5 m
(d) 13.5 m

Answer 1

(c) 1.5 m

Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF = ?
Now, in right-angled triangles ABC and DEF, we have:
$$\begin{gathered} \angle BAC=\angle EDF=90° \\ \angle ACB=\angle DFE(\mathrm{Angular}\mathrm{elevation}\mathrm{of}\mathrm{the}\mathrm{Sun}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{time}) \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ △ABC~△DEF \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{AB}{AC}=\frac{DE}{DF} \\ \Rightarrow \frac{1.8}{0.45}=\frac{6}{DF} \\ \Rightarrow DF=\frac{6\times 0.45}{1.8}=1.5\mathrm{m} \end{gathered}$$