Question

If two vertical poles $20$m and $80$m high stand apart on a horizontal plane, then the height(in m) of the point of intersection of the lines joining the top of each pole to the foot of other is?

• A. $16$
• B. $18$
• C. $50$
• D. $15$

Answer 1

The correct option is A $16$

Let, $AB$ be the first pole whose height $=20m$

$EF$ be the second pole whose height $=80m$

$BD=a$,$DF=b$

Let, the height of the point $=h$

Clearly, $AB||CD||EF$

$△BCD\sim △BEF$

$\frac{CD}{EF}=\frac{BD}{BF}$

$\Rightarrow \frac{h}{80}=\frac{a}{a+b}..........\left(1\right)$

Similarly, from $△FCD\sim △FAB$

$\frac{CD}{AB}=\frac{FD}{FB}$

$\Rightarrow \frac{h}{20}=\frac{b}{a+b}..........\left(2\right)$

Dividing eq $\left(1\right)$ by $\left(2\right)$, we get,

$\Rightarrow \frac{\frac{h}{80}}{\frac{h}{20}}=\frac{\frac{a}{a+b}}{\frac{b}{a+b}}$

$\Rightarrow \frac{1}{4}=\frac{a}{b}$

$\Rightarrow b=4a...........\left(3\right)$

Putting the value of $b$ in equation $\left(1\right)$,we get

$\frac{h}{80}=\frac{a}{a+4a}$

$\Rightarrow \frac{h}{80}=\frac{1}{5}$

$\Rightarrow h=\frac{80}{5}$

$\Rightarrow h=16$

Hence, the required height is $16m$