Two vertical poles A B=15
Question:

Two vertical poles $A B=15 \mathrm{~m}$ and $C D=10 \mathrm{~m}$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $B C$ and $A D$, then the height of $P($ in $\mathrm{m})$ above the line $A C$ is :

  1. (1) $20 / 3$

  2. (2) 5

  3. (3) $10 / 3$

  4. (4) 6


Correct Option: , 4

Solution:

Let $P E \perp A C$ and $\frac{A E}{E C}=\frac{m}{n}$

$\because \Delta A E P \sim \Delta A C D, \frac{m}{P E}=\frac{m+n}{10}$

$\Rightarrow P E=\frac{10 m}{m+n}$……(i)

$\because \Delta C E P \sim \Delta C A B, \frac{n}{P E}=\frac{m+n}{15}$

$\Rightarrow P E=\frac{15 n}{m+n}$…..(ii)

From (i) and (ii),

$10 m=15 n \Rightarrow m=\frac{3}{2} n$

So, $P E=6$

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