Question

two poles 20m and 80m high are 100m apart. find the height of the intersection of the lines joining the top of each pole to the base of opposite pole. this is my sum pls say and help me pls​

Answer 1

please see the attachment

Answer 2

Given : two poles 20m and 80m high are 100m apart.

To find : height

Explanation:

In triangle ABC

AB ║EF

therefore ,

$\bigtriangleup ABC \sim \bigtriangleup EFC \\\\\text{by AA criterion}\\\\\frac{AB}{EF}=\frac{BC}{FC}\\\\\frac{a}{h}=\frac{p}{FC}\\\\FC= \frac{ph}{a}..(1)$

in triangle BCD

EF║DC

$\bigtriangleup DCB \sim \bigtriangleup EFB \\\\\text{by AA criterion}\\\\\frac{DC}{EF}=\frac{BC}{FB}\\\\\frac{B}{h}=\frac{p}{FB}\\\\FB= \frac{ph}{b}..(2)$

adding 1 and 2 we get

$FC+ FB = \frac{ph}{a}+\frac{ph}{b}\\\\p = ph(\frac{a+b}{ab})\\\\h = \frac{ab}{a+b}$

here  , a= 80 m and b = 20 m

then

height is

$h = \frac{ab}{a+b}\\\\h =\frac{80\times 20}{80+20}\\\\h =\frac{1600}{100} \\\\h= 16 m$

hence , the height is 16 m

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