Question

if two poles 5m and 15m high are 100m apart. then find height of point of intersection of line joining top of each pole to foot of opposite pole

Answer 1

Solution :-

Assuming that 5 m high pole is at the origin and the 15 m high pole is at x = 100 m

Then, the line from the base of the 5 m pole to the top of 15 m high pole is 

⇒ y = 15x/100 ..........(1)

The line from the base of 15 m high pole to the top of 5 m high is

⇒ y = - 5x/100 + 5 ............(2)

Equating these two equations. 

⇒ 15x/100 = - 5x/100 + 5 

⇒ 0.15x = - 0.05x + 5

⇒ 0.15x + 0.05x = 5

⇒ 0.2x = 5

⇒ x = 5/0.2

⇒ x = 25

Putting the value of x = 25 in (1).

⇒ y = 15x/100

⇒ y = (15*25)/100

⇒ y = 375/100

⇒ y = 3.75 m

So, height of point of intersection is 3.75 m

Answer.

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Let

$\textbf{\underline{Origin=5m\;high\;pole}}$

$\textbf{\underline{15m\;high\;pole\;is\;at\;p}}$

= 100 m

$\textbf{\underline{Line\;from\;base\;15m\;pole\;to\;top\;5m\;high\;=\;c}}$

1st situation :-

$\tt{\rightarrow c=\dfrac{15p}{100}....(1)}$

2nd Situation :-

$\tt{\rightarrow -c=\dfrac{-5p}{100}+5.....(2)}$

Solving (1) and (2)

$\tt{\rightarrow\dfrac{15p}{100}=\dfrac{-5p}{100}+5}$

0.15p = -0.05p + 5

0.15p + 0.05p = 5

0.2p = 5

$\tt{\rightarrow p=\dfrac{5}{0.2}}$

p = 25

Substitute the value of p in (1)

$\tt{\rightarrow c=\dfrac{15p}{100}}$

$\tt{\rightarrow c=\dfrac{15\times 25}{100}}$

$\tt{\rightarrow c=\dfrac{375}{100}}$

c = 3.75 m

Hence,

${\boxed{\sf\:{Height\;of\;point\;of\; Intersection=3.75m}}}$