There is a tree between houses of A and B. If the tree leans on A’s House, the tree top rests on his window which is 12 m from ground. If the tree leans on B’s House, the tree top rests on his window which is 9 m from ground. If the height of the tree is 15 m, what is distance between A’s and B’s house?

A. 21 m

B. 25 m

C. 16 m

D. 12 m

Answer: Option A

Solution(By Examveda Team)

In $$\Delta $$STR, by Pythagoras theorem
$$\eqalign{ & R{T^2} = S{T^2} + R{S^2} \cr & \therefore S{T^2} = {15^2} - {9^2} = 144 \cr & \therefore ST = 12\,{\text{m}} \cr} $$
In $$\Delta $$TQP, by Pythagoras theorem
$$\eqalign{ & P{T^2} = T{Q^2} + P{Q^2} \cr & \therefore T{Q^2} = {15^2} - {12^2} = 81 \cr & \therefore TQ = 9\,{\text{m}} \cr} $$
Distance between houses
⇒ SQ = ST + TQ
⇒ SQ = 12 + 9
⇒ SQ = 21 m