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?

Answer: Option A

Solution

Answer: Option A

Solution:

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