In the given figure, ABCD is a rectangle and P is a point on DC such that BC = 24 cm, DP = 10 cm and CD = 15 cm. If AP produced intersects BC produced at Q, then the length of AQ.

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let the length of }}CQ = x\,\& \,\angle PQC = \theta \cr & {\text{In }}\Delta ABQ, \cr & \tan \theta = \frac{{15}}{{24 + x}}.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{Again in }}\Delta PCQ \cr & \tan \theta = \frac{5}{x}\,.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & {\text{From equation}}\left( {\text{i}} \right)\,\& \,\left( {{\text{ii}}} \right) \cr & \frac{{15}}{{24 + x}} = \frac{5}{x} \cr & 3x = 24 + x \cr & 2x + 24 \cr & x = 12 \cr & {\text{In }}\Delta ABQ, \cr & {\text{By pythagoras theorem}} \cr & A{Q^2} = A{B^2} + B{Q^2} \cr & = {15^2} + {36^2} \cr & = 225 + 1296 \cr & = 1521 \cr & AQ = 39 \cr} $$