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.

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.

Answer: Option C

Solution

Answer: Option C

Solution:

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