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 find the length of AQ.

Given : 

A rectangle ABCD in which BC = 24, DP = 10 cm and CD = 15 cm 

CP = 15 - 10 = 5  

Concept used :

The similarity of a triangle  

If values of corresponding angles between two triangles are the same then triangles are said to be similar 

If two triangles are similar then the ratio of their corresponding sides are the same 

Calculations :

In ∆ADP and ∆QCP 

∠ADP = ∠QCP (both are 90°)

∠APD = ∠QPC (vertically opposite angles)

As AD||BQ, then 

∠PAD = ∠PQC (alternate angles) 

So, ∆ADP and ∆QCP are similar triangles 

Now,

AD/QC = DP/CP = AP/QP 

In ∆APD 

AD² + DP² = PA² 

24² + 10² = PA2   (AD = BC opposite side of a rectangle)

PA = 26 

⇒ 24/QC = 10/5 = 26/QP 

⇒ QP = 13 

Now, 

AQ = AP + PQ = 26 + 13  

⇒ 39 

∴ AQ will be 39 cm