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.

Question

Question

This question was previously asked in

SSC CGL Tier 2 Quant Previous Paper 3 (Held On: 18 November 2020)

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Answer 1

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

F1 Savita SSC 08-4-22 D3

∠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

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