Q22 of 56 Page 6

In Δ ABC, P divides the side AB such that AP: PB = 1: 2. Q is a point in AC such that PQ∥BC. Find the ratio of the areas of Δ APQ and trapezium BPQC.

Given: AP: PB = 1: 2


To find:


Theorem Used:


1.) If two corresponding angles of two triangles are equal, the triangles are said to be similar.


2.) The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides.


Explanation:



We know PQ∥BC,



In ∆APQ and ∆ABC


∠A=∠A [Common]


∠APQ=∠B [Corresponding angle]


∆ABC ~ ∆APQ


From theorem 2.) stated above,




⇒ 9 ar(∆APQ) = ar(∆APQ) +ar (trapezium BPQC)


⇒ 9 ar(∆APQ)- ar(∆APQ) = ar(trapezium BPQC)


⇒ 8 ar(∆APQ) =ar (trapezium BPQC)



More from this chapter

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In Δ ABC, ∠A is obtuse, PB⊥AC and QC⊥AB. Prove that:

(i) AB×AQ=AC×AP


(ii) BC2 =(AC×CP+AB×BQ)

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