Q. 224.4( 9 Votes )

In ΔABC, P divides the side AB such that AP: PB = 1: 2. Q is a point in AC such that PQBC. Find the ratio of the areas of ΔAPQand trapezium BPQC.

Answer :

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 PQBC,



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)



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