Question

In the given figure, $PQ \parallel BC$ and AP : PB = 1 : 2. Find $\frac {area \triangle APQ} {area\triangle ABC}$.

Answer 1

Answer:

The ar(ΔAPQ)/ar( ΔABC) = 1/9

Step-by-step explanation:

Given:

PQ || BC, AP : PB = 1 : 2  

AP / PB = 1 / 2  

PB = 2 AP …………(1)

In ΔABC, PQ || BC

According to BASIC PROPORTIONALITY THEOREM (BPT) :  

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

AP/AB = AQ/AC  = PQ/BC

AP/AB = AP /(AP + PB)  

AP/AB = AP / (AP + 2 AP)

[From eq 1]

AP/AB = AP / 3AP  

AP/AB = ⅓  ……………..(2)

ΔAPQ ~ ΔABC  

[Two triangles are said to be similar, if their corresponding sides are proportional]

ar(ΔAPQ)/ar( ΔABC) = (AP/AB)²

[The ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.]

ar(ΔAPQ)/ar( ΔABC) = (1/3)²

[From eq 2]

ar(ΔAPQ)/ar( ΔABC) = 1/9

Hence, the ar(ΔAPQ)/ar( ΔABC) = 1/9

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