sq. units. In A ABC, points Pand Q are on sides AB and AC respectively such that PQ|| BC. If AP: PB =1:2 and ar(APQ) = 6 sq. units, then ar(PBCQ) = (a) 12 (b) 18 (c) 36 (d) 48
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Step-by-step explanation:
If a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same ratio.
As PQ∥BC
So
PB
AP
=
QC
AQ
∠AQP=∠ACB
∠APQ=∠ABC
So by AAA △AQP∼△ACB
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Hence
Area(ABC)
Area(APQ)
=
(AB)
2
(AP)
2
Area(ABC)
Area(APQ)
=
(AP+PB)
2
(AP)
2
Area(ABC)
Area(APQ)
= (3x) 2 (x) 2
Area(ABC)
Area(APQ)
= 91
Let Area(APQ)=k
Area(ABC)=9k
Area(BPQC)=Area(ABC)−Area(APQ)=9k−k=8k
Area(BPQC)
Area(APQ)
= 81
∴ the ratio of the △APQ and trapezium BPQC = 81
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