Question

In the figure BP:PQ:QC =1:2:1. Area of ∆APQ is 8sqcm.what is the area of ∆ABPand ∆ABC

Answer 1

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)

=

9

1

Let Area(APQ)=k

Area(ABC)=9k

Area(BPQC)=Area(ABC)−Area(APQ)=9k−k=8k

Area(BPQC)

Area(APQ)

=

8

1

∴ the ratio of the △APQ and trapezium BPQC =

8

1

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