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Class 10 > In a triangle ABC, P divides the sides AB such that...
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In a triangle ABC, P divides the sides AB such that AP : PB = 1 : 2, Q is a point on AC such that PQ ? BC. Find the ratio of the areas of ?APQ and trapezium BPQC.

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Since AP/PB=1/2 So, AP/AB=1/3 Triangles APQ and ABC are similar (AA rule) because angle APQ=angle ABC... corresponding angles & angle A =angle A.. common so the ratio of areas of triangles APQ and ABC=square of the ratio of the corresponding sides=square of 1/3=1/9 so if area of triangle...
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Since AP/PB=1/2 So, AP/AB=1/3 Triangles APQ and ABC are similar (AA rule) because angle APQ=angle ABC... corresponding angles & angle A =angle A.. common so the ratio of areas of triangles APQ and ABC=square of the ratio of the corresponding sides=square of 1/3=1/9 so if area of triangle APQ=x then the area of triangle ABC=9x From this we get, area of trapezium BPQC= area of triangle ABC- the area of triangle APQ = 9x-x= 8x so required area=area triangle ABC/area trapezium= x/8x= 1/8 read less
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