Question

In a triangle ABC P divides the side AB such that AP: PB is equal to 2 : 3.Q is a point on AC such that PQ is parallel to BC. The ratio of the areas of triangle A P Q and trapezium BPQC is

Answer 1

HELLO DEAR,

In ΔAPQ  and ΔABC

∠APQ =∠B         [corresponding angles]

∠PAQ =∠BAC    [common]

ΔAPQ∼ΔABC  [By AA Similarity criterion]

• Using the thales theorem

ar(ΔAPQ) / ar(ΔABC) = AP²/AB²

Let AP = 2x and PB = 3x

AB = AP + PB  

AB = 2x + 3x

AB = 5x

ar(ΔAPQ) / ar(ΔABC) = (2x)²/ (5x)²

ar(ΔAPQ) / ar(ΔABC) = 4x²/25x²

ar(ΔAPQ) / ar(ΔABC) = 4/25

Let Area of ΔAPQ  = 2x sq. units

and Area of ΔABC = 25x sq.units

ar[trap.BPQC] = ar(ΔABC  ) – ar(ΔAPQ)

ar[trap.BPQC] = 25x - 4x

ar[trap.BPQC] = 21x sq units

Now,

arΔAPQ/ar(trap.BCED) = x sq.units/21x sq.units

arΔAPQ/ar(trap.BCED) = 4/21

arΔAPQ : ar(trap.BCED) = 4 : 21  

 Hence, the ratio of the areas of ΔAPQ and the trapezium BPQC 4 : 21 

I HOPE IT'S HELP YOU DEAR,

THANKS