Question

in triangle ABC , p.divides the side AB such that AP: PB = 1 : 2 nd Q is a point in Ac such that pq parallel bc . find ratio of the areas of triangle APQ ND trap. bpqc ​

Answer 1

Answer:

The ratio of the areas of ΔAPQ and the trapezium BPQC 1: 8  

Step-by-step explanation:

Given :  

In ΔABC , P divides the side AB such that AP : PB = 1: 2, Q is a point on AC on such that PQ || BC.

To prove : The ratio of the areas of ΔAPQ and the trapezium BPQC.

Proof :  

In ΔAPQ  and ΔABC

∠APQ =∠B         [corresponding angles]

∠PAQ =∠BAC    [common]

ΔAPQ∼ΔABC  

[By AA Similarity criterion]

By Using the Theorem, the ratio of the areas of two triangles is equal to the square of the ratio of their corresponding sides.

ar(ΔAPQ) / ar(ΔABC) = (AP/AB)²

ar(ΔAPQ) / ar(ΔABC) = AP²/AB²

Let AP = 1x and PB = 2x

AB = AP + PB  

AB = 1x + 2x

AB = 3x

ar(ΔAPQ) / ar(ΔABC) = (1x)²/ (3x)²

ar(ΔAPQ) / ar(ΔABC) = 1x²/9x²

ar(ΔAPQ) / ar(ΔABC) = 1/9

Let Area of ΔAPQ  = 1x sq. units and Area of ΔABC = 9x sq.units

ar[trap.BPQC] = ar(ΔABC  ) – ar(ΔAPQ)

ar[trap.BPQC] = 9x - 1x

ar[trap.BPQC] = 8x sq units

Now,

arΔAPQ/ar(trap.BCED) = x sq.units/8x sq.units

arΔAPQ/ar(trap.BCED) = ⅛

arΔAPQ : ar(trap.BCED) = 1 : 8  

Hence, the ratio of the areas of ΔAPQ and the trapezium BPQC 1:8