Q. 224.4( 9 Votes )

# In Δ ABC, P divides the side AB such that AP: PB = 1: 2. Q is a point in AC such that PQ∥BC. Find the ratio of the areas of Δ APQ and trapezium BPQC.

Answer :

**Given:** AP: PB = 1: 2

**To find:**

**Theorem Used:**

1.) If two corresponding angles of two triangles are equal, the triangles are said to be similar.

2.) The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides.

**Explanation:**

We know PQ∥BC,

In ∆APQ and ∆ABC

∠A=∠A [Common]

∠APQ=∠B [Corresponding angle]

**∆ABC ~ ∆APQ**

From theorem 2.) stated above,

⇒ 9 ar(∆APQ) = ar(∆APQ) +ar (trapezium BPQC)

⇒ 9 ar(∆APQ)- ar(∆APQ) = ar(trapezium BPQC)

⇒ 8 ar(∆APQ) =ar (trapezium BPQC)

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PREVIOUSIn an isosceles Δ ABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC2. Prove that Δ APC∼Δ BCQ.NEXTABC is a right-angled triangle right angled at B. A circle is inscribed in it in the lengths of the two sides containing the right angle are 5 cm and 12 cm. Find the radius of the circle.

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