Question

# ABCD is a parallelogram of area 162 sq., P is the point on AB such that AP:PB = 1:2, calculate the area of Triangle APD and the ratio PQ:QD, where Q is the point of intersection of AC and PD

Hint:

### we know that the height between two parallel lines is always the same .

So the height of triangle APD is equal to height of parallelogram. Where as base of

triangle AP = 1/(1+2) of Base of parallelogram AB . find the area using this relation.

PQ: QD ratio can be found by using similarity of triangles APQ and CDQ .

## The correct answer is: PQ:QD = 1: 3

### Ans :- 27 sq.cm ; 1:3

Explanation :-

Given ,AP : PB = 1:2 AP : (AP+PB) = 1 : (1+2) AP : AB = 1 : 3

Step 1 :- Find the area of parallelogram

area of parallelogram = base × height = AB × h

Given , area of parallelogram = 162 sq.m = AB × h

Step 2:- Find the Area of triangle ADP

Area of triangle ADP = ½ × base AP × height h = ½ × AP × h

As AP = AB /3 we get, Area of triangle ADP = ½ × ⅓ × AB × h = ⅙ × 162 sq.m

We get ,Area of triangle APD = 27 sq.m

Step 3:-

Consider , triangle APQ and CDQ

We get, ∠PAQ = ∠DCQ ( alternate interior angles)

∠QPA = ∠QDC ( alternate interior angles)

∠PQA = ∠DQC ( vertically opposite angles )

So, triangle APQ and CDQ are similar by AAA property .

PQ:DQ = AP :AB = 1:3

Therefore ,We get PQ:QD = 1:3

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