Answer in Brief

*ABCD* is a parallelogram. *P* is the mid-point of *AB*. *BD* and *CP* intersect at *Q* such that *CQ*: *QP* = 3.1. If ar (Δ*PBQ*) = 10cm^{2}, find the area of parallelogram *ABCD*.

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#### Solution

It is given that CQ : QP = 3: 1 and Area (PBQ) = 10 cm^{2}

Let CQ = x and QP = 3x

We need to find area of the parallelogram ABCD.

From the figure,

Area (PBQ) = `1/2 xx x xx h = 10 ⇒ x xx h = 20`

And,

Area (BQC) = `1/2 xx 3x xx h = 30 cm^2`

Now, let H be the perpendicular distance between AP and CD. Therefore,

Area (PCB) = `1/2 PB xx H = 30 cm^2` …… (1)

Thus the area of the parallelogram ABCD is,

Area (ABCD) = AB × H

⇒ Area (ABCD) = 2BP × H

From equation (1), we get

Area (ABCD) = 4 × 30 = 120 cm^{2}

Hence, the area of the parallelogram ABCD is 120 cm^{2}

Concept: Concept of Area

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