# 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 (ΔPBQ) = 10cm2, find the area of parallelogram ABCD. - Mathematics

ABCD is a parallelogram. P is the mid-point of ABBD and CP intersect at Q such that CQQP = 3.1. If ar (ΔPBQ) = 10cm2, find the area of parallelogram ABCD.

#### Solution

It is given that CQ : QP = 3: 1 and Area (PBQ) = 10 cm2

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 cm2

Hence, the area of the parallelogram ABCD is 120 cm2

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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Q 9 | Page 60
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