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.
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`
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