Question

In the given figure, ABCD is a parallelogram. P and Q are midpoints of AB and AD respectively. Area $\Delta CPQ$ : Area of parallelogram ABCD is equal to

• A. 3 : 5
• B. 3 : 8
• C. 5 : 8
• D. 1 : 8

Answer 1

The correct option is B

3 : 8

(i) ABCD is a parallelogram ...... (given)

(ii) Area $\Delta ABD=\frac{1}{2}$ area parallelogram ABCD .... (diagonal divides a parallelogram into two triangles of equal area)

(iii) Area $\Delta ABQ=\frac{1}{2}$ Area $\Delta ABD$ ..... (median divides a triangle into two triangles of equal area)
$=\frac{1}{4}$ area parallelogram ABCD .... (from (ii))

(iv) Area $\Delta APQ=\frac{1}{2}$ Area $\Delta ABQ$ .... (median divides a triangle into two triangles of equal area)
$=\frac{1}{8}$ area parallelogram ABCD ……(from(iii))

(v) Area $\Delta CDQ=\frac{1}{2}$ area $\Delta ACD$ ..... (median divides triangle into two triangles of equal area)
$=\frac{1}{2}$ [ $\frac{1}{2}$ area parallelogram ABCD] .... (diagonal divides a parallelogram into two triangles of equal area)
$=\frac{1}{4}$ area parallelogram ABCD

(vi) Area $\Delta CBP=\frac{1}{2}$ Area $\Delta ACB$ .... (median divides triangle into two triangles of equal area)
$=\frac{1}{2}$ [$\frac{1}{2}$ area parallelogram ABCD] ....(diagonal divides a parallelogram into two triangles of equal area)
$=\frac{1}{4}$ area parallelogram ABCD

(viii) Using equations (iv), (v) and (vi) we get,

Area $\Delta APQ$ + Area $\Delta CDQ$ + Area $\Delta CBP=(\frac{1}{8}+\frac{1}{4}+\frac{1}{4})$ area parallelogram ABCD
$=\frac{5}{8}$ area parallelogram ABCD

Hence area $\Delta CPQ=(1-\frac{5}{8})$ area parallelogram ABCD
$=\frac{3}{8}$ area parallelogram ABCD
area $\Delta CPQ$ : area parallelogram ABCD = 3 : 8