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$A B C D$ is a parallelogram. $P$ is a parallelogram. is the mid-point of $A B$ is the mid-point of . $B D$ . and $C P$ and intersect at $Q$ intersect at such that $C Q : Q P = 3 : 1$ such that . If $a r (Δ P B Q) = 10 cm 2$ . If , then the area of parallelogram $A B C D$ , then the area of parallelogram from the choices is:

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80 c m 2

100 c m 2

120 c m 2

140 c m 2

answer is C.

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Detailed Solution

Given that,

A B C D

is a parallelogram.

P

is the mid-point of

A B

B D

and

C P

intersect at

Q

such that

C Q : Q P = 3 : 1

a r (Δ P B Q) = 10 cm 2

Assume that,

C Q = x

and

Q P = 3 x

. Area of the triangle

= 12 × (b a s e) × (h e i g h t)

Area of the triangle PBQ

= 12 × x × h

10 = 12 × x × h

x × h = 20 c m 2

Calculate the area of

△ B Q C .

Area (BQC) = 12 × 3 x × h = 12 × 3 × 20 Area (BQC) = 30 c m 2 .

Assume that

h

be the perpendicular distance and calculate the area of

△ P C B .

⇒ Area (Δ PCB) = 12 × PB × h ⇒ 30 = 12 × PB × h

⇒ PB × h = 60 c m 2

Calculate the area of parallelogram

A B C D

.
Area of the parallelogram

= b a s e × h e i g h t

a r (ABCD) = AB × H = 2 BP × H = 2 (60) ⇒ a r (ABCD) = 120 c m 2 .

The area of the parallelogram ABCD is

120 c m 2

.
Hence, option 3) is correct.

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