1.

ABCD is a parallelogram. P, Q are the midpoints of sides BC and CD respectively. If the area of Δ ABC is 24 cm2, then the area of Δ APQ is1). 12 cm2). 9 cm23). 20 cm24). 18 cm2

Answer»

Solution:

Given: ABCD is a parallelogram and P and Q are midpoints on side BC and CD.

Let us take as the midpoint of side AD and as the midpoint of side AB. Let us take the POINT of intersection of PM and QN as O.

It is also given that the area of ∆ABC is 24cm² then the area of WHOLE parallelogram will be,

Area of the parallelogram ABCD = 2 × 24 = 48 cm².

Now we have to find the area of ∆AQP and 

Area of ∆AQP = Area of ABCD - Area of PCQ,QDM,ABP

Area of PCQ = ⅛ × 48 = 6 cm²

Area of QDM = ¼ × 48 = 12 cm²

Area of ABP = ¼ × 48 = 12 cm²

Area of ∆AQP = Area of ABCD - (6 + 12 + 12)

Area of ∆AQP = 48 - 30

Area of ∆AQP = 18 cm²

So, the correct option is 4).18 cm²



Discussion

No Comment Found

Related InterviewSolutions