InterviewSolution
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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 |
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Answer» Solution: Given: ABCD is a parallelogram and P and Q are midpoints on side BC and CD. Let us take M as the midpoint of side AD and N 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 |
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