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P and Q are the points on AB and BC respectively of ∥gm ABCD, then
- a)ar (∆PDC) = 1/2 ar (∆AQD)
- b)ar(∆PDC) = ar (∆AQD) = 1/2 ar (parallelogram ABCD)
- c)ar(∆PDC) = ar (∆AQD) = 2/3 ar(parallelogram ABCD)
- d)ar (∆PDC) = 2/3 ar(∆AQD)
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
P and Q are the points on AB and BCrespectively of gm ABCD, thena)ar (...
∵ ABCD and ∆PDC lie on same base, i.e., CD and between the same parallels, i.e., AB and CD.
∴
Similarly,
⇒ ar (∆PDC) = ar (∆AQD) = 1/2 ar (quad. ABCD)
∴
Similarly,
⇒ ar (∆PDC) = ar (∆AQD) = 1/2 ar (quad. ABCD)
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P and Q are the points on AB and BCrespectively of gm ABCD, thena)ar (PDC) = 1/2ar (AQD)b)ar(PDC) = ar (AQD)= 1/2ar (parallelogram ABCD)c)ar(PDC) = ar (AQD) = 2/3ar(parallelogram ABCD)d)ar (PDC) = 2/3ar(AQD)Correct answer is option 'B'. Can you explain this answer?
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