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P and Q are any 2 points lying on the sides DC and AD respectively of a parallelogram ABCD. show that ar(APB)=ar(BQC).
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Can you draw a rough figure for this?
|dw:1424166282048:dw|
Good, so, now we need to prove Ar(APB) = Ar(BQC) We know that Area of a Triangle = \(\cfrac{1}{2} \times \text{Base} \times \text{Height} \)
Also, area of a parallelogram = Base * Corresponding Height So, area of parallelogram ABCD = AB * PQ |dw:1424166594017:dw|
|dw:1424166645213:dw| Sorry, you can use R instead of Q. So, for AB as the base you've Area of Parallelogram = AB * PR
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