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Question

Prove by vector method that the parallelograms on the same base and between same parallels are equal in area.

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Solution

Let AB=aandAD=b
Area of parallelogram ABCD =a×b1
(Cross product, Since Area= baseXheight)
=a(bsinθ)
Now in parallelogram ABB'A
AB=a and AD=ma (let)
( A'D is parallel to AB)
Consider triangle ADA'
By triangular law of vectors
AA=ma+b
Vector area of ABB'A' =a×(ma+b)=(a×ma)+(a×b)
(a×(ma)=0, both are parallel, sin0=0)
=0+(a×b)
=a×b2
(1)=(2),
Hence proved

989019_1003625_ans_c4fc7b45706c4305b788eb63263be809.png

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