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Centre of gravity of a trapezium of height h and parallel sides a and b, measured from the side b is at a distance of

Ah(b+2a)/(2(b +a))

Bh(b+2a)/(2(b+a))

Ch(b+2a)/(3(b+a))

Dh(b+2a)/(3(b+a))

Answer:

C. h(b+2a)/(3(b+a))


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