If each of the dimensions of a rectangle is increased by 100%, then the area is increased by

400%

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100%

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200%

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300%

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Solution

The correct option is D 300%
b'If l and b are the dimensions of a rectangle, then l increases by 100%of l and b increases by 100% of b. Therefore
$A r e a = (l + \frac{100}{100} l) \times (b + \frac{100}{100} b) = (2 l) \times (2 b) = 4 l b$
$I n c r e a s e = 4 l b - l b = 3 l b I n c r e a s e p e r c e n t = \frac{3 l b}{l b} \times 100 = 300$ '

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If each of the dimensions of a rectangle is increased by 100%, then the area is increased by

The correct option is D 300%b'If l and b are the dimensions of a rectangle, then l increases by 100%of l and b increases by 100% of b. ThereforeArea=(l+100100l)×(b+100100b) =(2l)×(2b)=4lbIncrease=4lb−lb=3lbIncrease percent=3lblb×100=300'