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Difference in Areas of Rectangle.

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Difference in Areas of Rectangle.

by amanpreet » Thu Mar 22, 2012 2:23 pm
The length of each side of square A is increased by 100% to make square B.If the length of the side of square B is increased by 50% to make square C, by what percent is the area of square C greater than the sum of the Areas of squares A and B?
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by pemdas » Thu Mar 22, 2012 2:41 pm
amanpreet wrote:The length of each side of square A is increased by 100% to make square B.If the length of the side of square B is increased by 50% to make square C, by what percent is the area of square C greater than the sum of the Areas of squares A and B?
square A has 4 sides and each one is increased by 2 to make square B. Then sides of square B are increased by 1.5 times to make square C. Area square C=(1.5B)^2 and B=2A, hence area of square C=(1.5*2A)^2 or 9A^2. The areas of square A=A^2 and square B=(2A)^2=4A^2. The required will be: (area of sq.C- Sum of areas of sq.A&B)/Sum of areas sq.A&B = (9A^2 - A^2 - 4A^2)/(A^2+4A^2)=4/5 or 80%
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by seal4913 » Thu Mar 22, 2012 2:47 pm
You can also plug in values and slove

Square A length = 2. Area is 4
Square B length is 100% so its doubled A. Which is 4 so area is 16.
Square C length is 50% B so 1.5 x 4 = 6. Which makes the area 36.

So now you take the sum of the area of A and B which is 4 and 16 so that is 20.
The area of C is 36.

The difference is 36-20 = 16. So % change is 16/20 (the change over the starting amount). You get 4/5 which is .8. Times that by 100 to get percent and it's 80%.
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