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Question 231456: If the length of each edge of a cube is increased by 75%, by what percent is the surface area of the original cube increased?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Surface area of the original cube is equal to 6 * s^2
If you increase each s by 75%, then each s becomes 1.75 * s.
The surface area then becomes 6 * (1.75 * s)^2
This is equivalent to 6 * 1.75^2 * s^2.
This equals to 18.375 * s^2
Since the original surface area equals 6 * s^2, and the new surface area = 18.375 * s^2, then the new surface area is equal to 3.0625 times the original surface area.
This would be 306.25% of the original surface area which would be an increase of 206.5% of the original surface area.
To see if this is accurate, use some numbers and see if the percentages work out.
Let s = 15
6 * 15^2 = a surface area of 1350.
Increase s by 75%.
s goes from 15 to 15 + .75 * 15 = 26.25
New Surface area is 26.25^2 * 6 = 4134.375
Divide 4134.375 by 1350 to get 3.0625
4134.375 is equal to 3.0625 times 1350 which is an increase of 206.25%
Take 206.25% of 1350 and add it to 1350 to get 4134.375.
It all checks out.
Your answer is that the surface area of the original cube is increased by 206.25%
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