# The length of each side of square A is increased by 100 percent to make square B. Then each side of square is increased by 50 percent to make square C. By what percent is the area of square C greater than the sum of the areas of square A and B?

Jan 27, 2016

Area of C is 80% greater than area of A $+$ area of B

#### Explanation:

Define as a unit of measurement the length of one side of A.

Area of A $= {1}^{2} = 1$ sq.unit

Length of sides of B is 100% more than length of sides of A
$\rightarrow$ Length of sides of B $= 2$ units
Area of B $= {2}^{2} = 4$ sq.units.

Length of sides of C is 50% more than the length of sides of B
$\rightarrow$ Length of sides of C $= 3$ units
Area of C $= {3}^{2} = 9$ sq.units

Area of C is $9 - \left(1 + 4\right) = 4$ sq.units greater than the combined areas of A and B.

$4$ sq.units represents $\frac{4}{1 + 4} = \frac{4}{5}$ of the combined area of A and B.

4/5 = 80%