## Algebra 1

a) original is 4x4, new is 10% larger on each side original area: $4*4 = 16$ new area: $4*1.1*4*1.1 = 4.4*4.4 = 19.36$ $(19.36-16)/16 = 3.36/16 = .21$ = 21% b) original is 6x6, new is 10% larger on each side original area: $6*6 = 36$ new area: $6*1.1*6*1.1 = 6.6 * 6.6 = 43.56$ $(43.56-36)/36 = 7.56/36 = .21 =$ 21% c) original is 8x8, new is 10% larger on each side original area: $8*8 = 64$ new area: $8*1.1*8*1.1 = 77.44$ $(77.44-64)/64 = 13.44/64 = .21 =$ 21% Let $x$ be the original side length of a square, and let us increase the side lengths by 10%. (The new side length is $1.1*x$, or $1.1x$.) The original area of the square would be $x^{2}$, and the new area would be $(1.1*x)^{2}$, or $1.21x^{2}$. Change in area: $(1.21x^{2} - x^{2})/x^{2} = .21x^{2}/x^{2} = .21$ = 21%