The edge of a cube is increased by 20%. What is the percentage increase in its surface area?
The suface area = 6s^2 where s is the side of the cube
So....now with the increase....the side = 1.2s
So
The new surface area = 6 [ (1.2)s ] ^2 = 8.64s^2
So....the percentage increase =
[ 8.64 - 6 ] / 6 x 100% = 0.44 x 100% = 44%
