If the width and length of a rectangle are increased by 25%, by what percentage does the area increase?

1. Problem Description

If the width and length of a rectangle are increased by 25%, by what percentage does the area increase?

2. Solution Steps

Let $w$ be the original width and $l$ be the original length of the rectangle. The original area $A_1$ is given by

$A_1 = w \times l$.

The new width $w'$ is $w + 0.25w = 1.25w = \frac{5}{4}w$.

The new length $l'$ is $l + 0.25l = 1.25l = \frac{5}{4}l$.

The new area $A_2$ is given by

$A_2 = w' \times l' = (\frac{5}{4}w) \times (\frac{5}{4}l) = \frac{25}{16}wl$.

The increase in area is $A_2 - A_1 = \frac{25}{16}wl - wl = \frac{25-16}{16}wl = \frac{9}{16}wl$.

The percentage increase in area is

$\frac{A_2 - A_1}{A_1} \times 100 = \frac{\frac{9}{16}wl}{wl} \times 100 = \frac{9}{16} \times 100 = \frac{900}{16} = \frac{225}{4} = 56.25\%$.

3. Final Answer

The area increases by 56.25%.