, 2025

Solution 1:

Since $10\%$ of $5 \text{ cm}$ is $\dfrac{10}{100}\times5\text{ cm}=0.1\times5\text{ cm}=0.5\text{ cm}$, then the length of the resulting

rectangle is $5\text{ cm}+0.5\text{ cm}=5.5\text{ cm}$, and its area is $5.5\text{ cm}\times4\text{ cm}=22\text{ cm}^2$.

Solution 2:

When $5 \text{ cm}$ is increased by $10\%$, the resulting length is $\$\left(1+\dfrac{10}{100}\right)\times5\text{}$
cm}$or$$1.1\times5\text{}$ cm}$which is equal to$5.5 $\text{}$ cm}$.

Thus, the area of the resulting rectangle is $5.5\text{ cm}\times4\text{ cm}=22\text{ cm}^2$.
Solution 1:

A square with area $100 \text{ cm}^2$ has both length and width equal to $\sqrt{100\text{ cm}^2}=10\text{ cm}$.

Since $30\%$ of $10\text{ cm}$ is $\dfrac{30}{100}\times10\text{ cm}=0.3\times10\text{ cm}=3\text{ cm}$, then the length of the
resulting rectangle is $10\text{ cm}+3\text{ cm}=13\text{ cm}$, and its width is $10\text{ cm}-3\text{ cm}=7\text{ cm}$.

The area of the resulting rectangle is $13\text{ cm}\times7\text{ cm}=91\text{ cm}^2$ which is $\dfrac{91\text{ cm}^2}{100\text{ cm}^2}\times100\%=91\%$ of the area of the
original square.

Therefore, the area decreased by $100\%-91\%=9\%$. Solution 2: A square with area $100 $\text{}$
cm}^2 has both length and width equal to

$\sqrt{100\text{ cm}^2}=10\text{}$ cm}$.

When $10 \text{ cm}$ is increased by $30\%$, the resulting length is $\left(1+\dfrac{30}{100}\right)\times10\text{ cm}$ or $1.3\times10\text{ cm}$ which is equal to $13 \text{ cm}$.

When $10 \text{ cm}$ is decreased by $30\%$, the resulting length is $\left(1-\dfrac{30}{100}\right)\times10\text{ cm}$ or $0.7\times10\text{ cm}$ which is equal to $7 \text{ cm}$.

Thus, the area of the resulting rectangle is $13\text{ cm}\times7\text{ cm}=91\text{ cm}^2$ which is $100\text{ cm}^2-91\text{ cm}^2=9\text{ cm}^2$ less than the area of the
original square.

Therefore, the area decreased by $\dfrac{9\text{ cm}^2}{100\text{ cm}^2}\times100\%=9\%$.
Suppose the length of the original rectangle is $\ell$ and its width is $w$. Then the length of the resulting rectangle is $\left(1+\dfrac{x}{100}\right)\times\ell$, and its width is $\left(1-\dfrac{20}{100}\right)\times w$ or $\dfrac{8}{10}w$. The area of the original rectangle, $\$\ell$
w, is equal to the area of the resulting rectangle

$\left(1+\dfrac{x}{100}\right)\times\ell \times \dfrac{8}{10}w\$.$

Setting the areas equal and simplifying, we get the following equivalent
equations: $$\begin{align*} \left(1+\dfrac{x}{100}\right)\times\ell \times \dfrac{8}{10}w &= \ell w \\ \left(1+\dfrac{x}{100}\right)\times\dfrac{8}{10}\times \ell w &= \ell w \\ \left(1+\dfrac{x}{100}\right)\times\dfrac{8}{10} &= 1 \ \ \text{ (since $\ell w>0$)}\\ 1+\dfrac{x}{100} &= \dfrac{10}{8} \\ \dfrac{x}{100} &= \dfrac{10}{8} -\dfrac{8}{8}\\ \dfrac{x}{100} &= \dfrac{2}{8}\\ x&= \dfrac{1}{4}\times100\end{align*}$$ and so $x=25$.