1. In the diagram, square $E F G H$ is centred inside square $A B C D$ and the region between the two squares is shaded.
(a) Determine the area of the shaded region if the length of $A B$ is $10 \mathrm{~cm}$ and the length of $E F$ is $8 \mathrm{~cm}$.
(b) Determine the length of $E F$ if the length of $A B$ is $13 \mathrm{~cm}$ and the area of the shaded region is $120 \mathrm{~cm}^{2}$.
(c) Determine the length of $E F$ if the length of $A B$ is $18 \mathrm{~cm}$ and the area of the shaded region is $\frac{3}{4}$ of the area of $A B C D$.
(d) Suppose that the area of the shaded region is $64 \%$ the area of $A B C D$. If the length of $A B$ is $a \mathrm{~cm}$ and the length of $E F$ is $b \mathrm{~cm}$, determine the value of $\frac{a}{b}$.
1. (a) Since $A B C D$ is as square with side length $A B=10 \mathrm{~cm}$, its area is $(10 \mathrm{~cm})^{2}=100 \mathrm{~cm}^{2}$. Since $E F G H$ is a square with side length $E F=8 \mathrm{~cm}$, is area is $(8 \mathrm{~cm})^{2}=64 \mathrm{~cm}^{2}$. The shaded region is the region inside $A B C D$ that lies outside square $E F G H$, so its area is the difference between the two areas, or $100 \mathrm{~cm}^{2}-64 \mathrm{~cm}^{2}=36 \mathrm{~cm}^{2}$.
(b) The area of $A B C D$ is $(13 \mathrm{~cm})^{2}=169 \mathrm{~cm}^{2}$ and the area of the shaded region is $120 \mathrm{~cm}^{2}$. The area of $E F G H$ is $169 \mathrm{~cm}^{2}-120 \mathrm{~cm}^{2}=49 \mathrm{~cm}^{2}$.
Since $E F G H$ is a square, its side length is $\sqrt{49 \mathrm{~cm}^{2}}=7 \mathrm{~cm}$.
(c) The area of the shaded region is given to be $\frac{3}{4}$ of the total area of $A B C D$, so the area of $E F G H$ is $1-\frac{3}{4}=\frac{1}{4}$ of the area of $A B C D$.
The area of $A B C D$ is $(18 \mathrm{~cm})^{2}=324 \mathrm{~cm}^{2}$.
The area of $E F G H$ is $\frac{1}{4}$ of the area of $A B C D$, so the area of $E F G H$ is $\frac{324}{4} \mathrm{~cm}^{2}=81 \mathrm{~cm}^{2}$. Since $E F G H$ is a square, its side length is $\sqrt{81 \mathrm{~cm}^{2}}=9 \mathrm{~cm}$.
(d) The area of $A B C D$ is $a^{2} \mathrm{~cm}^{2}$ and the area of $E F G H$ is $b^{2} \mathrm{~cm}^{2}$.
The given condition meanss that $a^{2}-b^{2}=\frac{64}{100} a^{2}$, which can be rearranged to $\frac{36}{100} a^{2}=b^{2}$. Taking square roots of both sides and noting that $a>0$ and $b>0$, we have $\sqrt{\frac{36}{100}} a^{2}=\sqrt{b^{2}}$ or $\frac{6}{10} a=b$, and so $\frac{3}{5} a=b$.
Multiplying both sides of this equation by $\frac{5}{3}$ gives $a=\frac{5}{3} b$. Dividing both sides of this equation by $b$ gives $\frac{a}{b}=\frac{5}{3}$.