2. Let $P$ and $Q$ be the midpoints of sides $\overline{B C}$ and $\overline{C D}$ of rectangle $A B C D$. What percentage of the area of rectangle $A B C D$ is the area of triangle $A P Q$?
2. Let $P$ and $Q$ be the midpoints of sides $\overline{B C}$ and $\overline{C D}$ of rectangle $A B C D$. What percentage of the area of rectangle $A B C D$ is the area of triangle $A P Q$?

## Solution.

Half of the length $a$ is $\frac{1}{2} a$.

Half of the length $b$ is $\frac{1}{2} b$.

In the diagram, the midpoints of the sides of the rectangle and the lengths $a, \frac{1}{2} a, b, \frac{1}{2} b$ as well as the areas $P_{1}, P_{2}$, and $P_{3}$ are marked.

![](https://cdn.mathpix.com/cropped/2024_05_30_2e894aa57f8bdfbd97cfg-15.jpg?height=465&width=1102&top_left_y=704&top_left_x=431)

Diagram: 2 POINTS

The area of the rectangle: $P_{p}=a \cdot b$.

The area of a right triangle is half the area of a rectangle with the same side lengths as the legs of the right triangle, so we have:

$P_{1}=\frac{1}{2} \cdot \frac{1}{2} a \cdot b=\frac{a b}{4}$

$P_{2}=\frac{1}{2} \cdot a \cdot \frac{1}{2} b=\frac{a b}{4}$

$P_{3}=\frac{1}{2} \cdot \frac{1}{2} a \cdot \frac{1}{2} b=\frac{a b}{8}$

The area of triangle $A P Q$ is obtained by subtracting the areas of the right triangles from the area of the rectangle:

$P_{\Delta}=a b-\frac{a b}{4}-\frac{a b}{4}-\frac{a b}{8}=\frac{8 a b-2 a b-2 a b-a b}{8}=\frac{3 a b}{8}$.

The percentage of the area of triangle $A P Q$ relative to the total area of the rectangle is

$P_{\Delta}: P_{p} \cdot 100=\frac{3 a b}{8}: a b \cdot 100=\frac{3 a b}{8} \cdot \frac{1}{a b} \cdot 100=300: 8=37.5 \%$.

The product $a b$ can be placed in the denominator and canceled out because $a \neq 0, b \neq 0$. Otherwise, the rectangle would not exist.