$P Q R S$ is a square with side length $8 . X$ is the midpoint of side $P Q$, and $Y$ and $Z$ are the midpoints of $X S$ and $X R$, respectively, as shown. The area of trapezoid $Y Z R S$ is
(A) 24
(B) 16
(C) 20
(D) 28
(E) 32

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-090.jpg?height=363&width=328&top_left_y=1729&top_left_x=1321)
$P Q R S$ is a square with side length 8 . $X$ is the midpoint of side $P Q$, and $Y$ and $Z$ are the midpoints of $X S$ and $X R$, respectively, as shown. The area of trapezoid YZRS is
(A) 24
(D) 28
(B) 16
(C) 20
(E) 32

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-389.jpg?height=448&width=424&top_left_y=256&top_left_x=1317)

## Solution

If $P Q R S$ is a square with side length 8 , it must have an area of 64 square units. The area of DXSR is thus $\frac{1}{2}(8)(8)=32$. If we take the point $T$ to be the midpoint of $S R$ and join $Y$ and $Z$ to $T$, we would have the following diagram.

Each of the four smaller triangles contained within DXSR has an equal area, which is therefore $\frac{1}{4}(32)=8$. Since the area of trapezoid $Y Z R S$ is made up of three of these triangles, it has an area of $3 ¥ 8=24$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-389.jpg?height=453&width=396&top_left_y=814&top_left_x=1339)

ANswer: (A)