In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel. Let $M$ be the midpoint of diagonal $AC$, $N$ the midpoint of diagonal $BD$, and $P$ the midpoint of side $AB$. We know that $AB=15 \text{ cm}$, $CD=24 \text{ cm}$, and the height of the trapezoid is $h=14 \text{ cm}$.

a) Calculate the length of segment $MN$.

b) Calculate the area of triangle $MNP$.

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-02.jpg?height=320&width=532&top_left_y=760&top_left_x=654)
Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-02.jpg?height=317&width=532&top_left_y=1388&top_left_x=654)

a) If $P$ is the midpoint of $A B$ and $L$ is the midpoint of $A D$, it follows that $M L$ is the midsegment of triangle $A C D$ and thus $L M=12 \text{ cm}$. Since $L N$ is the midsegment of triangle $A B D$, it follows that $L N=7.5 \text{ cm}$. Therefore, $M N=L M-L N=4.5 \text{ cm}$

b) Since $M N$ lies on the midsegment, its distance to the line $C D$ is $h / 2=7 \text{ cm}$. Therefore, the height of triangle $M N P$ relative to side $M N$ is $7 \text{ cm}$, and thus

$$
\text{Area}(M N P)=1 / 2 \cdot 7 \cdot 4.5=15.75 \text{ cm}^2
$$