In the diagram, square $A B C D$ has side length 2, with $M$ the midpoint of $B C$ and $N$ the midpoint of $C D$. The area of the shaded region $B M N D$ is
(A) 1
(B) $2 \sqrt{2}$
(C) $\frac{4}{3}$
(D) $\frac{3}{2}$
(E) $4-\frac{3}{2} \sqrt{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-133.jpg?height=439&width=442&top_left_y=1897&top_left_x=1313)
In the diagram, square $A B C D$ has side length 2 , with $M$ the midpoint of $B C$ and $N$ the midpoint of $C D$. The area of the shaded region $B M N D$ is
(A) 1
(B) $2 \sqrt{2}$
(C) $\frac{4}{3}$
(D) $\frac{3}{2}$
(E) $4-\frac{3}{2} \sqrt{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-260.jpg?height=477&width=471&top_left_y=713&top_left_x=1380)

## Solution

The area of $\triangle M N C$ is $\frac{1}{2}(1)(1)=\frac{1}{2}$. Since $\triangle B D C$ is half the square, it will have an area of 2.

Since the shaded region has an area equal to that of $\triangle B D C$ minus the area of $\triangle M N C$, its area will be $2-\frac{1}{2}=\frac{3}{2}$.

ANSWER: (D)