$A B C D$ is a rectangle, with $M$ the midpoint of $B C$ and $N$ the midpoint of $C D$. If $C M=4$ and $N C=5$, what percent of the area of the rectangle is shaded?
(A) 70
(B) 78
(C) 80
(D) 87.5
(E) 75

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-111.jpg?height=345&width=461&top_left_y=215&top_left_x=1293)
Solution 1

Since $M$ is the midpoint of $B C$ and $C M=4$, then $B C=8$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-202.jpg?height=347&width=466&top_left_y=664&top_left_x=878)

Since $N$ is the midpoint of $C D$ and $N C=5$, then $C D=10$.

Since $A B C D$ is a rectangle, its area is $10 \times 8=80$.

Also, the area of $\triangle N C M$ is $\frac{1}{2}(4)(5)=10$, so the shaded area of the rectangle is the area of the whole rectangle minus the area of $\triangle N C M$, or 70 .

Thus, the fraction of the rectangle that is shaded is $\frac{70}{80}=0.875$, so $87.5 \%$ of the area is shaded.

## Solution 2

The calculation from Solution 1 can also be done more generally.

Suppose $B C=2 x$ and $C D=2 y$.

Since $M$ is the midpoint of $B C$, then $C M=x$.

Since $N$ is the midpoint of $C D$, then $N C=y$.

Since $A B C D$ is a rectangle, its area is $(2 x)(2 y)=4 x y$.

Also, the area of $\triangle N C M$ is $\frac{1}{2}(N C)(M C)=\frac{1}{2} x y$, so the shaded area of the rectangle is the area of the whole rectangle minus the area of $\triangle N C M$, or $4 x y-\frac{1}{2} x y=\frac{7}{2} x y$.

Thus, the fraction of the rectangle that is shaded is $\frac{\frac{7}{2} x y}{4 x y}=\frac{7}{8}=0.875$, so $87.5 \%$ of the area is shaded.

ANSWER: (D)