In the diagram, the area of rectangle $A B C D$ is 40 . The area of $M B C N$ is
(A) 15
(B) 10
(C) 30
(D) 12
(E) 16

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-098.jpg?height=328&width=504&top_left_y=2254&top_left_x=1212)
Solution 1

Since the area of rectangle $A B C D$ is 40 and $A B=8$, then $B C=5$.

Therefore, $M B C N$ is a trapezoid with height 5 and parallel bases of lengths 4 and 2 , so has area $\frac{1}{2}(5)(4+2)=15$.

## Solution 2

Since the area of rectangle $A B C D$ is 40 and $A B=8$, then $B C=5$.

Draw a line from $N$ to $A B$ parallel to $B C$ (and so perpendicular to $A B$ ) meeting $A B$ at $P$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-186.jpg?height=336&width=504&top_left_y=160&top_left_x=867)

This line divides $M B C N$ into a rectangle $P B C N$ of width 2 and height 5 , and a triangle $M P N$ with base $M P$ of length 2 and height $P N$ of height 5.

The area of $M B C N$ is sum of the areas of these two parts, or $2(5)+\frac{1}{2}(2)(5)=10+5=15$.

ANSWER: (A)

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