A quadrilateral is bounded by the lines with equations $x=0, x=4, y=x-2$, and $y=x+3$. The area of this quadrilateral is
(A) 16
(B) 24
(C) 4
(D) $20 \sqrt{2}$
(E) 20
Solution 1

The lines with equations $x=0$ and $x=4$ are parallel vertical lines.

The lines with equations $y=x-2$ and $y=x+3$ are parallel lines with slope 1 .

Since the quadrilateral has two sets of parallel sides, it is a parallelogram.

Thus, its area equals the length of its base times its height.

We consider the vertical side along the $y$-axis as its base.

Since the two sides of slope 1 have $y$-intercepts -2 and 3 , then the length of the vertical base is $3-(-2)=5$.

Since the parallel vertical sides lie along the lines with equations $x=0$ and $x=4$, then these sides are a distance of 4 apart, which means that the height of the parallelogram is 4 .

Therefore, the area of the quadrilateral is $5 \times 4=20$.

## Solution 2

The lines with equations $x=0$ and $x=4$ are parallel vertical lines.

The lines with equations $y=x-2$ and $y=x+3$ are parallel lines with slope 1 .

The lines with equations $y=x-2$ and $y=x+3$ intersect the line with equation $x=0$ at their $y$-intercepts, namely -2 and 3 , respectivley.

The lines with equations $y=x-2$ and $y=x+3$ intersect the line with equation $x=4$ at $(4,2)$ and $(4,7)$, respectively (the point on each line with $x$-coordinate 4 ).

We draw horizontal lines through $(0,-2)$, intersecting $x=4$ at $(4,-2)$, and through $(4,7)$, intersecting $x=0$ at $(0,7)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-069.jpg?height=542&width=550&top_left_y=247&top_left_x=1378)

The area of the quadrilateral equals the area of the large rectangle with vertices $(0,7),(4,7)$, $(4,-2),(0,-2)$ minus the combined area of the two triangles.

This rectangle has side lengths $4-0=4$ and $7-(-2)=9$ and so has area $4 \times 9=36$.

The two triangles are right-angled and have bases of length $4-0=4$ and heights of length $7-3=4$ and $2-(-2)=4$. This means that they can be combined to form a square with side length 4 , and so have combined area $4^{2}=16$.

Therefore, the area of the quadrilateral is $36-16=20$.

ANSWER: (E)