In the diagram, the four points have coordinates $A(0,1)$, $B(1,3), C(5,2)$, and $D(4,0)$. What is the area of quadrilateral $A B C D$ ?
(A) 9
(B) 3
(C) 6
(D) $\sqrt{85}$
(E) $2 \sqrt{5}+2 \sqrt{17}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-100.jpg?height=366&width=423&top_left_y=825&top_left_x=1293)
We draw a horizontal line through $B$ (meeting the $y$-axis at $P$ ) and a vertical line through $C$ (meeting the $x$-axis at $Q$ ). Suppose the point of intersection of these two lines is $R$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-188.jpg?height=364&width=423&top_left_y=2046&top_left_x=905)

We know that $P$ has coordinates $(0,3)$ (since $B$ has $y$-coordinate 3 ) and $Q$ has coordinates $(5,0)$ (since $C$ has $x$-coordinate 5 ), so $R$ has coordinates $(5,3)$.

Using the given coordinates, $O A=1, A P=2, P B=1, B R=4, R C=1, C Q=2, Q D=1$,
and $D O=4$.

The area of $A B C D$ equals the area of $P R Q O$ minus the areas of triangles $A P B, B R C, C Q D$, and $D O A$.

$P R Q O$ is a rectangle, so has area $3 \times 5=15$.

Triangles $A P B$ and $C Q D$ have bases $P B$ and $Q D$ of length 1 and heights $A P$ and $C Q$ of length 2 , so each has area $\frac{1}{2}(1)(2)=1$.

Triangles $B R C$ and $D O A$ have bases $B R$ and $D O$ of length 4 and heights $C R$ and $A O$ of length 1 , so each has area $\frac{1}{2}(4)(1)=2$.

Thus, the area of $A B C D$ is $15-1-1-2-2=9$.

(Alternatively, we could notice that $A B C D$ is a parallelogram. Therefore, if we draw the diagonal $A C$, the area is split into two equal pieces. Dropping a perpendicular from $C$ to $Q$ on the $x$-axis produces a trapezoid $A C Q O$ from which only two triangles need to be removed to determine half of the area of $A B C D$.)

ANSWER: (A)