2011 AMC 10A Problem 9

Draw a quick sketch. All four letters are positive numbers, so the minus signs tell you which side of each axis a line is on.

1. The width. The line $x = d$ is to the right of the $y$-axis, a distance $d$ away. The line $x = -c$ is to the left, a distance $c$ away. They are on opposite sides, so the width is $c + d$. (For example, from $x = -2$ to $x = 3$ is $2 + 3 = 5$ units.)
2. The height. The line $y = a$ is above the $x$-axis, a distance $a$ away. The line $y = -b$ is below it, a distance $b$ away. So the height is $a + b$.
3. The area. Multiply width by height and expand:
$$(a + b)(c + d) = ac + ad + bc + bd.$$

The answer is $\boxed{\textbf{(A)}\ ac+ad+bc+bd}$.

The distance between two parallel lines is found by subtracting their positions: from $-c$ to $d$ is $d - (-c) = d + c$. Subtracting a negative number turns into adding. A rough sketch around the origin makes this obvious, and then the area is just a product to multiply out.

Plug in easy numbers. Let $a = b = c = d = 1$. The rectangle then runs from $x = -1$ to $x = 1$ and from $y = -1$ to $y = 1$, a $2 \times 2$ square with area $4$.

Now test the choices with all letters equal to $1$: (A) gives $1 + 1 + 1 + 1 = 4$, while (B), (C), (D) and (E) each have two plus signs and two minus signs, so they give $0$. Only (A) matches.

The vertical sides are the lines $x = -c$ and $x = d$. Since $c$ and $d$ are positive, one lies left of the $y$-axis and the other to the right, so the width is $d - (-c) = c + d$.

The horizontal sides are $y = a$ (above the $x$-axis) and $y = -b$ (below), so the height is $a - (-b) = a + b$.

The area is
$$(a + b)(c + d) = ac + ad + bc + bd.$$

The answer is $\boxed{\textbf{(A)}\ ac+ad+bc+bd}$.