Printable

2011 AMC 10A Problem 9

2011 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Coordinate Geometry · difficulty 2 of 5 · about 2 min · position P1-10

A rectangular region is bounded by the graphs of the equations $y=a, y=-b, x=-c,$ and $x=d$ , where $a,b,c,$ and $d$ are all positive numbers. Which of the following represents the area of this region?

A)
$ac+ad+bc+bd$
B)
ac-ad+bc-bd
C)
ac+ad-bc-bd
D)
-ac-ad+bc+bd
E)
$ac-ad$

Answer

Key insight

The lines sit on opposite sides of the axes, so the rectangle is (c + d) by (a + b); expanding gives four positive terms.

Solution

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

Why this works

The distance between the parallel lines $x = p$ and $x = q$ is $|p - q|$ ; with one coordinate negative, the subtraction becomes an addition. Sketching the four lines around the origin makes the signs obvious, and then the area is a simple product to expand.

Alternative approach

Plug in $a = b = c = d = 1$ . The rectangle runs from $x = -1$ to $1$ and $y = -1$ to $1$ , so its area is $4$ . Only choice (A) evaluates to $4$ ; the other four each give $0$ .

The trap

Computing the width as d - c or the height as a - b, forgetting that x = -c and y = -b lie on the negative side of the axes.

Common mistakes

Computing the width as d - c or the height as a - b, forgetting that x = -c and y = -b lie on the negative side of the axes.
Sign errors when expanding $(a + b)(c + d)$ , landing on one of the mixed-sign distractors.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer