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2012 AMC 10A Problem 2

2012 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 1 of 5 · about 1 min · position P1-10

A square with side length $8$ is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?

A)
$2 \text{by} 4$
B)
$2 \text{by} 6$
C)
$2 \text{by} 8$
D)
$4 \text{by} 4$
E)
$4 \text{by} 8$

Answer

Key insight

Cutting a square in half along a midline halves one side and leaves the other side untouched: 4 by 8.

Solution

To produce two congruent rectangles, the cut must be a straight line through the middle of the square, parallel to a pair of sides. That cut splits one side of length $8$ into two pieces of length $4$ , while the perpendicular side keeps its full length $8$ .

Each rectangle is therefore $4$ by $8$ . As a check, its area is $32$ , exactly half of the square's area $64$ .

The answer is $\boxed{\textbf{(E)}\ 4\ \text{by}\ 8}$ .

Why this works

Halving a figure with a single straight cut changes only the dimension that the cut crosses. A quick area check (half of $64$ is $32$ , and only $4 \times 8 = 32$ among the choices) confirms the picture.

The trap

Halving both dimensions (4 by 4), which quarters the square rather than halving it.

Common mistakes

Halving both dimensions (4 by 4), which quarters the square rather than halving it.
Picking $2$ by $8$ by dividing $8$ by $4$ instead of by $2$ .

Techniques

Set up the equation/formula and compute; no special trick needed