2012 AMC 10A Problem 2

P1-10~1 minPrint

Quadrilaterals & Polygon Areas Set up the equation/formula and compute; no special trick needed

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

Show solution without answering

Key insight

A cut straight through the middle splits one side of 8 into two 4s and leaves the other side at 8.

Solution

To get two matching rectangles from one square, cut straight across the middle, parallel to two of the sides.
The cut crosses one pair of sides. Each of those sides, length $8$ , is split into two pieces of length $4$ .
The other pair of sides is not touched by the cut, so they keep their full length $8$ .
So each rectangle is $4$ by $8$ .
Check with area: the square has area $8 \times 8 = 64$ , half of that is $32$ , and $4 \times 8 = 32$ . It matches.

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

Why this works

One straight cut only changes the length it runs across; the length it runs along stays the same. An area check is a quick way to catch a mistake: each half must have half the area, and among the choices only $4 \times 8$ gives $32$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 9: Polygons (§9.3)