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2012 AMC 10B Problem 2

2012 AMC 10B 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 circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?

A)
$50$
B)
$100$
C)
$125$
D)
$150$
E)
$200$

Answer

Key insight

The inscribed circle touches both long sides, so the rectangle's width is the diameter 10; the length is twice that.

Solution

The circle is tangent to the top and bottom sides of the rectangle, so the distance between those sides, the width, equals the circle's diameter: $2 \cdot 5 = 10$ .

The length is twice the width, so it is $20$ . The area is
$20 \cdot 10 = 200.$

The answer is $\boxed{\textbf{(E)}\ 200}$ .

Why this works

A circle inscribed in a figure is tangent to its sides, and two parallel tangent lines are exactly one diameter apart. Once one dimension is pinned down, the given ratio supplies the other. Always convert "radius" to "diameter" before comparing with side lengths.

The trap

Using the radius 5 as the width (area 50) or as the length, instead of the diameter 10.

Common mistakes

Using the radius 5 as the width (area 50) or as the length, instead of the diameter 10.
Taking the ratio backwards and making the length $5$ or the width $20$ .

Techniques

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