AMC 10 Step by Step

2012 AMC 10B Problem 2

P1-10~1 minPrint

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?

figure

Log in to save your progress.
Show solution without answering

Key insight

The circle touches the top and bottom of the rectangle, so the width equals the circle's diameter, 10. The length is twice that, 20.

Solution

  1. The circle fits snugly inside the rectangle, touching the top side and the bottom side. So the distance from top to bottom (the width) is the distance straight across the circle, which is the diameter.
  2. The radius is 55, so the diameter is 2×5=102 \times 5 = 10. The width is 1010.
  3. The length is twice the width: 2×10=202 \times 10 = 20.
  4. The area is length times width: 20×10=20020 \times 10 = 200.

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

Why this works

When a circle touches two parallel sides, those sides are exactly one diameter apart. Whenever a problem gives a radius and asks about side lengths, turn the radius into a diameter first. After that, the ratio 2:12:1 gives the other side right away.

Concise solution

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

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⋅5=102 \cdot 5 = 10.

The length is twice the width, so it is 2020. The area is
20⋅10=200.20 \cdot 10 = 200.

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

Review in the AoPS books

Similar problems

Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.