2010 AMC 10B Problem 7

P1-10~2 minPrint

Triangles: Area & Pythagorean Quadrilaterals & Polygon Areas Add construction lines/points (drop altitudes, extend segments, connect centers)Set up the equation/formula and compute; no special trick needed

A triangle has side lengths $10$ , $10$ , and $12$ . A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

Show solution without answering

Key insight

Cut the 10-10-12 triangle down the middle. Each half is a right triangle with sides 6, 8, 10, so the height is 8 and the area is 48.

Solution

To use the rectangle we need its area, which is the triangle's area. The triangle has two equal sides, so it splits neatly into two right triangles.

1. Draw the height from the top corner (where the two sides of length $10$ meet) straight down to the side of length $12$ . Because the two slanted sides are equal, this height lands exactly in the middle, cutting the $12$ into $6$ and $6$ .
2. Each half is a right triangle with long side $10$ and one short side $6$ . By the Pythagorean theorem the height is
$\sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8.$
(This is the familiar $6$ - $8$ - $10$ right triangle.)
3. The triangle's area is $\frac12 \cdot \text{base} \cdot \text{height} = \frac12 \cdot 12 \cdot 8 = 48$ .
4. The rectangle also has area $48$ and width $4$ , so its length is $48 \div 4 = 12$ .
5. Its perimeter is $4 + 12 + 4 + 12 = 32$ .

The answer is $\boxed{\textbf{(D)}\ 32}$ .

Why this works

A triangle with two equal sides always splits into two matching right triangles when you draw the height to the third side. That gives the height by the Pythagorean theorem, and from there the area. Read the last line of the question carefully: it asks for the perimeter, not the area or the length.

Concise solution

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

Drop the altitude from the apex of the isosceles triangle to the side of length $12$ . It bisects that side into two segments of length $6$ , creating two right triangles with hypotenuse $10$ and leg $6$ . The altitude is therefore $\sqrt{10^2 - 6^2} = 8$ (a $6$ - $8$ - $10$ triangle).

Area of the triangle: $\frac{1}{2}\cdot 12 \cdot 8 = 48$ .

The rectangle has width $4$ and area $48$ , so its length is $48/4 = 12$ . Its perimeter is $2(4 + 12) = 32$ .

The answer is $\boxed{\textbf{(D)}\ 32}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)