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2010 AMC 10B Problem 7

2010 AMC 10B Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 2 min · position P1-10

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?

A)
$16$
B)
$24$
C)
$28$
D)
$32$
E)
$36$

Answer

Key insight

The altitude of the 10-10-12 isosceles triangle splits it into two 6-8-10 right triangles, so its area is 48.

Solution

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 $\boxed{\textbf{(D)}\ 32}$ .

Why this works

An isosceles triangle always hides two congruent right triangles; the altitude to the base gives you a Pythagorean computation and hence the area. After that, the rectangle is a one-line conversion: area over width gives length, and the perimeter follows.

The trap

Reporting the rectangle's length 12 or its area 48 instead of the perimeter, or using 12 as the height of the triangle.

Common mistakes

Reporting the rectangle's length 12 or its area 48 instead of the perimeter, or using 12 as the height of the triangle.
Computing the triangle's area as $\frac12 \cdot 10 \cdot 12 = 60$ by treating two sides as base and height.

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Set up the equation/formula and compute; no special trick needed