2010 AMC 10B Problems/Problem 7

Problem

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?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$

Solution 1

The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - (	\dfrac{12}{2})^2} =  \sqrt{64} = 8$

Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$

We have that the area of the rectangle is the same as the area of the triangle, namely $48$. We also have the width of the rectangle: $4$.

The length of the rectangle therefore is: $l = \dfrac{48}{4} = 12$

The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$

The answer is:

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

Solution 2

An alternative way to find the area of the triangle is by using Heron's formula, $A=\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$. Thus the area equals $\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48.$ We know that the width of the rectangle is $4$, so $48/4 = 12$, which is the length. The perimeter of the rectangle is $2(4+12) =$ $\boxed{\textbf{(D)}\ 32}$.

(Solution by Flamedragon)

Solution 3

Note that a triangle with side lengths $10,10,12$ is essentially $2$ “6,8,10” right triangles stuck together. Hence, the height is $8$, and our area is $48$.

So, the length of the rectangle is $\frac{48}{4}=12$, and our perimeter $P=2(4+12)=\boxed{\textbf{(D)}\ 32}$

Video Solution

https://youtu.be/I3yihAO87CE?t=89

~IceMatrix

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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