2011 AMC 8 Problems/Problem 16

Let $A$ be the area of the triangle with sides of length $25, 25$ , and $30$ . Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$ . What is the relationship between $A$ and $B$ ?

$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad\textbf{(D) } A = \dfrac43B \\ \\ \textbf{(E) }A = \dfrac{16}9B$

Solution

Using the Pythagorean Theorem, the height of the first triangle is $\sqrt{25^2-15^2}=20$ . Using this, its area is $A=\frac{20\cdot 30}{2}=300$ . Similarly, the height of the second triangle is $\sqrt{25^2-20^2}=15$ , and the area is $B=\frac{15\cdot 40}{2}=300$ . Therefore, $A=B$ and the answer is $\boxed{\textbf{(C)}}$ .

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2011 AMC 8 Problems/Problem 16

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