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

25-25-30

We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have $15^2 + x^2 =25^2$

\[\x^2 = 25^2 - 15^2\] (Error compiling LaTeX. Unknown error_msg)

$x^2 = (25 + 15)(25-15)$ $x^2= 40\cdot 10$ $x^2= 400$ $x = \sqrt{400}$ $x= 20$

Thus we have two 15-20-25 right triangles.

25-25-40

We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles. Let the area of a 15-20-25 right triangle be $x$ . $a = 2x$ $b = 2x$ $\textbf{C) }\boxed{a = b}$

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

Solution

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