2011 AMC 8 Problems/Problem 16
Contents
Problem
Let be the area of the triangle with sides of length , and . Let be the area of the triangle with sides of length and . What is the relationship between and ?
Solution 1
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
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 .
Solution 2
Using Heron's formula, we can calculate the area of the two triangles. The formula states that
\[\begin{center} A = \sqrt{s(s - a)(s - b)(s - c)} \end{center}\] (Error compiling LaTeX. Unknown error_msg)
where is the semiperimeter of a triangle with side lengths , , and .
For the triangle, . Therefore,
\[\begin{center} A = \sqrt{40 \cdot 15 \cdot 15 \cdot 10} = 300 \end{center}\] (Error compiling LaTeX. Unknown error_msg)
For the triangle, . Therefore,
\[\begin{center} B = \sqrt{45 \cdot 20 \cdot 20 \cdot 5} = 300 \end{center}\] (Error compiling LaTeX. Unknown error_msg)
Hence,
\[A = B \longrightarrow \boxed{textbf{(C) }\] (Error compiling LaTeX. Unknown error_msg)
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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