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 (Using the Pythagorean Theorem)
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)
Using Heron's formula, we can calculate the area of the two triangles. The formula states that where is the semiperimeter of a triangle with side lengths , , and .
For the 25-25-30 triangle, Therefore,
For the 25-25-40 triangle, Therefore,
Hence,
Video Solution 1
https://www.youtube.com/watch?v=mYn6tNxrWBU
~ by ==SpreadtheMathLove==
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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