2019 AMC 10B Problems/Problem 15
Problem
Right triangles and , have areas of 1 and 2, respectively. A side of is congruent to a side of , and a different side of is congruent to a different side of . What is the square of the product of the lengths of the other (third) sides of and ?
Solution 1
First of all, let the two sides which are congruent be and , where . The only way that the conditions of the problem can be satisfied is if is the shorter leg of and the longer leg of , and is the longer leg of and the hypotenuse of .
Notice that this means the value we are looking for is the square of , which is just .
The area conditions give us two equations: and .
This means that and that .
Taking the second equation, we get , so since , .
Since , we get .
The value we are looking for is just so the answer is .
Solution 2
Like in Solution 1, we have and .
Squaring both equations yields and .
Let and . Then , and , so .
We are looking for the value of , so the answer is .
Solution 3
Firstly, let the right triangles be and , with being the smaller triangle. As in Solution 1, let and . Additionally, let and .
We are given that and , so using , we have and . Dividing the two equations, we get = , so .
Thus is a right triangle, meaning that . Now by the Pythagorean Theorem in , .
The problem requires the square of the product of the third side lengths of each triangle, which is . By substitution, we see that = . We also know .
Since we want , multiplying both sides by gets us . Now squaring gives .
[Note: there is a mismatch of variables near the beginning that someone can fix: xw/2 is the area of the small triangle, which is actually the 30-60-90.]
Video Solution
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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