2019 AMC 10B Problems/Problem 15
Contents
Problem
Two right triangles, and
, have areas of 1 and 2, respectively. One side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other. What is the square of the product of the third side lengths of
and
?
Solution
First of all, name the two sides which are congruent to be and
, where
. The only way that the conditions of the problem can be satisfied is if
was 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
.
We have 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
First, construct right triangles △ABC and △EDF, with △ABC being the smaller triangle. We are given that one side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other.
So, ≅
, call this length
, and
≅
, call this length
Additionally, call the length
, and call the length
Recapping our variables, we have =
=
,
=
=
,
=
, and
=
We are given that and
Since area = , this gives
and
Dividing the two equations, we get =
From this, we get
We see that △EDF is a right triangle, meaning that
In △ABC, and
are the legs. By the Pythagorean Theorem,
The question asks for the square of the product of the third side lengths of each triangle, which is
Using substitution, we see that =
) =
We know
Dividing both sides by , we get
Since we want , multiplying both sides by
gets us
Squaring this,
.
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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