2015 AMC 10A Problems/Problem 19
Contents
[hide]Problem
The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?
Solution
can be split into a right triangle and a right triangle by dropping a perpendicular from to side . Let be where that perpendicular intersects .
Because the side lengths of a right triangle are in ratio , .
Because the side lengths of a right triangle are in ratio and + , .
Setting the two equations for equal to each other, .
Solving gives .
The area of .
is congruent to , so their areas are equal.
A triangle's area can be written as the sum of the figures that make it up, so .
.
Solving gives , so the answer is .
Solution 2
The area of is 12.5, and so the leg length of 45-45-90 is 5. Thus, the altitude to hypotenuse , , has length by 45-45-90 right triangles. Now, it is clear that , and so by the Exterior Angle Theorem, is an isosceles 30-75-75 triangle. Thus, , and so the area of is . The answer is .
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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