Difference between revisions of "2015 AMC 10A Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
The area of <math>ABC</math> is 12.5, and so the leg length of 45-45-90 <math>ABC</math> is 5. Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by 45-45-90 right triangles. Now, it is clear that <math>\angle{ACD} = \angle{BCE} = 30^\circ</math>, and so by the Exterior Angle Theorem, <math>\triangle{CDE}</math> is an isosceles 30-75-75 triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math>, and so the area of <math>CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is <math>\textbf{(D)}</math>. | The area of <math>ABC</math> is 12.5, and so the leg length of 45-45-90 <math>ABC</math> is 5. Thus, the altitude to hypotenuse <math>AB</math>, <math>CF</math>, has length <math>\dfrac{5}{\sqrt{2}}</math> by 45-45-90 right triangles. Now, it is clear that <math>\angle{ACD} = \angle{BCE} = 30^\circ</math>, and so by the Exterior Angle Theorem, <math>\triangle{CDE}</math> is an isosceles 30-75-75 triangle. Thus, <math>DF = CF \tan 15^\circ = \dfrac{5}{\sqrt{2}} (2 - \sqrt{3})</math>, and so the area of <math>CDE</math> is <math>DF \cdot CF = \dfrac{25}{2} (2 - \sqrt{3})</math>. The answer is <math>\textbf{(D)}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Revision as of 21:05, 4 February 2015
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 | |
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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