Difference between revisions of "2017 AMC 10B Problems/Problem 15"
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==Solution== | ==Solution== | ||
+ | |||
+ | <asy> | ||
+ | pair A,B,C,D,E; | ||
+ | A=(0,4); | ||
+ | B=(3,4); | ||
+ | C=(3,0); | ||
+ | D=(0,0); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,N); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,S); | ||
+ | E=foot(B,A,C); | ||
+ | draw(E--B); | ||
+ | draw(A--C); | ||
+ | draw(rightanglemark(B,E,C)); | ||
+ | label("$E$",E,N); | ||
+ | draw(D--E); | ||
+ | </asy> | ||
First, note that <math>AC=5</math> because <math>ABC</math> is a right triangle. In addition, we have <math>AB\cdot BC=2[ABC]=AC\cdot BE</math>, so <math>BE=\frac{12}{5}</math>. Using similar triangles within <math>ABC</math>, we get that <math>AE=\frac{9}{5}</math> and <math>CE=\frac{16}{5}</math>. | First, note that <math>AC=5</math> because <math>ABC</math> is a right triangle. In addition, we have <math>AB\cdot BC=2[ABC]=AC\cdot BE</math>, so <math>BE=\frac{12}{5}</math>. Using similar triangles within <math>ABC</math>, we get that <math>AE=\frac{9}{5}</math> and <math>CE=\frac{16}{5}</math>. |
Revision as of 15:01, 13 August 2017
Problem
Rectangle has and . Point is the foot of the perpendicular from to diagonal . What is the area of ?
Solution
First, note that because is a right triangle. In addition, we have , so . Using similar triangles within , we get that and .
Let be the foot of the perpendicular from to . Since and are parallel, is similar to . Therefore, we have . Since , . Note that is an altitude of from , which has length . Therefore, the area of is
See Also
2017 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.