Difference between revisions of "2009 AMC 10A Problems/Problem 10"
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== Solution == | == Solution == | ||
− | It is a well-known fact that in any right triangle <math>ABC</math> with the right angle at <math>B</math> and <math>D</math> the foot of the altitude from <math>B</math> onto <math>AC</math> we have <math>BD^2 = AD\cdot CD</math>. (See below for a proof.) Then <math>BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac{AC\cdot BD}2 = \boxed{ | + | It is a well-known fact that in any right triangle <math>ABC</math> with the right angle at <math>B</math> and <math>D</math> the foot of the altitude from <math>B</math> onto <math>AC</math> we have <math>BD^2 = AD\cdot CD</math>. (See below for a proof.) Then <math>BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3</math>, and the area of the triangle <math>ABC</math> is <math>\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\boxed{\text{(B)}}</math>. |
''Proof'': Consider the [[Pythagorean theorem]] for each of the triangles <math>ABC</math>, <math>ABD</math>, and <math>CBD</math>. We get: | ''Proof'': Consider the [[Pythagorean theorem]] for each of the triangles <math>ABC</math>, <math>ABD</math>, and <math>CBD</math>. We get: |
Revision as of 20:24, 2 November 2012
Problem
Triangle has a right angle at . Point is the foot of the altitude from , , and . What is the area of ?
Solution
It is a well-known fact that in any right triangle with the right angle at and the foot of the altitude from onto we have . (See below for a proof.) Then , and the area of the triangle is .
Proof: Consider the Pythagorean theorem for each of the triangles , , and . We get:
- .
Substituting equations 2 and 3 into the left hand side of equation 1, we get .
Alternatively, note that .
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |