Difference between revisions of "2009 AMC 10A Problems/Problem 10"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | == Solution == | + | == Solution 1== |
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>. | 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>. | ||
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Alternatively, note that <math>\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}</math>. | Alternatively, note that <math>\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}</math>. | ||
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+ | == Solution 2 == | ||
+ | |||
+ | For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas as well to solve the problem. | ||
+ | |||
+ | Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have, | ||
+ | |||
+ | <cmath>AB^2 = h^2 + 9 \Rightarrow AB = \sqrt{h^2 + 9}</cmath> | ||
+ | <cmath>BC^2 = h^2 + 16 \Rightarrow BC = \sqrt{h^2 + 16}</cmath> | ||
+ | |||
+ | Then, by area formulas, we know that | ||
+ | |||
+ | <cmath>\frac{1}{2}(\sqrt{(h^2+9)(h^2+16}) = \frac{1}{2}(7)(h)</cmath> | ||
+ | |||
+ | Squaring and solving the subsequent equation yields our solution that <math>h^2 = 12 \Rightarrow h = 2\sqrt{3}.</math> Since the area of the triangle is half of this quantity into the base, we have | ||
+ | <cmath>\text{area} = \frac{1}{2}(7)(2\sqrt{3})\Rightarrow \boxed{7\sqrt{3}}</cmath> | ||
== See Also == | == See Also == |
Revision as of 20:18, 29 December 2020
Contents
[hide]Problem
Triangle has a right angle at . Point is the foot of the altitude from , , and . What is the area of ?
Solution 1
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 .
Solution 2
For those looking for a dumber solution, we can use Pythagoras and manipulation of area formulas as well to solve the problem.
Assume the length of is equal to . Then, by Pythagoras, we have,
Then, by area formulas, we know that
Squaring and solving the subsequent equation yields our solution that Since the area of the triangle is half of this quantity into the base, we have
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.