Difference between revisions of "2009 AMC 12B Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | Let <math>D</math> be the foot of the altitude to <math>\overline BC</math>. Then <math>BD = \sqrt {13^2 - 12^2} = 5</math> and <math>DC = \sqrt {15^2 - 12^2} = 9</math>. Thus <math>BC = BD + BC = 5 + 9 = 14</math> or <math>BC = DC - BD = 9 -5 = 4</math>. The sum of the two possible values is <math>14 + 4 = \boxed{18}</math>. The answer is <math>\mathrm{(D)}</math>. | + | Let <math>D</math> be the foot of the altitude to <math>\overline BC</math>. Then <math>BD = \sqrt {13^2 - 12^2} = 5</math> and <math>DC = \sqrt {15^2 - 12^2} = 9</math>. Thus <math>BC = BD + BC = 5 + 9 = 14</math> or assume that the triangle is obtuse at angle <math>B</math> then <math>BC = DC - BD = 9 -5 = 4</math>. The sum of the two possible values is <math>14 + 4 = \boxed{18}</math>. The answer is <math>\mathrm{(D)}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2009|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2009|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:42, 23 February 2017
Problem
Triangle 
has 
and 
, and the altitude to 
has length 
. What is the sum of the two possible values of 
?
Solution
Let 
be the foot of the altitude to 
. Then 
and 
. Thus 
or assume that the triangle is obtuse at angle 
then 
. The sum of the two possible values is 
. The answer is 
.
See also
| 2009 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
