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Difference between revisions of "2009 AMC 12B Problems/Problem 13"

(New page: == Problem == Triangle <math>ABC</math> has <math>AB = 13</math> and <math>AC = 15</math>, and the altitude to <math>\overline{BC}</math> has length <math>12</math>. What is the sum of the...)
 
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== Solution ==
 
== Solution ==
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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>.
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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}}
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{{MAA Notice}}

Latest revision as of 17:42, 23 February 2017

Problem

Triangle $ABC$ has $AB = 13$ and $AC = 15$, and the altitude to $\overline{BC}$ has length $12$. What is the sum of the two possible values of $BC$?

$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$

Solution

Let $D$ be the foot of the altitude to $\overline BC$. Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$. Thus $BC = BD + BC = 5 + 9 = 14$ or assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$. The sum of the two possible values is $14 + 4 = \boxed{18}$. The answer is $\mathrm{(D)}$.

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

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