2017 AMC 10B Problems/Problem 21

Revision as of 13:30, 16 February 2017 by Willwin4sure (talk | contribs) (Solution)

Problem

In $\triangle ABC$, $AB=6$, $AC=8$, $BC=10$, and $D$ is the midpoint of $\overline{BC}$. What is the sum of the radii of the circles inscibed in $\triangle ADB$ and $\triangle ADC$?

$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3$

Solution

We can use the formula that states that the area of a triangle is equal to the inradius times the semiperimeter. We know that $AD=BD=CD=5$, and that $[ABD]=[ACD]=12$, where $[P]$ is the area of polygon $P$. We can determine the semiperimeters of $ABD$ and $ACD$ as $\frac{5+5+6}{2}=8$ and $\frac{5+5+8}{2}=9$, respectively. Thus, the sum of the inradii is $\frac{12}{8}+\frac{12}{9}=\boxed{\frac{17}{6}(\text{D})}$.

~willwin4sure

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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. AMC logo.png