2017 AMC 10B Problems/Problem 21

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

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2017 AMC 10B Problems/Problem 21

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