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Difference between revisions of "2017 AMC 10B Problems/Problem 21"
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<math>\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</math>
 
<math>\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</math>
 
==Solution==
 
==Solution==
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Placeholder
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We can use the formula that states that the area of a triangle is equal to the inradius times the semiperimeter. We know that <math>AD=BD=CD=5</math>, and that <math>[ABD]=[ACD]=12</math>, where <math>[P]</math> is the area of polygon <math>P</math>. We can determine the semiperimeters of <math>ABD</math> and <math>ACD</math> as <math>\frac{5+5+6}{2}=8</math> and <math>\frac{5+5+8}{2}=9</math>, respectively. Thus, the sum of the inradii is <math>\frac{12}{8}+\frac{12}{9}=\boxed{\frac{17}{6}(\text{D})}</math>.
 +
 
 +
~willwin4sure
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:30, 16 February 2017

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 (Problems • Answer Key • Resources)
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

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