Difference between revisions of "2017 AMC 10B Problems/Problem 21"

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

Revision as of 12:12, 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

Placeholder

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

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