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

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edit by Lej: Area of Incircle = (incircle radius)(Semiperimeter)
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edit by Lej: Area of Incircle = (inradius)(Semiperimeter)
  
 
==See Also==
 
==See Also==
 
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{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}
 
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{{MAA Notice}}

Revision as of 09:23, 16 January 2019

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 inscribed 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 note that by the converse of the Pythagorean Theorem, $\triangle ABC$ is a right triangle with a right angle at $A$. Therefore, $AD = BD = CD = 5$, and $[ADB] = [ADC] = 12$. Since $A = rs$, the inradius of $\triangle ADB$ is $\frac{12}{(5+5+6)/2} = \frac 32$, and the inradius of $\triangle ADC$ is $\frac{12}{(5+5+8)/2} = \frac 43$. Adding the two together, we have $\boxed{\textbf{(D) } \frac{17}6}$.


edit by asdf334: what do you mean by A = rs?

edit by Lej: Area of Incircle = (inradius)(Semiperimeter)

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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