2016 AMC 8 Problem 25 Geometry

Understand the problem

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle? [asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy] $\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$
Source of the problem
2016 AMC (American Mathematical Contests) 8  problem 25
Topic
Number Theory
Difficulty Level
Easy

Start with hints

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Hint figure :   [asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW);[/asy]
  [asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW);[/asy]  Verify \( AD = 8  \) and recall the concepts of Similar Triangles. Then proceed .    
    [asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW);[/asy]   Verify \( \triangle AED \sim \triangle ACD \) . And proceed to find the radius of the semicircle i.e. DE . [NOTE : As DE is the radius and AC be the tangent so \( DE \perp AC \)  .]
We have \( \triangle AED \sim \triangle ACD \) . This similarity means that we can create a proportion: \( \frac {AD}{AC} = \frac {DE}{CD} \) . We plug in \( AD = 8 , AC = 17 \ and \ CD = 15 \) . So \( DE = \frac {8}{17}  \times 15   = \frac {120}{17} \)  .

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