Difference between revisions of "2016 AMC 8 Problems/Problem 25"
Reaganchoi (talk | contribs) m |
Reaganchoi (talk | contribs) m |
||
| Line 1: | Line 1: | ||
25. A semicircle is inscribed in an isosceles triangle with base <math>16</math> and height <math>15</math> so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle? | 25. A semicircle is inscribed in an isosceles triangle with base <math>16</math> and height <math>15</math> 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> | ||
<math>\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}</math> | <math>\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}</math> | ||
Revision as of 09:44, 23 November 2016
25. A semicircle is inscribed in an isosceles triangle with base 
and height 
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]](http://latex.artofproblemsolving.com/5/8/5/585fca3c7a910c5787429738a441c0c2accc394d.png)
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
