Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}</math> | <math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}</math> | ||
| − | ==Solution | + | ==Solution 1 (Pythagorean Theorem)== |
We can draw another radius from the center to the point of tangency. This angle, <math>\angle{ODB}</math>, is <math>90^\circ</math>. Label the center <math>O</math>, the point of tangency <math>D</math>, and the radius <math>r</math>. | We can draw another radius from the center to the point of tangency. This angle, <math>\angle{ODB}</math>, is <math>90^\circ</math>. Label the center <math>O</math>, the point of tangency <math>D</math>, and the radius <math>r</math>. | ||
<asy> | <asy> | ||
Revision as of 09:51, 24 July 2024
Contents
Problem
In the right triangle 
, 
, 
, and angle 
is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
![[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("$A$", (0,0), W); label("$C$", (12,0), E); label("$B$", (12,5), NE); label("$12$", (6, 0), S); label("$5$", (12, 2.5), E);[/asy]](http://latex.artofproblemsolving.com/1/7/f/17f66d4cef8b689eb626ec8c226380e1e72d8719.png)
Solution 1 (Pythagorean Theorem)
We can draw another radius from the center to the point of tangency. This angle, 
, is 
. Label the center 
, the point of tangency 
, and the radius 
.
![[asy] draw((0,0)--(12,0)--(12,5)--(0,0)); draw(arc((8.67,0),(12,0),(5.33,0))); label("$A$", (0,0), W); label("$C$", (12,0), E); label("$B$", (12,5), NE); label("$12$", (6, 0), S); label("$5$", (12, 2.5), E); draw((8.665,0)--(7.4,3.07)); label("$O$", (8.665, 0), S); label("$D$", (7.4, 3.1), NW); label("$r$", (11, 0), S); label("$r$", (7.6, 1), W); [/asy]](http://latex.artofproblemsolving.com/3/8/f/38fe1daab146801175ebc91b54d9343afa54e6b3.png)
Since 
is a kite, then 
. Also, 
. By the Pythagorean Theorem, 
. Solving, 
.
~MrThinker
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solutions
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See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
