Difference between revisions of "2017 AMC 8 Problems/Problem 22"
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+ | ==Solution 3== | ||
+ | Like solution 2, we reflect <math>\triangle ABC</math> over line <math>\overline{AC}</math> and label the reflection of point <math>B</math> as <math>D</math>. As <math>AB = AD = 13</math> by the Pythagorean Theorem, we use the formula <math>rs=A</math>, where <math>r</math> is the inradius (what we're trying to find), <math>s</math> is the semiperimeter (<math>\frac{\overline{AB}+\overline{AD}+\overline{BD}}{2}</math>), and <math>A</math> is the area of the triangle in which the incircle is inscribed in. Substitution gives: <cmath>r=\frac{\frac{10\cdot12}{2}}{\frac{13+13+10}{2}}</cmath> | ||
+ | <cmath>r=\frac{60}{18}</cmath> | ||
+ | <cmath>r=\boxed{\textbf{(D) }\frac{10}{3}}</cmath> | ||
+ | |||
+ | ~megaboy6679 | ||
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== |
Revision as of 19:39, 2 November 2024
Contents
[hide]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?
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 .
Since is a kite, then . Also, . By the Pythagorean Theorem, . Solving, .
~MrThinker
Solution 2 (Basic Trigonometry)
If we reflect triangle over line , we will get isosceles triangle . By the Pythagorean Theorem, we are capable of finding out that the . Hence, . Therefore, as of triangle , the radius of its inscribed circle
Solution 3
Like solution 2, we reflect over line and label the reflection of point as . As by the Pythagorean Theorem, we use the formula , where is the inradius (what we're trying to find), is the semiperimeter (), and is the area of the triangle in which the incircle is inscribed in. Substitution gives:
~megaboy6679
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solutions
- savannahsolver
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.