Difference between revisions of "2006 iTest Problems/Problem 11"
Rockmanex3 (talk | contribs) (Solution to Problem 11 -- Inscribed Circle of a Triangle) |
Shurong.ge (talk | contribs) (→Solution) |
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==Solution== | ==Solution== | ||
− | Since we know three sides, we can calculate the semiperimeter and area and use the results to find the radius of the [[incircle]]. The semiperimeter of the triangle is <math>\tfrac12 \cdot (13 + 30 + 37) = 40</math>. By [[Heron's Formula]], the area of the triangle is <math>\sqrt{40 \cdot 27 \cdot 10 \cdot 3} = 180</math>. Since the radius of the incircle times the semiperimeter equals the area, the radius of the incircle must be <math>\tfrac{180}{40} = \boxed{\textbf{(A) }\tfrac{9}{2}}</math>. | + | Since we know three sides, we can calculate the semiperimeter and area and use the results to find the radius of the [[incircle]]. The semiperimeter of the triangle is <math>\tfrac12 \cdot (13 + 30 + 37) = 40</math>. By [[Heron's Formula]], the area of the triangle is <math>\sqrt{40 \cdot 27 \cdot 10 \cdot 3} = 180</math>. Since the radius of the incircle times the semiperimeter equals the area, <math>A = sr</math> the radius of the incircle must be <math>\tfrac{180}{40} = \boxed{\textbf{(A) }\tfrac{9}{2}}</math>. |
==See Also== | ==See Also== |
Latest revision as of 20:49, 21 January 2020
Problem
Find the radius of the inscribed circle of a triangle with sides of length , , and .
Solution
Since we know three sides, we can calculate the semiperimeter and area and use the results to find the radius of the incircle. The semiperimeter of the triangle is . By Heron's Formula, the area of the triangle is . Since the radius of the incircle times the semiperimeter equals the area, the radius of the incircle must be .
See Also
2006 iTest (Problems) | ||
Preceded by: Problem 10 |
Followed by: Problem 12 | |
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