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Difference between revisions of "2006 iTest Problems/Problem 11"

(Solution to Problem 11 -- Inscribed Circle of a Triangle)
 
(→‎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 $13$, $30$, and $37$.

$\text{(A) }\frac{9}{2}\qquad \text{(B) }\frac{7}{2}\qquad \text{(C) }4\qquad \text{(D) }-\sqrt{2}\qquad \text{(E) }4\sqrt{5}\qquad \text{(F) }6\qquad \\ \text{(G) }\frac{11}{2}\qquad \text{(H) }\frac{13}{2}\qquad \text{(I) }\text{none of the above}\qquad \text{(J) }1\qquad \text{(K) }\text{no triangle exists}\qquad$

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 $\tfrac12 \cdot (13 + 30 + 37) = 40$. By Heron's Formula, the area of the triangle is $\sqrt{40 \cdot 27 \cdot 10 \cdot 3} = 180$. Since the radius of the incircle times the semiperimeter equals the area, $A = sr$ the radius of the incircle must be $\tfrac{180}{40} = \boxed{\textbf{(A) }\tfrac{9}{2}}$.

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 10 		Followed by:
Problem 12
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