# 2006 iTest Problems/Problem 11

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## 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, the radius of the incircle must be $\tfrac{180}{40} = \boxed{\textbf{(A) }\tfrac{9}{2}}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 10 Followed by:Problem 12 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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