1950 AHSME Problems/Problem 35

Problem

In triangle $ABC$ , $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:

$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$

Solution

The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\boxed{\textbf{(B)}\ 4}$ .

Solution 2

Since this is a right triangle, we have $\frac{a+b-c}{2}=\boxed{4}$

- kante314

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 34	Followed by Problem 36
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1950 AHSME Problems/Problem 35

Contents

Problem

Solution

Solution 2

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