1950 AMC 12 Problem 35

Question

In triangle

ABC,

AC=24

inches,

BC=10

inches,

AB=26

inches. The radius of the inscribed circle is:

26

in

4

in

13

in

8

in

None of these

The side lengths form a right triangle

For a right triangle with legs

a,b

and hypotenuse

c,

the inradius is

\frac{a+b-c}{2}

Since

10^2+24^2=26^2,

the triangle is right. Its area is

\tfrac12(10)(24)=120,

and its semiperimeter is

\tfrac{10+24+26}{2}=30.

Using

K=rs,

the inradius, in inches, is

r=\frac{K}{s}=\frac{120}{30}=4.

Thus, the correct answer is B.

Problem 35 in other years