1954 AMC 12 Problem 43

Question

The hypotenuse of a right triangle is

10

inches and the radius of the inscribed circle is

1

inch. The perimeter of the triangle in inches is:

15

22

24

26

30

For a right triangle with legs

a,b

and hypotenuse

c,

the inradius is

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

Substitute

r=1

and

c=10

to find

a+b

For a right triangle,

r=\frac{a+b-c}{2}.

With

r=1

and

c=10,

this gives

a+b=12.

Hence the perimeter is

a+b+c=12+10=22.

Thus, the correct answer is B.

Problem 43 in other years