1956 AMC 12 Problem 39

Attempt Problem 39 of the 1956 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1956 AMC 12 solutions, or check the answer key.

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39.

The hypotenuse

c

and one arm

a

of a right triangle are consecutive integers. The square of the second arm is:

ca

\dfrac ca

c+a

c-a

none of these

Answer: C

Concepts:Pythagorean theorem difference of squares consecutive integers

Difficulty rating: 1470

Small hint:

If the second arm is

b,

then

b^2=c^2-a^2

Big hint:

Factor the difference of squares and use

c-a=1

Solution:

By the Pythagorean theorem,

b^2=c^2-a^2=(c-a)(c+a).

Since

c

and

a

are consecutive and

c\gt a,

we have

c-a=1.

Hence

b^2=c+a.

Thus, the correct answer is C.

Problem 39 in other years