1951 AMC 12 Problem 49

Question

The medians of a right triangle which are drawn from the vertices of the acute angles are

5

and

\sqrt{40}.

The value of the hypotenuse is:

10

2\sqrt{40}

\sqrt{13}

2\sqrt{13}

None of these

Let the legs be

a,b

and place the right angle at their common endpoint

The two median lengths give

a^2+\frac{b^2}{4}=25

and

\frac{a^2}{4}+b^2=40

Let the legs be

a,b.

The medians from their opposite acute vertices have squared lengths

a^2+\frac{b^2}{4}=25,\qquad \frac{a^2}{4}+b^2=40.

Solving gives

a^2=16

and

b^2=36.

Thus the hypotenuse

c

satisfies

c^2=a^2+b^2=52,

so

c=2\sqrt{13}.

Thus, the correct answer is D.

Problem 49 in other years