1958 AMC 12 Problem 43

Question

\overline{AB}

is the hypotenuse of a right triangle

ABC.

Median

AD=7

and median

BE=4.

The length of

AB

is:

10

5\sqrt3

5\sqrt2

2\sqrt{13}

2\sqrt{15}

Let the legs opposite

A

and

B

have squared lengths

u

and

v

Use the two median formulas together with the Pythagorean relation for the hypotenuse

Let

a=BC,

b=CA,

and

c=AB.

Since the right angle is at

C,

c^2=a^2+b^2.

The median formulas give

\begin{aligned} 4(7^2) &=a^2+4b^2,\\ 4(4^2) &=4a^2+b^2. \end{aligned}

Solving yields

a^2=4

and

b^2=48.

Therefore

\begin{aligned} AB=c &=\sqrt{a^2+b^2}\\ &=\sqrt{52}=2\sqrt{13}. \end{aligned}

Thus, the correct answer is D.

Problem 43 in other years