1958 AMC 12 Problem 36

Question

The sides of a triangle are

30,

70,

and

80

units. If an altitude is dropped upon the side of length

80,

the larger segment cut off on this side is:

62

63

64

65

66

Let the altitude divide the side of length

80

into

x

and

80-x

Equate the two expressions for the square of the altitude

Let

x

be the segment adjacent to the side of length

30.

If the altitude has length

h,

then

\begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2. \end{aligned}

Cancelling

x^2

and solving gives

x=15.

The other segment is

80-15=65,

which is the larger one.

Thus, the correct answer is D.

Problem 36 in other years