1984 AMC 12 Problem 17

Attempt Problem 17 of the 1984 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 1984 AMC 12 solutions, or check the answer key.

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17.

A right triangle

ABC

with hypotenuse

AB

has side

AC=15.

Altitude

CH

divides

AB

into segments

AH

and

HB,

with

HB=16.

The area of

\triangle ABC

120

144

150

216

144\sqrt5

Use the right-triangle projection relation

AC^2=AH\cdot AB

After finding

AH,

use

CH^2=AH\cdot HB

Let

AH=h.

Similarity in the right triangle gives

\begin{aligned} AC^2&=AH\cdot AB,\\ 225&=h(h+16), \end{aligned}

h=9.

Hence

AB=25.

Also

CH=\sqrt{AH\cdot HB},

CH=\sqrt{9\cdot16}=12.

The area is

\frac12\cdot25\cdot12=150.

Therefore, the correct answer is C.

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