1984 AHSME Problems/Problem 17

Problem

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 $\triangle ABC$ is:

$\mathrm{(A) \ }120 \qquad \mathrm{(B) \ }144 \qquad \mathrm{(C) \ } 150 \qquad \mathrm{(D) \ }216 \qquad \mathrm{(E) \ } 144\sqrt{5}$

Solution

$[asy] unitsize(.4cm); draw((0,0)--(0,12)); draw((9,0)--(-16,0)); draw((9,0)--(0,12)); draw((-16,0)--(0,12)); label("$A$",(9,0),ENE); label("$B$",(-16,0),WNW); label("$C$",(0,12),ENE); label("$H$",(0,0),S); label("$15$",(4.5,6),NE); label("$16$",(-8,0),S); [/asy]$ $AHC\sim ACB$ by $AA$ , so $\frac{AH}{AC}=\frac{AC}{AH+16}$ . Since $AC=15$ , we have $\frac{AH}{15}=\frac{15}{AH+16}$ . Cross mutliplying, we have $AH(AH+16)=225$ . Solving this quadratic yields $AH=9$ . Also, $AHC\sim CHB$ , so $\frac{AH} {HC}=\frac{HC}{HB}$ . Substituting in known values, we have $\frac{9}{HC}=\frac{HC}{16}$ , so $HC^2=144$ and $HC=12$ .

The area of $\triangle ABC$ is $\frac{1}{2}(AB)(HC)=\frac{1}{2}(25)(12)=150, \boxed{\text{C}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1984 AHSME Problems/Problem 17

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