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 $\frac{1}{2}(AB)(HC)=\frac{1}{2}(25)(12)=150, \boxed{\text{C}}$.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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