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Difference between revisions of "1984 AHSME Problems/Problem 17"

(Created page with "==Problem== A right triangle <math> ABC </math> with hypotenuse <math> AB </math> has side <math> AC=15 </math>. Altitude <math> CH </math> divides <math> AB </math> ...")
 
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</math>. Cross mutliplying, we have <math> AH(AH+16)=225 </math>. Solving this [[quadratic]] yields <math> AH=9 </math>. Also, <math> AHC\sim CHB </math>, so <math> \frac{AH}
 
</math>. Cross mutliplying, we have <math> AH(AH+16)=225 </math>. Solving this [[quadratic]] yields <math> AH=9 </math>. Also, <math> AHC\sim CHB </math>, so <math> \frac{AH}
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{HC}=\frac{HC}{HB} </math>. Substituting in known values, we have <math> \frac{9}{HC}=\frac{HC}{16} </math>, so <math> HC^2=144 </math> and <math> HC=12 </math>.  
 
{HC}=\frac{HC}{HB} </math>. Substituting in known values, we have <math> \frac{9}{HC}=\frac{HC}{16} </math>, so <math> HC^2=144 </math> and <math> HC=12 </math>.  
  
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==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=16|num-a=18}}
 
{{AHSME box|year=1984|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 17:13, 12 March 2018

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 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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