Difference between revisions of "1984 AHSME Problems/Problem 17"
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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>. | ||
Latest revision as of 18:13, 12 March 2018
Problem
A right triangle 
with hypotenuse 
has side 
. Altitude 
divides into segments 
and 
, with 
. The area of 
is:
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]](http://latex.artofproblemsolving.com/9/f/d/9fd2d6da5d64a48e003fd0fd9198da3736be361e.png)
by 
, so 
. Since , we have 
. Cross mutliplying, we have 
. Solving this quadratic yields 
. Also, 
, so 
. Substituting in known values, we have 
, so 
and 
.
The area of is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
