Difference between revisions of "1958 AHSME Problems/Problem 36"
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, <cmath>30^2+y^2=x^2 \qquad(1)</cmath> | , <cmath>30^2+y^2=x^2 \qquad(1)</cmath> | ||
and <cmath>(80-x)^2=70^2+y^2 \qquad(2)</cmath> | and <cmath>(80-x)^2=70^2+y^2 \qquad(2)</cmath> | ||
| − | Adding <math>(1)</math> and <math>(2)</math> and simplifying gives <math>x=15</math>. Therefore, the answer is <math>80-15=\boxed{(D)~65}</math> | + | Adding <math>(1)</math> and <math>(2)</math> and simplifying gives <math>x=15</math>. Therefore, the answer is <math>80-15=\boxed{\textbf{(D)}~65}</math> |
~megaboy6679 | ~megaboy6679 | ||
Revision as of 20:20, 27 January 2023
Problem
The sides of a triangle are 
, 
, and 
units. If an altitude is dropped upon the side of length , the larger segment cut off on this side is:
Solution
Let the shorter segment be x. The larger segment is therefore 80-x. Let the altitude be y. By the Pythagorean Theorem
, ![\[30^2+y^2=x^2 \qquad(1)\]](http://latex.artofproblemsolving.com/2/0/0/20062b5bedf933da2221ed124b7073921e02b17f.png)
and ![\[(80-x)^2=70^2+y^2 \qquad(2)\]](http://latex.artofproblemsolving.com/e/2/0/e20002d9536b22208b458dd6879349b0fa2d19d9.png)
Adding 
and 
and simplifying gives 
. Therefore, the answer is 
~megaboy6679
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 35 |
Followed by Problem 37 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
