Difference between revisions of "1958 AHSME Problems/Problem 19"
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== Problem == | == Problem == | ||
| − | The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b | + | The sides of a right triangle are <math> a</math> and <math> b</math> and the hypotenuse is <math> c</math>. A perpendicular from the vertex divides <math> c</math> into segments <math> r</math> and <math> s</math>, adjacent respectively to <math> a</math> and <math> b</math>. If <math> a : b = 1 : 3</math>, then the ratio of <math> r</math> to <math> s</math> is: |
<math> \textbf{(A)}\ 1 : 3\qquad | <math> \textbf{(A)}\ 1 : 3\qquad | ||
Latest revision as of 03:30, 29 June 2017
Problem
The sides of a right triangle are 
and 
and the hypotenuse is 
. A perpendicular from the vertex divides into segments 
and 
, adjacent respectively to and . If 
, then the ratio of to is:
Solution
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
