Difference between revisions of "1958 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
| + | [asy] | ||
| + | size(300); | ||
| + | defaultpen(linewidth(0.8)); | ||
| + | pair A=(-1,0),C=(1,0),B=dir(40),D=origin; | ||
| + | draw(A--B--C--A); | ||
| + | draw(D--B); | ||
| + | dot("<math>A</math>", A, SW); | ||
| + | dot("<math>B</math>", B, NE); | ||
| + | dot("<math>C</math>", C, SE); | ||
| + | dot("<math>D</math>", D, S); | ||
| + | label("<math>70^\circ</math>",C,2*dir(180-35)); | ||
| + | [/asy] | ||
<math>\fbox{}</math> | <math>\fbox{}</math> | ||
Revision as of 02:04, 22 December 2015
Problem
The altitude drawn to the base of an isosceles triangle is 
, and the perimeter 
. The area of the triangle is:
Solution
[asy]
size(300);
defaultpen(linewidth(0.8));
pair A=(-1,0),C=(1,0),B=dir(40),D=origin;
draw(A--B--C--A);
draw(D--B);
dot("
", A, SW);
dot("
", B, NE);
dot("
", C, SE);
dot("
", D, S);
label("
",C,2*dir(180-35));
[/asy]
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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. 
