1958 AHSME Problems/Problem 31: Difference between revisions
| Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
[asy] | <math>[asy] | ||
size(300); | size(300); | ||
defaultpen(linewidth(0.8)); | defaultpen(linewidth(0.8)); | ||
| Line 15: | Line 15: | ||
draw(A--B--C--A); | draw(A--B--C--A); | ||
draw(D--B); | draw(D--B); | ||
dot("<math>A< | dot("</math>A<math>", A, SW); | ||
dot("<math>B< | dot("</math>B<math>", B, NE); | ||
dot("<math>C< | dot("</math>C<math>", C, SE); | ||
dot("<math>D< | dot("</math>D<math>", D, S); | ||
label("<math>70^\circ< | label("</math>70^\circ<math>",C,2*dir(180-35)); | ||
[/asy] | [/asy]</math> | ||
<math>\fbox{}</math> | <math>\fbox{}</math> | ||
Revision as of 01:04, 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
![$[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$](http://latex.artofproblemsolving.com/0/e/f/0ef66380c2667b5769ce869b56ae97caa8b4387f.png)
A
B
C
D
70^\circ![$",C,2*dir(180-35)); [/asy]$](http://latex.artofproblemsolving.com/8/8/f/88f6522245d426d5f7b374e1c74e9bb02fb738e9.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
