Difference between revisions of "1958 AHSME Problems/Problem 31"

Revision as of 01:06, 22 December 2015

The altitude drawn to the base of an isosceles triangle is $8$ , and the perimeter $32$ . The area of the triangle is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 32\qquad \textbf{(E)}\ 24$

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

@@ Line 7: / Line 7: @@
 \textbf{(D)}\ 32\qquad
 \textbf{(E)}\ 24</math>
-== Solution ==
-<math>[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>
-<math>\fbox{}</math>
 == See Also ==