Difference between revisions of "1997 AHSME Problems/Problem 15"

(See also)
Line 22: Line 22:
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1997|num-b=10|num-a=12}}
+
{{AHSME box|year=1997|num-b=14|num-a=16}}

Revision as of 10:12, 9 August 2011

Problem

Medians $BD$ and $AE$ of triangle $ABC$ are perpendicular, $BD=8$, and $CE=12$. The area of triangle $ABC$ is

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]

$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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
All AHSME Problems and Solutions