Difference between revisions of "1997 AHSME Problems/Problem 15"
Talkinaway (talk | contribs) (Created page with "==Problem== Medians <math>BD</math> and <math>AE</math> of triangle <math>ABC</math> are perpendicular, <math>BD=8</math>, and <math>CE=12</math>. The area of triangle <math>AB...") |
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Medians <math>BD</math> and <math>AE</math> of triangle <math>ABC</math> are perpendicular, <math>BD=8</math>, and <math>CE=12</math>. The area of triangle <math>ABC</math> is | Medians <math>BD</math> and <math>AE</math> of triangle <math>ABC</math> are perpendicular, <math>BD=8</math>, and <math>CE=12</math>. The area of triangle <math>ABC</math> is | ||
− | + | <asy> | |
defaultpen(linewidth(.8pt)); | defaultpen(linewidth(.8pt)); | ||
dotfactor=4; | dotfactor=4; | ||
Line 13: | Line 13: | ||
pair G = intersectionpoint(E--C,B--D); | pair G = intersectionpoint(E--C,B--D); | ||
dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); | dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); | ||
− | label(" | + | 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(A--B--C--cycle); | ||
draw(B--D); | draw(B--D); | ||
draw(E--C); | draw(E--C); | ||
− | draw(rightanglemark(C,G,D,3)); | + | draw(rightanglemark(C,G,D,3));</asy> |
<math> \textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96 </math> | <math> \textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96 </math> | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1997|num-b=10|num-a=12}} |
Revision as of 09:12, 9 August 2011
Problem
Medians and of triangle are perpendicular, , and . The area of triangle is
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |