Difference between revisions of "1997 AHSME Problems/Problem 15"
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<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> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | <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 F = midpoint(B--C); | ||
| + | pair G = intersectionpoint(E--C,B--D); | ||
| + | dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); | ||
| + | label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE); | ||
| + | draw(A--B--C--cycle); | ||
| + | draw(B--D); | ||
| + | draw(E--C); | ||
| + | draw(A--F); | ||
| + | draw(rightanglemark(B,G,E,3));</asy> | ||
| + | |||
| + | One median divides a triangle into <math>2</math> equal areas, so all three medians will divide a triangle into <math>6</math> equal areas. | ||
| + | |||
| + | The median <math>CE</math> is divided into a <math>2:1</math> ratio at centroid <math>G</math>, so <math>GE = \frac{1}{3}\cdot CE = \frac{1}{3}\cdot 12 = 4</math> | ||
| + | |||
| + | Similarly, <math>BG = \frac{2}{3}\cdot 8 = \frac{16}{3}</math> | ||
| + | |||
| + | The area of the right triangle <math>\triangle BEG</math> is <math>\frac{1}{2}\cdot\frac{16}{3}\cdot 4</math> | ||
| + | |||
| + | The area of the whole figure is <math>6\cdot \frac{1}{2}\cdot\frac{16}{3}\cdot 4 = 64</math>, and the correct answer is <math>\boxed{D}</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=14|num-a=16}} | {{AHSME box|year=1997|num-b=14|num-a=16}} | ||
Revision as of 11:43, 9 August 2011
Problem
Medians 
and 
of triangle 
are perpendicular, 
, and 
. The area of triangle 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]](http://latex.artofproblemsolving.com/a/0/f/a0fa86e3c4a1430fc804dcaddd4a346d89ca0c0f.png)
Solution
![[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 F = midpoint(B--C); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(A--F); draw(rightanglemark(B,G,E,3));[/asy]](http://latex.artofproblemsolving.com/a/c/c/acccc291b5626d64877734020e9e2e0750df4263.png)
One median divides a triangle into 
equal areas, so all three medians will divide a triangle into 
equal areas.
The median 
is divided into a 
ratio at centroid 
, so 
Similarly, 
The area of the right triangle 
is 
The area of the whole figure is 
, and the correct answer is 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
