Difference between revisions of "1995 AHSME Problems/Problem 8"
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Line 8: | Line 8: | ||
size(120); | size(120); | ||
defaultpen(0.7); | defaultpen(0.7); | ||
− | pair A = (0,6), B = (8,0), C= (0,0), D = ( | + | pair A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6); |
draw(A--B--E--D--E--B--C--A--B--cycle); | draw(A--B--E--D--E--B--C--A--B--cycle); | ||
label("\(A\)",A,W); | label("\(A\)",A,W); | ||
Line 19: | Line 19: | ||
</asy> | </asy> | ||
− | Since <math>\triangle | + | Since <math>\triangle BDE</math> is similar to <math>\triangle ABC</math>, and thus <math>BD=10\cdot \dfrac{2}{3}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}</math>. |
==See also== | ==See also== |
Revision as of 20:34, 4 August 2008
Problem
In , and . Points and are on and , respectively, and . If , then
Solution
Since is similar to , and thus .
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |