Difference between revisions of "2000 AMC 10 Problems/Problem 7"
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− | <math>\ | + | <math>\textbf{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\textbf{(C)}\ 2+2\sqrt{2} \qquad\textbf{(D)}\ \frac{3+3\sqrt{5}}{2} \qquad\textbf{(E)}\ 2+\frac{5\sqrt{3}}{3}</math> |
+ | ==Solution== | ||
<asy> | <asy> | ||
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<math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>. | <math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>. | ||
− | Adding, we get <math>2+\frac{4\sqrt{3}}{3 | + | Adding, we get <math>\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}</math>. |
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==See Also== | ==See Also== | ||
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{{AMC10 box|year=2000|num-b=6|num-a=8}} | {{AMC10 box|year=2000|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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+ | [[Category:Introductory Geometry Problems]] |
Revision as of 09:47, 8 November 2021
Problem
In rectangle , , is on , and and trisect . What is the perimeter of ?
Solution
.
Since is trisected, .
Thus,
.
Adding, we get .
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.