Difference between revisions of "2023 AMC 12B Problems/Problem 11"
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+ | ==Problem== | ||
+ | What is the maximum area of an isosceles trapezoid that has legs of length <math>1</math> and one base twice as long as the other? | ||
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+ | <math>\textbf{(A) }\frac 54 \qquad \textbf{(B) } \frac 87 \qquad \textbf{(C)} \frac{5\sqrt2}4 \qquad \textbf{(D) } \frac 32 \qquad \textbf{(E) } \frac{3\sqrt3}4</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 19:38, 15 November 2023
Problem
What is the maximum area of an isosceles trapezoid that has legs of length and one base twice as long as the other?
Solution
Denote by the length of the shorter base. Thus, the height of the trapezoid is
Thus, the area of the trapezoid is
where the inequality follows from the AM-GM inequality and it is binding if and only if .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.