Difference between revisions of "2008 AMC 12B Problems/Problem 25"
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Let <math>ABCD</math> be a trapezoid with <math>AB||CD, AB=11, BC=5, CD=19,</math> and <math>DA=7</math>. Bisectors of <math>\angle A</math> and <math>\angle D</math> meet at <math>P</math>, and bisectors of <math>\angle B</math> and <math>\angle C</math> meet at <math>Q</math>. What is the area of hexagon <math>ABQCDP</math>? | Let <math>ABCD</math> be a trapezoid with <math>AB||CD, AB=11, BC=5, CD=19,</math> and <math>DA=7</math>. Bisectors of <math>\angle A</math> and <math>\angle D</math> meet at <math>P</math>, and bisectors of <math>\angle B</math> and <math>\angle C</math> meet at <math>Q</math>. What is the area of hexagon <math>ABQCDP</math>? | ||
− | <math>\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3\qquad \textbf{(D)}\ 35\ | + | <math>\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}</math> |
==Solution== | ==Solution== | ||
==See Also== | ==See Also== | ||
− | {{AMC12 box|year=2008|ab=B|num-b= | + | {{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}} |
Revision as of 16:09, 2 March 2008
Problem 25
Let be a trapezoid with and . Bisectors of and meet at , and bisectors of and meet at . What is the area of hexagon ?
Solution
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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 |