MathDB
1993 AMC 12 #14 - Convex Pentagon

Source:

January 2, 2012
geometryrectangleAMC

Problem Statement

The convex pentagon ABCDEABCDE has A=B=120\angle A=\angle B=120^{\circ}, EA=AB=BC=2EA=AB=BC=2 and CD=DE=4CD=DE=4. What is the area of ABCDEABCDE? [asy] draw((0,0)--(1,0)--(1.5,sqrt(3)/2)--(0.5,3sqrt(3)/2)--(-0.5,sqrt(3)/2)--cycle); dot((0,0)); dot((1,0)); dot((1.5,sqrt(3)/2)); dot((0.5,3sqrt(3)/2)); dot((-0.5,sqrt(3)/2)); label("A", (0,0), SW); label("B", (1,0), SE); label("C", (1.5,sqrt(3)/2), E); label("D", (0.5,3sqrt(3)/2), N); label("E", (-.5, sqrt(3)/2), W); [/asy] <spanclass=latexbold>(A)</span> 10<spanclass=latexbold>(B)</span> 73<spanclass=latexbold>(C)</span> 15<spanclass=latexbold>(D)</span> 93<spanclass=latexbold>(E)</span> 125 <span class='latex-bold'>(A)</span>\ 10 \qquad<span class='latex-bold'>(B)</span>\ 7\sqrt{3} \qquad<span class='latex-bold'>(C)</span>\ 15 \qquad<span class='latex-bold'>(D)</span>\ 9\sqrt{3} \qquad<span class='latex-bold'>(E)</span>\ 12\sqrt{5}