Problem:
The convex pentagon ABCDE has β A=β B=120β,EA=AB=BC=2 and CD=DE=4. What is the area of ABCDE?
Answer Choices:
A. 10
B. 73β
C. 15
D. 93β
E. 125β
Solution:
Draw CE. Since EA=BC and β A=β B, it follows that ABCE is an isosceles trapezoid. Let F be the foot of the perpendicular from A to CΛEΛ, and G be the foot of the perpendicular from B to CE. Then EF=CG. Since β GBC=30β, we have CG=21β(BC)=1 and BG=23ββ(BC)=3β. Now CE=CG+GF+FE=1+2+1=4, so CDE is an equilateral triangle. Thus β ,
ABCE] and [CDE]β=21β(BG)(AB+CE)=21β3β(2+4)=33β,=43ββ(CE)2=43ββ(16)=43ββ
Therefore, [ABCDE]=[ABCE]+[CDE]=73β.
OR
Draw HI where H is the midpoint of ED and I is the midpoint of CD. Then ABCIHE is a regular hexagon and β³HDI is congruent to any of the six equilateral triangles of side 2 that make up ABCIHE. Thus, the area of ABCDE is the sum of the areas of 7 equilateral triangles of side 2, so it is 7(2243ββ)=73β.
Note. If DE,DC and AB are extended to meet at J and K as in the figure, then we can compute [ABCDE] as 7/9 of the area of equilateral triangle DJK.
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