Problem:
The convex pentagon ABCDE has ∠A=∠B=120∘,EA=AB=BC=2 and CD=DE=4. What is the area of ABCDE?
Answer Choices:
A. 10
B. 73​
C. 15
D. 93​
E. 125​
Solution:
Draw CE. Since EA=BC and ∠A=∠B, it follows that ABCE is an isosceles trapezoid. Let F be the foot of the perpendicular from A to CˉEˉ, and G be the foot of the perpendicular from B to CE. Then EF=CG. Since ∠GBC=30∘, we have CG=21​(BC)=1 and BG=23​​(BC)=3​. Now CE=CG+GF+FE=1+2+1=4, so CDE is an equilateral triangle. Thus †,
ABCE] and [CDE]​=21​(BG)(AB+CE)=21​3​(2+4)=33​,=43​​(CE)2=43​​(16)=43​​
Therefore, [ABCDE]=[ABCE]+[CDE]=73​.
OR
Draw HI where H is the midpoint of ED and I is the midpoint of CD. Then ABCIHE is a regular hexagon and △HDI 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(2243​​)=73​.
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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