In isosceles trapezoid EFGH, FG parallel EH, FG=10, GH=12, and ∠E=60. Find the area of EFGH.
1 answer:
Answer:
22√3 units squared
Step-by-step explanation:
1) Draw the trapezoid
2) Notice you have to find the height to find the area of the trapezoid
3) Notice it is a 30 60 90 triangle when you draw the line to find the height
4) Solve for the 30 60 90 triangle and you get the height to be 2√3
5) Solve for the area of the trapezoid using this formula A= 1/2 h ( b1 + b2)
6) Fill in the numbers for the equation A= 1/2∙ 2√3 ( 10 + 12)
7) Keep aside the 1/2 for now
8) Simplify equation 2√3 + 22 or 2√3 + (10 + 12)
9) Simplify further 44√3
10) Now bring back the 1/2 and put it in the equation 1/2∙ 44√3
11) Simplify to get 22√3
12) 22√3 is the area of the trapezoid
13) 22√3 units squared is the final answer
I will put a my drawing of the trapezoid for reference and this is what I used to solve the question.
If you want to know how I solved the 30 60 90 triangle please let me know.
Sorry I was one week late I just saw it today.
:D
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