In isosceles trapezoid EFGH, FG parallel EH, FG=10, GH=12, and ∠E=60. Find the area of EFGH.

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

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

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step-by-step explanation:

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