Question

Please help me i offered all my points and this is really important!!! The question is attached.

Answer 1

Answer:

  60 +25√3

Step-by-step explanation:

In the figure of the isosceles trapezoid below, the angles at C and D are supplementary to the given angle, so are 60°. That makes triangle BDE a 30°-60°-90° right triangle, which has side length ratios ...

  DE : BE : BD = 1 : √3 : 2 = 5 : 5√3 : 10

Triangle BDE can be relocated to the other end of the figure to become triangle CAD'. Then the area of concern is that of the rectangle with height 5√3 and length 5+4√3. The area is then ...

  Area = lh = (5√3)(5 +4√3) = 5·5√3 +5·4·3

  Area = 60 +25√3 . . . square units

_____

In the figure, 6.93 = 4√3, and 8.66 = 5√3, 16.93 = 10+4√3.

Answer 2

Answer:

25$\sqrt{3}$ +60

Step-by-step explanation: The first thing you need to do is realize that, this figure is a isosceles trapezoid due to the markings on each side.

So now we know both sides are 10.

We also know the the top two angles are congruent to each other and so are the bottom two angles due to the trapezoid being isosceles.

So the top two angles are 120 degrees and bottom two angles are 60 degrees.

It seems like we can't find the sides, let's try drawing two lines from each top angle all the way down to form two right triangles.

Wow, these two triangles are special right triangles in the form of

30 - 60 - 90 degrees.

shorter side = n

longer side = n$\sqrt{3}$

hypotenuse = 2n

So, 2n = 10

n = 5 for the short side

The bottom base is 4$\sqrt{3}$ + 5 + 5 = 10 + 4$\sqrt{3}$

The longer side is  5$\sqrt{3}$.

The area of trapezoid = (base1 + base2)/2 * height

= (4$\sqrt{3}$ + 10 + 4$\sqrt{3}$)/2 * 5$\sqrt{3}$ = (10 + 8$\sqrt{3}$)/2 * 5$\sqrt{3}$ = (5+4$\sqrt{3}$)*5$\sqrt{3}$ = 25$\sqrt{3}$ +60

So, 25$\sqrt{3}$ + 60 is our answer.