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The problems didn't come with pictures sry. i just snipped the problems :p

NoWorkNeeded May 30, 2020

#1**+1 **

1)

We can find the area of △BXA this way:

area of △BXA = area of △ABC - area of △BXC

△ABC is a " 30-60-90 " triangle, and in all " 30-60-90 " triangles,

the side opposite the 30° angle is half the length of the hypotenuse.

BC = 12 / 2 = 6

Also, the side opposite the 60° is √3 times the side opposite the 30° angle.

AC = BC * √3

AC = 6√3

Now we can find the area of △ABC:

area of △ABC = (1/2) * AC * BC

area of △ABC = (1/2) * 6√3 * 6

area of △ABC = 18√3

△BXC is also a " 30-60-90 " triangle. So we can say:

BC = XC * √3

XC = BC / √3

XC = 6 / √3

XC = 2√3

Now we can find the area of △BXC:

area of △BXC = (1/2) * XC * BC

area of △BXC = (1/2) * 2√3 * 6

area of △BXC = 6√3

area of △BXA = area of △ABC - area of △BXC

area of △BXA = 18√3 - 6√3

area of △BXA = 12√3

hectictar May 30, 2020