+107 1)
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The problems didn't come with pictures sry. i just snipped the problems :p
+9466 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
