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# Geometry Problems

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1) 2) 3) The problems didn't come with pictures sry. i just snipped the problems :p

May 30, 2020

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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

May 30, 2020