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avatar+107 

1)

2)

 

3)

 

 

 

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

 May 30, 2020
 #1
avatar+9015 
+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

 May 30, 2020

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