+219 Let G be the center of the equilateral triangle XYZ. A dilation centered at G with scale factor -2/3 is applied to triangle XYZ, to obtain triangle X'Y'Z'. Let A be the area of the region that is contained in both triangles XYZ and X'Y'Z'. Find A/the area of XYZ.
Thank you!!!!!!!!!
+118609 I think I have thought this through but I have not completed all the maths.
I let XYZ be circumscribed by a circle of radius 3 (I find it easier to use actual numbers)
area Triangle XGY = 0.5*3*3sin120 =(9sqrt3)/4
area Triangle X'GY' = 2/3 of that = (3sqrt3)/2
Now find the area of one little triangle.
the equation of XZ is y=sqrt3x+3
point A has a y value of 1 so the x value is (-2sqrt3)/3
Y' is the intersection of x^2+y^2=4 and y=1 so I think it is (-sqrt3,1)
Distance AY'= sqrt3/3
So from that you can work out the area of each of the little triangles.
Then you can work out the rest.
(maybe there is a better way, but that will work)
You will need to check all my logic and arithmetic.
