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!!!!!!!!!

Noori Aug 25, 2020

#1

#4**+4 **

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.

Melody Aug 31, 2020