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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!!!!!!!!!

 Aug 25, 2020
 #1
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A/[XYZ] = 5/9.

 Aug 26, 2020
 #2
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Could I have some more help? I do not know the process, and I do not understand your process.

Noori  Aug 26, 2020
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As far as I figured, It looks like you squared to get. That is indeed the area of X'Y'Z', but the problem asks for the region that is contained in both. I think this is true but, not sure what to do?

Noori  Aug 29, 2020
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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.

 

 Aug 31, 2020

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