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Let G be the center of the equilateral triangle XYZ. A dilation centered at G with scale factor -3/4 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.

 Apr 26, 2023
 #1
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Let the side length of equilateral triangle XYZ be s. Then, the side length of equilateral triangle X'Y'Z' is 3/4s.

The area of triangle XYZ is s^2*sqrt(3)​/4.

The area of triangle X'Y'Z' is (3/4)^2*s^2*sqrt(3)​/4=9/16s^2*sqrt(3)​.

The area of the region that is contained in both triangles XYZ and X'Y'Z' is s^2*sqrt(3)​/4−9/16s^2*sqrt(3)​=5/16s^2*sqrt(3)​.

Therefore, A/area of XYZ=5/16*s^2*sqrt(3)​/s*2*sqrt(3)/4 = 5/16​.

 Apr 26, 2023
 #2
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Not correct :(

Guest Apr 26, 2023

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