Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{4}{3}$ 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 $\frac{A}{[XYZ]}.$

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The area of  the required triangle within the ratio form will be 16/9

A closed shape with both inner and exterior angles, a triangular is created by three lines.  the fundamental geometric shapes. a vertex-filled triangle. There is only one distinct triangle if the product of two fades on the sides is bigger than the longest side. A closed shape with both inner and exterior angles, a triangular is created by three lines.

A dilation centered with the scale factor is given at -4/3

The area that is for the triangle XYZ will be

A = [tex]r^{2}[/tex] /  ΔABC

[tex]r^{2}[/tex] = A / ΔABC

Adding the valure of r in the above equation that is being presented

(-4/3)² = A / ΔABC

16/9 = A / ΔABC

the center of equilateral triangle with the scale factor will be 16/9. THis will b edone with the scalor factor.

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