Question
sorry for being dumb pls help:
Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{3}{4}$ 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 $\frac{A}{[XYZ]}.$
Let $G$ be the center of equilateral triangle $XYZ.$ A dilation centered at $G$ with scale factor $-\frac{3}{4}$ 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 $\frac{A}{[XYZ]}.$
Answers
oobleck
area grows as the square of the scale factor.
So, the area of X'Y'Z' is (3/4)^2 = 9/16 that of XYZ.
A/XYZ = 25/16
So, the area of X'Y'Z' is (3/4)^2 = 9/16 that of XYZ.
A/XYZ = 25/16