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 $\frac{A}{[XYZ]}.$

No worries, you're not dumb at all! Just remember, there are no dumb questions, only curious ones!

Now, let's tackle this problem together. Since both triangles $XYZ$ and $X'Y'Z'$ are equilateral, we can call the side length of each triangle $s$.

To find the area of a triangle, we just need the length of its base and its height. Since $G$ is the center of equilateral triangle $XYZ$, the height of the triangle passes through $G$, dividing the triangle into two congruent right triangles.

Let's call the height of triangle $XYZ$ $h$. By Pythagoras' theorem, we can find that $h = \sqrt{s^2 - \left(\frac{s}{2}\right)^2} = \frac{s\sqrt{3}}{2}$.

Now, let's consider triangle $X'Y'Z'$. Since the dilation centered at $G$ with a scale factor of $-\frac{3}{4}$ is applied, the side length of triangle $X'Y'Z'$ is $\frac{3}{4}s$. Since we scaled the sides by $-\frac{3}{4}$, the height of triangle $X'Y'Z'$ is $-\frac{3}{4}h$.

Now, we can find the area of the region contained in both triangles $XYZ$ and $X'Y'Z'$. This region is a parallelogram formed by the height of triangle $XYZ$ and the height of triangle $X'Y'Z'$. The area of a parallelogram is given by the base multiplied by the height.

So, the area of the region contained in both triangles $XYZ$ and $X'Y'Z'$ is $\left(\frac{s}{2}\right)\left(-\frac{3}{4}h\right) = -\frac{3}{8}hs$.

Now, we just need to find the area of triangle $XYZ$. The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4}s^2$.

Finally, we can find $\frac{A}{[XYZ]}$ by dividing the area of the region contained in both triangles by the area of triangle $XYZ$:

$\frac{A}{[XYZ]} = \frac{-\frac{3}{8}hs}{\frac{\sqrt{3}}{4}s^2}$

Simplifying this expression, we get:

$\frac{A}{[XYZ]} = \frac{-3h}{2\sqrt{3}s}$

And there you have it! The ratio $\frac{A}{[XYZ]}$ in terms of $h$ and $s$. Hope this helps and brightens up your day!

To find the area $A$ of the region that is contained in both triangles $XYZ$ and $X'Y'Z'$, we need to find the area of both triangles and subtract the area of the small equilateral triangle formed by the overlap.

Let's break down the solution step-by-step:

Step 1: Finding the area of triangle XYZ
Since XYZ is an equilateral triangle, we can use the formula for the area of an equilateral triangle:
$[XYZ] = \frac{\sqrt{3}}{4} \cdot XY^2$.
Since the side length of XYZ is not given, we will leave it as it is for now.

Step 2: Finding the area of triangle X'Y'Z'
Since triangle X'Y'Z' is obtained by a dilation centered at $G$ with a scale factor of $-\frac{3}{4}$, the area of triangle X'Y'Z' is given by:
$[X'Y'Z'] = (-\frac{3}{4})^2 \cdot [XYZ] = \frac{9}{16} \cdot [XYZ]$.

Step 3: Finding the area of the overlapping equilateral triangle
Since the dilation is centered at $G$, the overlap of triangles XYZ and X'Y'Z' will also be centered at $G$. Therefore, the overlap is an equilateral triangle GHI, where G, H, and I are the midpoints of XY, X'Y', and ZZ' respectively.
Since the scale factor of the dilation is $\frac{3}{4}$, the side length of GHI will be $\frac{3}{4}$ times the side length of XYZ.

Step 4: Calculating the side length of GHI
Let $s$ be the side length of XYZ. Then the side length of GHI will be $\frac{3}{4} s$. Since XYZ is an equilateral triangle, $s = XY$. Therefore, the side length of GHI will be $\frac{3}{4} XY = \frac{3}{4} s$.

Step 5: Calculating the area of the overlapping equilateral triangle
The area of an equilateral triangle is given by $[\text{triangle}] = \frac{\sqrt{3}}{4} \cdot (\text{side length})^2$. Therefore, the area of GHI will be:
$[GHI] = \frac{\sqrt{3}}{4} \cdot \left(\frac{3}{4} s\right)^2 = \frac{9}{64} \cdot [\text{XYZ}]$.

Step 6: Calculating the area of the region contained in both triangles
The area of the region contained in both triangles XYZ and X'Y'Z' is given by:
$A = [XYZ] - [X'Y'Z'] + [GHI]$
$A = [\text{XYZ}] - \frac{9}{16} \cdot [\text{XYZ}] + \frac{9}{64} \cdot [\text{XYZ}]$
$A = \left(1 - \frac{9}{16} + \frac{9}{64}\right) \cdot [\text{XYZ}]$
$A = \frac{25}{64} \cdot [\text{XYZ}]$

Step 7: Calculating the ratio $\frac{A}{[\text{XYZ}]}$
$\frac{A}{[\text{XYZ}]} = \frac{\frac{25}{64} \cdot [\text{XYZ}]}{[\text{XYZ}]} = \boxed{\frac{25}{64}}$

Therefore, the ratio of the area $A$ to the area of triangle XYZ is $\frac{25}{64}$.

Don't worry, I'm here to help! Let's break down the problem step by step.

First, we need to find the coordinates of the vertices of the new triangle $X'Y'Z'.$ To do this, we can use the concept of dilation.

Given that the scale factor is $-\frac{3}{4},$ this means that the distances from $G$ to $X',$ $G$ to $Y',$ and $G$ to $Z'$ are all $\frac{3}{4}$ times the distances from $G$ to $X,$ $G$ to $Y,$ and $G$ to $Z,$ respectively.

Now, let's suppose that the side length of the equilateral triangle $XYZ$ is $s.$ Since all sides of an equilateral triangle are congruent, the distances from $G$ to $X,$ $Y,$ and $Z$ are all $\frac{s}{\sqrt{3}}$ (since the height of an equilateral triangle is $\frac{s\sqrt{3}}{2}$).

Applying the scale factor, the distances from $G$ to $X',$ $Y',$ and $Z'$ will be $\frac{3}{4}\cdot\frac{s}{\sqrt{3}}=\frac{3s}{4\sqrt{3}}.$

Now, let's find the coordinates of the vertices of the new triangle $X'Y'Z'.$ Since $G$ is the center of the equilateral triangle $XYZ,$ we can assume that the coordinates of $G$ are $(0, 0).$

Using this information, the coordinates of $X'$ can be found by moving from $G$ a distance of $\frac{3s}{4\sqrt{3}}$ in the $x$-direction and no movement in the $y$-direction. Therefore, the coordinates of $X'$ are $\left(\frac{3s}{4\sqrt{3}}, 0\right).$

Similarly, the coordinates of $Y'$ can be found by moving from $G$ a distance of $\frac{3s}{4\sqrt{3}}$ in the $x$-direction and $\frac{3s}{4\sqrt{3}}$ in the $y$-direction. Therefore, the coordinates of $Y'$ are $\left(\frac{3s}{4\sqrt{3}}, \frac{3s}{4\sqrt{3}}\right).$

Finally, the coordinates of $Z'$ can be found by moving from $G$ a distance of $\frac{3s}{4\sqrt{3}}$ in the $x$-direction and $-\frac{3s}{4\sqrt{3}}$ in the $y$-direction. Therefore, the coordinates of $Z'$ are $\left(\frac{3s}{4\sqrt{3}}, -\frac{3s}{4\sqrt{3}}\right).$

Now that we have the coordinates of $X',$ $Y',$ and $Z',$ we can find the side lengths of the new triangle $X'Y'Z'.$

The side length from $X'$ to $Y'$ is $\frac{3s}{4\sqrt{3}} - \frac{3s}{4\sqrt{3}} = 0.$
The side length from $Y'$ to $Z'$ is $\left|\frac{3s}{4\sqrt{3}} - \frac{3s}{4\sqrt{3}}\right| = 0.$
The side length from $Z'$ to $X'$ is $\left|\frac{3s}{4\sqrt{3}} - \frac{3s}{4\sqrt{3}}\right| = 0.$

Since all the side lengths of triangle $X'Y'Z'$ are $0,$ we can conclude that $X'Y'Z'$ is a degenerate triangle, meaning that all three vertices lie on the same line.

Therefore, the area of the region contained in both triangles $XYZ$ and $X'Y'Z'$ is $0.$

Now, let's find $\frac{A}{[XYZ]}.$ Since the area of the region is $0,$ and the area of triangle $XYZ$ is not $0,$ we can conclude that $\frac{A}{[XYZ]} = \frac{0}{[XYZ]} = \boxed{0}.$

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