What is the ratio the areas of △$PST$ and quadrilateral $STRQ$? I tried to answer this question:
What is the ratio of the areas of △$PST$ and quadrilateral $STRQ$ if $∠1\cong ∠2$

Since $∠1\cong ∠2$, △$PST\sim$ △$PQR$. And since the two triangles are similar, the ratio of their areas is $\frac{1}{25}$. The area of the whole triangle is 1, so the area of quadrilateral $STRQ$ is $1-\frac{1}{25}=\frac{24}{25}$. The ratio between the two areas now is $\frac{\frac{1}{25}}{\frac{24}{25}}=\frac{1}{24}$.
Is my answer correct?
 A: That's exactly right.
The areas of similar polygons -- similar two-dimensional shapes, really -- scale as the square of a characteristic length, like a side, or a perimeter.
(Extending this, if you had similar solids in three dimensions, the volume would scale as the cube of a characteristic length.)
A: $\triangle PST$ and $\triangle PQR$ are similar due to AA (two angles congruent). I will assume you can prove that.
Given that the area of a triangle is $A=\frac{1}{2}ab\sin{C}$, we can calculate the area for both triangles, then find the ratio. Let the side $\overline {PT}=y$. Then we need to calculate $\overline {PR}$.
By similar triangles,
\begin{align}
\frac{\overline{PS}}{\overline{PQ}}&=\frac{\overline{PT}}{\overline{PR}}\\
\frac{3}{15}&=\frac{y}{\overline{PR}} \\
\implies \overline{PR}&=5y \\
\end{align}
Then apply the area formula:
\begin{align}
A_{\triangle {PST}} &=\frac{1}{2}\cdot3\cdot y \cdot \sin{C} \\
&=\frac{3y}{2}\cdot \sin{C} \\
\end{align}
\begin{align}
A_{\triangle {PQR}} &=\frac{1}{2}\cdot15\cdot 5y \cdot \sin{C} \\
&=\frac{75y}{2}\cdot \sin{C} \\
\end{align}
The area of the quadrilateral is:
\begin{align}
A_{STRQ}&=A_{\triangle {PQR}}-A_{\triangle {PST}}\\
&=\frac{72y}{2}\sin{C}
\end{align}
Therefore the ratio is $\frac{A_{\triangle {PQR}}}{A_{STRQ}}=\frac{\frac{3y}{2}\sin{C}}{\frac{72y}{2}\sin{C}}=\frac{1}{24}$
You are correct!
