Notes
a square and a rectangle overlapping solution

A Square and a Rectangle Overlapping

Solution by Pythagoras' Theorem and Area of a Triangle and Rectangles

With the points labelled as above, the fact that $75\%$ of the purple square is shaded means that the length of $C D$ is one quarter of the length of $A D$. Since the purple quadrilateral is a square, $E D$ is the same length as $A D$. Let $x$ be the length of $C D$, so $E D$ has length $4 x$. Applying Pythagoras' theorem to triangle $E D C$ shows that the length of $E C$ is $\sqrt{17} x$. The ratios of the sides of triangle $E D C$ is therefore $1 : 4 : \sqrt{17}$.

Triangles $C B A$ and $C D E$ are both right-angled triangles and angles $A \hat{C} B$ and $D \hat{C} E$ are equal as they are vertically opposite so triangles $C B A$ and $C D E$ are similar. As the length of $C A$ is $3 x$, the lengths of triangle $C B A$ are $\frac{3}{\sqrt{17}} x$, $\frac{12}{\sqrt{17}}x$, and $3x$. The length of $E B$ is therefore $\sqrt{17} x + \frac{3}{\sqrt{17}}x = \frac{20}{\sqrt{17}} x$.

The ratio of the lengths of $C B$ and $E B$ is therefore $3 : 20$ and so the percentage of the red rectangle that is shaded is $85\%$.