Solution
$11 \mathrm{~cm}^{2}$

Let the perpendicular from $U$ to $Q R$ meet $Q R$ at $V$ and let the point where $U T$ and $S R$ intersect be $W$, as shown.
Let the lengths of $U V$ and $Q V$ be $h$ and $x$ respectively.
Square $P Q R S$ has area $25 \mathrm{~cm}^{2}$, so its side-length is $5 \mathrm{~cm}$.
Hence rhombus $Q R T U$ has side-length $5 \mathrm{~cm}$ and area $20 \mathrm{~cm}^{2}$.
Therefore $h$ is $\frac{20}{5} \mathrm{~cm}=4 \mathrm{~cm}$.
By Pythagoras' Theorem, $Q U^{2}=Q V^{2}+V U^{2}$. Therefore $x$ is $\sqrt{5^{2}-4^{2}} \mathrm{~cm}=3 \mathrm{~cm}$. So $U W=V R=(5-3) \mathrm{cm}=2 \mathrm{~cm}$.
Hence the area of trapezium $U Q R W$ is $\frac{1}{2} \times(2+5) \times 4 \mathrm{~cm}^{2}=14 \mathrm{~cm}^{2}$.
Therefore the area of the shaded region is $(25-14) \mathrm{cm}^{2}=11 \mathrm{~cm}^{2}$.
A9. The diagram shows the square $P Q R S$, which has area $25 \mathrm{~cm}^{2}$, and the rhombus $Q R T U$, which has area $20 \mathrm{~cm}^{2}$. What is the area of the shaded region?