In the diagram, $A B C D$ is a square with area $25 \mathrm{~cm}^{2}$. If $P Q C D$ is a rhombus with area $20 \mathrm{~cm}^{2}$, the area of the shaded region, in $\mathrm{cm}^{2}$, is
(A) 12
(B) 10
(C) 11
(D) 12.5
(E) 9

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-085.jpg?height=266&width=328&top_left_y=1732&top_left_x=1386)
Since $A B C D$ is a square and has an area of $25 \mathrm{~cm}^{2}$, then the square has a side length of $5 \mathrm{~cm}$. Since $P Q C D$ is a rhombus, then it is a parallelogram, so its area is equal to the product of its base and its height.

Join point $P$ to $X$ on $A D$ so that $P X$ makes a right angle with $A D$, and to $Y$ on $D C$ so that $P Y$ makes a right angle with $D C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-378.jpg?height=358&width=485&top_left_y=1485&top_left_x=1297)

Then the area of the shaded region is the area of rectangle $A B Z X$ plus the area of triangle $P X D$. Since the area of $P Q C D$ is $20 \mathrm{~cm}^{2}$ and its base has length $5 \mathrm{~cm}$, then its height, $P Y$, must have length $4 \mathrm{~cm}$.

Therefore, we can now label $D X=4, D P=5$ (since $P Q C D$ is a rhombus), $A X=1$, and $A B=5$. So $A B Z X$ is a 1 by 5 rectangle, and so has area $5 \mathrm{~cm}^{2}$.

Triangle $P X D$ is right-angled at $D$, and has $D P=5$ and $D X=4$, so by Pythagoras' Theorem, $P X=3$. Therefore, the area of triangle $P X D$ is $\frac{1}{2}(3)(4)=6 \mathrm{~cm}^{2}$. So, in total, the area of the shaded region is $11 \mathrm{~cm}^{2}$.