A rectangular piece of paper $A B C D$ is folded so that edge $C D$ lies along edge $A D$, making a crease $D P$. It is unfolded, and then folded again so that edge $A B$ lies along edge $A D$, making a second crease $A Q$. The two creases meet at $R$, forming triangles $P Q R$ and $A D R$, as shown. If $A B=5 \mathrm{~cm}$ and $A D=8 \mathrm{~cm}$, the
![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-069.jpg?height=278&width=614&top_left_y=2049&top_left_x=1123)
area of quadrilateral $D R Q C$, in $\mathrm{cm}^{2}$, is
(A) 10.5
(B) 10
(C) 11
(D) 11.5
(E) 12
To find the area of quadrilateral $D R Q C$, we subtract the area of $\triangle P R Q$ from the area of $\triangle P D C$.

First, we calculate the area of $\triangle P D C$.

We know that $D C=A B=5 \mathrm{~cm}$ and that $\angle D C P=90^{\circ}$.

When the paper is first folded, $P C$ is parallel to $A B$ and lies across the entire width of the paper, so $P C=A B=5 \mathrm{~cm}$.

Therefore, the area of $\triangle P D C$ is $\frac{1}{2} \times 5 \times 5=\frac{25}{2}=12.5 \mathrm{~cm}^{2}$.

Next, we calculate the area of $\triangle P R Q$.

We know that $\triangle P D C$ has $P C=5 \mathrm{~cm}, \angle P C D=90^{\circ}$, and is isosceles with $P C=C D$.

Thus, $\angle D P C=45^{\circ}$.

Similarly, $\triangle A B Q$ has $A B=B Q=5 \mathrm{~cm}$ and $\angle B Q A=45^{\circ}$.

Therefore, since $B C=8 \mathrm{~cm}$ and $P B=B C-P C$, then $P B=3 \mathrm{~cm}$. Similarly, $A C=3 \mathrm{~cm}$. Since $P Q=B C-B P-Q C$, then $P Q=2 \mathrm{~cm}$.

Also, $\angle R P Q=\angle D P C=45^{\circ}$ and $\angle R Q P=\angle B Q A=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-313.jpg?height=287&width=201&top_left_y=911&top_left_x=1019)

Using four of these triangles, we can create a square of side length $2 \mathrm{~cm}$ (thus area $4 \mathrm{~cm}^{2}$ ).

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-313.jpg?height=220&width=249&top_left_y=1348&top_left_x=992)

The area of one of these triangles (for example, $\triangle P R Q$ ) is $\frac{1}{4}$ of the area of the square, or $1 \mathrm{~cm}^{2}$. So the area of quadrilateral $D R Q C$ is therefore $12.5-1=11.5 \mathrm{~cm}^{2}$.

ANSWER: (D)