6. In the quadrilateral $P Q R S$, the length of $P Q$ is $11 \mathrm{~cm}$, the length of $Q R$ is $7 \mathrm{~cm}$, the length of $R S$ is $9 \mathrm{~cm}$ and the length of $S P$ is $3 \mathrm{~cm}$. Both $\angle Q R S$ and $\angle S P Q$ are $90^{\circ}$. What is the area of the quadrilateral $P Q R S$ ?
A $30 \mathrm{~cm}^{2}$
B $48 \mathrm{~cm}^{2}$
C $50 \mathrm{~cm}^{2}$
D $52 \mathrm{~cm}^{2}$
E $60 \mathrm{~cm}^{2}$
6. B The information in the question tells us that both triangle $S P Q$ and triangle $Q R S$ are right-angled. The area of the quadrilateral $P Q R S$ is equal to the sum of the areas of triangle $S P Q$ and triangle $Q R S$.
Therefore the area of $P Q R S$ is $\frac{1}{2}(11 \times 3) \mathrm{cm}^{2}+\frac{1}{2}(7 \times 9) \mathrm{cm}^{2}=\frac{1}{2}(33+63) \mathrm{cm}^{2}=\frac{1}{2}(96) \mathrm{cm}^{2}=48 \mathrm{~cm}^{2}$.