9. A square has side-length $10 \mathrm{~m}$. It is divided into parts by three straight line segments, as shown. The areas of the two shaded triangles are $P \mathrm{~m}^{2}$ and $Q \mathrm{~m}^{2}$.
What is the value of $P-Q$ ?
A 0
B 1
C 2
D 5
E 10
Solution
Let $R \mathrm{~m}^{2}$ be the area of the triangle shown on the diagram. The triangle with area $(P+R) \mathrm{m}^{2}$ has base $10 \mathrm{~m}$ and height $10 \mathrm{~m}$. Similarly, the triangle with area $(Q+R) \mathrm{m}^{2}$ also has base $10 \mathrm{~m}$ and height $10 \mathrm{~m}$. Therefore the areas of these two triangles are equal and hence $P+R=Q+R$. Therefore $P=Q$ and hence the value of $P-Q$ is 0 .