B4. The diagram alongside shows two quartercircles and two triangles, $A B C$ and $C D E$. One quarter-circle has radius $A B$, where $A B=$ $3 \mathrm{~cm}$. The other quarter-circle has radius $D E$, where $D E=4 \mathrm{~cm}$.
The area enclosed by the line $A E$ and the arcs of the two quarter circles is shaded.
What is the total shaded area, in $\mathrm{cm}^{2}$ ?
Add a point $X$ to the diagram, directly below point $A$ such that the lines $A X$ and $X D$ are perpendicular. This whole new shape consists of two quarter-circles and one rectangle $(A B D X)$. The area, in $\mathrm{cm}^{2}$, of this new shape is
$$
\frac{1}{4} \pi \times 3^{2}+\frac{1}{4} \pi \times 4^{2}+3 \times 1=\frac{25}{4} \pi+3 .
$$

The shaded area is the area of the new shape minus the area of triangle $A X E$. Hence, the shaded area, in $\mathrm{cm}^{2}$, is
$$
\frac{25}{4} \pi+3-\frac{7}{2}=\frac{25}{4} \pi-\frac{1}{2} .
$$