Answer: Perimeter \(=10+2\sqrt3+\frac{4\sqrt3\pi}{3}\). Area \(=4\pi+12+3\sqrt3\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Since \(O\) is the midpoint of \(BD\), \(OB=OD=3\). Also \(AB=DE=\sqrt3\).

So

\(OA=\sqrt{3^2+(\sqrt3)^2}=\sqrt{12}=2\sqrt3.\)

Similarly, \(OE=2\sqrt3\), so the radius of the arc is \(2\sqrt3\).

Using the geometry of the two right triangles, the angle \(AOE\) is \(\frac{2\pi}{3}\).

Therefore the arc length \(AFE\) is

\((2\sqrt3)\left(\frac{2\pi}{3}\right)=\frac{4\sqrt3\pi}{3}.\)

The straight parts of the perimeter are

\(AB+BC+CD+DE=\sqrt3+5+5+\sqrt3=10+2\sqrt3.\)

Hence the perimeter is

\(10+2\sqrt3+\frac{4\sqrt3\pi}{3}.\)

For the area, add the sector \(AOE\), triangle \(BCD\), and the two right triangles \(ABO\) and \(ODE\).

The sector area is

\(\frac12(2\sqrt3)^2\left(\frac{2\pi}{3}\right)=4\pi.\)

The area of triangle \(BCD\) is \(\frac12\cdot6\cdot4=12\), since \(BC=CD=5\) and half the base is \(3\).

The two small right triangles have total area

\(2\left(\frac12\cdot3\cdot\sqrt3\right)=3\sqrt3.\)

So the total area is \(4\pi+12+3\sqrt3\).

This gives the required answer: Perimeter \(=10+2\sqrt3+\frac{4\sqrt3\pi}{3}\). Area \(=4\pi+12+3\sqrt3\).