Answer: perimeter \((7.5+2\sin0.75)r\); area \(3.11r^2\).

Since arc \(BC\) has length \(1.5r\) and radius \(r\), the angle subtended at the centre is

\(\theta=\frac{1.5r}{r}=1.5\)

radians.

The chord \(BC\) has length

\(2r\sin\frac{\theta}{2}=2r\sin0.75.\)

The perimeter of the shaded region is made from the two straight vertical sides, the two straight top parts, the bottom side, and the arc \(BC\). Therefore

\(P=r+r+2r+(2r+2r\sin0.75)+1.5r.\)

So

\(P=(7.5+2\sin0.75)r.\)

For the area, start with the rectangle:

\(\text{rectangle area}=r(2r+2r\sin0.75) =2r^2+2r^2\sin0.75.\)

The circular segment removed above chord \(BC\) has area

\(\text{sector}-\text{triangle} =\frac12r^2(1.5)-\frac12r^2\sin1.5.\)

So the shaded area is

\(2r^2+2r^2\sin0.75-\frac12r^2(1.5)+\frac12r^2\sin1.5.\)

Hence

\(\text{area}=r^2\left(2+2\sin0.75-0.75+\frac12\sin1.5\right).\)

Evaluating the coefficient,

\(2+2\sin0.75-0.75+\frac12\sin1.5=3.11196\ldots\)

Therefore the area is

\(3.11r^2\)

to 3 significant figures.