Answer: \(\angle AOB=1.45\); area \(0.973r^2\); \(r=2.91\).

(a) Arc length is given by

\(s=r\theta.\)

Since the arc \(AB\) has length \(1.45r\),

\(1.45r=r\theta.\)

Hence

\(\angle AOB=1.45\)

radians.

(b) The area of sector \(AOB\) is

\(\frac12r^2(1.45).\)

The angle \(AOC\) is

\(\pi-1.45.\)

The area of triangle \(AOC\) is

\(\frac12(r)(0.5r)\sin(\pi-1.45).\)

So the area of \(ABC\) is

\(\frac12r^2(1.45)+\frac12(r)(0.5r)\sin(\pi-1.45).\)

Evaluating the coefficient gives

\(0.973r^2\)

to 3 significant figures.

(c) Use the cosine rule in triangle \(AOC\):

\(AC^2=r^2+(0.5r)^2-2(r)(0.5r)\cos(\pi-1.45).\)

This gives

\(AC=1.17r\)

to 3 significant figures.

The perimeter is made from \(AC\), \(CB\) and arc \(AB\). Since

\(CB=CO+OB=0.5r+r=1.5r,\)

the perimeter is

\(1.17r+1.5r+1.45r=4.12r.\)

Given that this is \(12\),

\(4.12r=12.\)

Hence

\(r=2.91.\)