Answer: perimeter \(=19.2\) cm; shaded area \(=11.4\text{ cm}^2\).

(a) The area of sector \(OCD\) is

\(\frac12r^2\theta.\)

Here \(r=10\) and the area is \(35\), so

\(\frac12(10)^2\theta=35.\)

Thus

\(50\theta=35,\)

and hence

\(\theta=0.7.\)

(b) The arc length \(CD\) is

\(10(0.7)=7.\)

Also

\(AC=12-10=2,\qquad DB=12-10=2.\)

To find \(AB\), split the isosceles triangle into two right-angled triangles. Then

\(\sin(0.35)=\frac{AB/2}{12}.\)

So

\(AB=24\sin(0.35)\approx8.23.\)

Therefore the perimeter of the shaded region is

\(7+2+2+8.23=19.23\ldots.\)

So the perimeter is

\(19.2\text{ cm}\)

to 3 significant figures.

(c) The shaded area is the area of triangle \(OAB\) minus the area of sector \(OCD\).

The area of triangle \(OAB\) is

\(\frac12(12)(12)\sin0.7 =72\sin0.7 \approx46.4.\)

Therefore the shaded area is

\(46.4-35=11.4.\)

So the shaded area is

\(11.4\text{ cm}^2.\)