Answer: perimeter \(85.6\) cm; area \(417\text{ cm}^2\).

(a) Let the minor angle \(AOB\) be \(\theta\). Since \(AB=15\) is a chord and \(OA=OB=10\), half the chord has length \(7.5\). Hence

\(\sin\frac{\theta}{2}=\frac{7.5}{10}=0.75.\)

Therefore

\(\theta=2\sin^{-1}(0.75)=1.696\ldots.\)

So

\(\theta=1.70\)

radians correct to 2 decimal places.

(b) The shaded perimeter consists of the major arc \(AB\), plus \(AC\) and \(BC\).

The length of the major arc is

\(10(2\pi-\theta).\)

Using \(\theta=1.696\ldots\), this is approximately

\(10(2\pi-1.696\ldots)=45.9.\)

To find \(AC\), drop a perpendicular from \(O\) to \(AB\). This bisects \(AB\), so half the chord is \(7.5\). The distance from \(O\) to \(AB\) is

\(\sqrt{10^2-7.5^2}=\sqrt{43.75}.\)

Since \(OC=25\), the perpendicular distance from \(C\) to \(AB\) is

\(25-\sqrt{43.75}.\)

Thus

So the perimeter is

\(45.9+2(19.9\ldots)=85.6\text{ cm}.\)

(c) The shaded area is the area of the major sector \(AOB\) plus the area of the kite \(OACB\).

The major sector area is

\(\frac12(10)^2(2\pi-\theta)=229\ldots.\)

The kite area is

\(\frac12(15)(25)=187.5.\)

Therefore the total shaded area is approximately

\(229\ldots+187.5=417\text{ cm}^2.\)