Answer: (a) \(1.75\) radians; (b) \(35.3\) m; (c) \(74.4\text{ m}^2\).

Answer: (a) \(1.75\) radians; (b) \(35.3\) m; (c) \(74.4\text{ m}^2\).

(a) The length of arc \(AB\) is \(12.25\), and the radius is \(7\). Using

\(s=r\theta,\)

we get

\(12.25=7\theta.\)

Therefore

\(\theta=1.75.\)

(b) Since \(BC\) is tangent at \(B\), \(OB\) is perpendicular to \(BC\). In right triangle \(OBC\),

\(OB=7,\qquad BC=24.\)

Hence

\(OC=\sqrt{7^2+24^2}=25.\)

So

\(\cos\angle BOC=\frac7{25}.\)

Thus

\(\angle BOC\approx1.287.\)

The arc from \(B\) to the point where \(OC\) meets the circle has length

\(7(1.287)\approx9.009.\)

The perimeter of the shaded region is made from the arc \(AB\), this second arc, and two radii. Therefore

\(\text{perimeter}=12.25+9.009+14.\)

So

\(\text{perimeter}\approx35.3\text{ m}.\)

(c) The shaded area is the sum of two sectors: sector \(AOB\) and sector \(BOC\).

So

\(\text{area}=\frac12(7^2)(1.75)+\frac12(7^2)(1.287).\)

This gives

\(\text{area}\approx74.4\text{ m}^2.\)