(a) Calculate the area of the shaded segment of the circle shown in the diagram [Take \(\pi = \frac{22}{7}\)] (b) A tin has radius 3cm and height 6cm. Find ...

(a) Area of the shaded segment

The shaded segment is found by subtracting the area of the triangle formed by the two radii from the area of the sector:

\[\text{area of segment}=\text{area of sector}-\text{area of triangle}.\]

The angle at the centre is \(63^\circ\), not \(30^\circ\). Both sides of the triangle are radii of length \(10\text{ cm}\).

\[\text{Area of sector}=\frac{63}{360}\times\frac{22}{7}\times10^2=55\text{ cm}^2.\]

For the triangle, use \(\frac12 ab\sin C\):

\[\text{Area of triangle}=\frac12\times10\times10\times\sin63^\circ\]

\[=50\times0.891=44.55\text{ cm}^2.\]

Therefore,

\[\text{Area of shaded segment}=55-44.55=10.45\text{ cm}^2.\]

(b) Tin with radius \(3\text{ cm}\) and height \(6\text{ cm}\)

Total surface area includes the curved surface and both circular ends, so:

\[\text{TSA}=2\pi r(r+h)\]

\[=2\times\frac{22}{7}\times3\times(3+6)\]

\[=\frac{1188}{7}=169.71\text{ cm}^2.\]

Volume of the cylindrical tin is:

\[V=\pi r^2h\]

\[=\frac{22}{7}\times3^2\times6=\frac{1188}{7}=169.71\text{ cm}^3.\]

Since \(1000\text{ cm}^3=1\) litre:

\[169.71\div1000=0.16971\text{ litres}\]

Correct to two decimal places, the capacity is \(0.17\) litres.

Answers: shaded segment \(=10.45\text{ cm}^2\); total surface area \(=169.71\text{ cm}^2\); capacity \(=0.17\) litres.

Examination reminder: For a circular segment, use sector minus triangle, and take the central angle directly from the diagram. Here, using \(30^\circ\) instead of \(63^\circ\) changes both areas and gives the wrong segment area.