The radius is half the diameter: \(48 \div 2 = 24\) metres.

The arc of a semicircle is half the circumference of the whole circle.

\(\frac{1}{2} \times 2 \times \pi \times 24 = 24\pi\) metres

So the option to circle is \(24\pi\).

A sector's arc is the same fraction of the circumference as its angle is of a full turn.

\(\frac{ 60 }{360} \times 2 \times \pi \times 6 = 2\pi\)

\(2\pi\) cm

A sector's area is the same fraction of the circle's area as its angle is of a full turn.

\(\frac{ 135 }{360} \times \pi \times 4^2 = 6\pi\)

\(6\pi\) cm²

The two angles at \(O\) make a full turn.

\(360 - 135 = 225\)

The reflex angle is \(225^\circ\).

The unshaded part is the major sector. It can be found either as its own fraction of the circle, or by taking the shaded part away from the whole — the second is quicker.

The whole circle: \(\pi \times 8^2 = 64\pi\)

The shaded sector: \(\frac{ 135 }{360} \times 64\pi = 24\pi\)

\(64\pi - 24\pi = 40\pi\)

\(40\pi\) cm²

The perimeter of a sector is the arc and the two straight edges. Leaving the two radii out is the commonest mistake in this topic.

The arc:

\(\frac{ 90 }{360} \times 2 \times \pi \times 8 = 4\pi\)

The two straight edges are both radii:

\(8 + 8 = 16\)

\(P = 4\pi + 16\) cm

The whole circumference is

\(2 \times \pi \times 12 = 24\pi\) cm

The arc is \(4\pi\) of that, so the sector is the fraction

\(\frac{ 4\pi }{ 24\pi } = \frac{ 4 }{ 24 }\)

of the whole circle. A full turn is \(360^\circ\), so

\(x = \frac{ 4 }{ 24 } \times 360 = 60\)

\(x = 60^\circ\)

Circle A has area \(\pi \times 3^2 = 9\pi\) cm².

The sector is \(\frac{ 225 }{360}\) of a full circle, since a full turn is \(360^\circ\).

Sector B has area \(\frac{ 225 }{360} \times \pi \times 5^2 = \frac{ 225 }{360} \times 25\pi = 15.625\pi\) cm².

Comparing \(9\pi\) with \(15.625\pi\): \(15.625\pi > 9\pi\).

Shape B has the greater area.

The circumference is \(2\pi r\), so the radius is

\(r = \frac{ 11\pi }{2\pi} = 5.5\) cm

The shaded part is the part of the circle outside angle x, so its angle at the centre is \(240^\circ\) and it is \(\frac{ 240 }{360}\) of the circle.

Then

\(\text{area} = \frac{ 240 }{360} \times \pi \times 5.5^2 \approx 63.36\) cm²

Area is proportional to the angle at the centre, so a sector's area rises and falls with its angle.

The shaded part is the part of the circle outside angle x, and angle \(x\) is larger than \(120^\circ\), so the shaded area is smaller than the answer to part (a).

The arc is a quarter of the circumference of a circle with the same radius:

\(\frac{1}{4} \times 2 \times \pi \times r = 3\pi\)

\(\frac{1}{2}\pi r = 3\pi\)

Divide both sides by \(\pi\), then multiply by 2:

\(r = 2 \times 3 = 6\)

The radius of the quarter circle is \(6\) mm.

The circle needs its whole circumference:

\(2 \times \pi \times 20 = 40\pi\)

The quarter circle needs its arc and its two straight edges, which are both radii:

\(3\pi + 6 + 6\)

Add the circle and the quarter circle:

\(40\pi + 3\pi + 6 + 6\)

Collect the terms in \(\pi\) and the whole numbers:

\((43\pi + 12)\) mm

The angle of the sector is not given, so find it from the area. The sector is \(\frac{ x }{360}\) of a circle of radius \(6\) cm:

\(\frac{ x }{360} \times \pi \times 6^2 = 39.4\)

\(x = \frac{ 360 \times 39.4 }{ \pi \times 6^2 } \approx 125.4^\circ\)

The arc is the same fraction of the circumference:

\(\text{arc} = \frac{ x }{360} \times 2 \times \pi \times 6 \approx 13.13\) cm

(Check: the arc is also \(2 \times 39.4 \div 6\), since area \(= \frac{1}{2} \times \text{arc} \times r\).)

The perimeter is the arc plus the two straight edges, each a radius of \(6\) cm:

\(P \approx 13.13 + 6 + 6 \approx 25.13\) cm

\(P \approx 25.1\) cm to 3 significant figures.