GCSE Arcs and sectors Practice Questions

Write down the formula for the arc length of a sector.

An arc is part of the circumference. A sector with an angle of $\theta$ at the centre takes up $\frac{\theta}{360}$ of the full turn, so its arc takes up the same fraction of the circumference $2\pi r$.

Work out what fraction of the whole circle the sector is.

A full turn is $360^\circ$, so an angle of $60^\circ$ at the centre gives $\frac{1}{6}$ of the whole circle.

Take that fraction of the circumference and state the arc length.

The arc is $\frac{1}{6}$ of the circumference $2\pi r$, so it is $3\pi$ cm — exact, in terms of $\pi$.

Think about what a sector is part of.

A sector is a slice of the whole circle, so the answer must be built from the area $\pi r^2$, not from the circumference $2\pi r$.

Work out what fraction of the circle the sector is.

The angle $\theta$ is $\frac{\theta}{360}$ of the full turn, so the sector is that same fraction of the area.

Select the correct expression.

The sector is $\frac{\theta}{360}$ of the circle of area $\pi r^2$. Of the others, $\frac{\theta}{360} \times 2\pi r$ is the ARC LENGTH, $\frac{\theta}{180} \times \pi r^2$ doubles the answer, $\frac{360}{\theta} \times \pi r^2$ is bigger than the whole circle, and $\pi r^2$ is the whole circle.

Write down the formula for the arc length of a sector.

An arc is part of the circumference. A sector with an angle of $\theta$ at the centre takes up $\frac{\theta}{360}$ of the full turn, so its arc takes up the same fraction of the circumference $2\pi r$.

Work out what fraction of the whole circle the sector is.

A full turn is $360^\circ$, so an angle of $54^\circ$ at the centre gives $\frac{3}{20}$ of the whole circle.

Work out the circumference of the whole circle.

The circumference of a circle of radius $10$ cm is $2\pi r$, which is $20\pi$ cm.

Work out the arc length.

The arc is $\frac{3}{20}$ of the circumference, giving $3\pi$ cm.

Rule out the four wrong options.

The first is the AREA of the sector, in square cm; the second is the circumference of the whole circle, not an arc of it; and the last two halve or double the correct fraction of the circumference, which does not match an angle of $54^\circ$.

Select the correct arc length.

The exact arc length is $3\pi$ cm.

Write down the formula for the area of a sector.

A sector is part of the whole circle. It takes up $\frac{\theta}{360}$ of the full turn, so it takes up the same fraction of the area $\pi r^2$.

Work out what fraction of the whole circle the sector is.

A full turn is $360^\circ$, so an angle of $90^\circ$ at the centre gives $\frac{1}{4}$ of the whole circle.

Substitute what is known into the formula.

The sector is $\frac{1}{4}$ of the circle of area $\pi r^2$, and that equals $36\pi$ square cm.

Divide both sides by pi.

The $\pi$ cancels from both sides.

Make the square of the radius the subject.

Dividing by $\frac{1}{4}$ multiplies by $4$, giving $r^2 = 144$.

Take the square root.

The radius is $12$ cm.

Check the answer by working the area out again.

A radius of $12$ cm does give a sector area of $36\pi$ square cm.

Rule out the other options by testing them.

None of the other radii give $36\pi$ square cm — they are all out by a square factor.

Check the units of the answer.

A length is measured in cm — the same units as the radius, not squared.

Select the correct radius.

The radius of the circle is $12$ cm.

Say what the perimeter of the track is made of.

Going once round the outside you travel along the two long straight sides and round the two curved ends.

Put the two semicircular ends together.

The two ends are identical semicircles, so together they make one complete circle. That is much quicker than working each one out.

Work out the radius of each semicircular end.

The width of the rectangle is the DIAMETER of each end, so the radius is half of it, $25$ m.

Work out the circumference of that circle.

The two curved ends together are $50\pi$ m.

Work out one semicircular end on its own as a check.

Each end is $25\pi$ m, and two of them give $50\pi$ m, as expected.

Work out the total length of the straight sides.

There are two straight sides, each $80$ m long.

Add the curved part and the straight part.

The $\pi$ term and the number term cannot be combined, so the perimeter is $50\pi + 160$ m.

Check which lengths are actually on the outside.

The two short sides of the rectangle are inside the track, not on its edge, so the width is never added on.

Note the most common error here.

Using the width $50$ m as the radius instead of the diameter would double the curved part. The radius is half the width.

Compare the curved part with the straight part.

The two curved ends come to about $157$ m, and the two straights come to $160$ m, so the straights are the longer part of one lap.

Check the size of the answer using an approximate value for pi.

Taking $\pi \approx 3.14$ gives the perimeter as roughly $317$ m, which is a sensible size.

Check what happens if the track is made longer.

Stretching the rectangle changes only the straight part; the curved ends still make one circle of radius $25$ m, so only the number term would change.

Check the width is used only through the radius.

The width enters the answer only by fixing the radius of the ends; it is never added on as a straight edge.

Check the units of the answer.

A perimeter is a length, so the answer is in m.

State the perimeter of the track.

The perimeter of the track is $50\pi + 160$ m.