Challenge · Arc Length & Sector Area

19 marks

These problems combine ideas: you will need to form your own equation and solve it. There is a worked example to show you how to start; then it is over to you. Stuck? Hints are at the foot of the sheet.

Worked example: how to start

A sector has an angle of $90^\circ$ at the centre. The length of its arc is $12$ cm. Work out the radius.

The arc is $\dfrac{90}{360}$ of the circumference, so $\dfrac{90}{360}\times 2\pi r = 12$.

This simplifies to $\dfrac{\pi r}{2} = 12$, so $\pi r = 24$.

$r = \dfrac{24}{\pi} = 7.64$ cm (3 s.f.).

1The length of the arc of a sector is equal to its radius. Work out the angle at the centre.[3 marks]

Answer:

2A sector has an angle of $120^\circ$ at the centre. The perimeter of the sector is $40$ cm. Work out the radius.[4 marks]

Answer:

3The ratio of the arc length of a sector to its perimeter is $2 : 5$. Work out the angle at the centre.[4 marks]

Answer:

4A sector of radius $9$ cm has the same area as a semicircle of radius $6$ cm. Work out the angle of the sector.[3 marks]

Answer:

5A sector has a perimeter of $20$ cm and an area of $16$ cm$^2$. Work out the two possible values of the radius.[5 marks]

Answer:

★Extension: A sector’s angle is doubled and its radius is halved. Prove that the new area is exactly half of the original area.

Stuck? Hints (don't peek unless you need to)1. Write the arc-length formula and set it equal to $r$. The $r$ cancels.2. Perimeter $=$ arc length $+$ two radii. Write it all in terms of $r$, then solve.3. Perimeter $=$ arc $+ 2r$. Turn the ratio into an equation; the radius cancels.4. Find the semicircle’s area first, then set the sector’s area equal to it.5. The area of a sector is $\tfrac12\times$ arc length $\times$ radius. Write the arc in terms of $r$ using the perimeter.

Solutions & mark scheme · Arc Length & Sector Area

Total: 19 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1The length of the arc of a sector is equal to its radius. Work out the angle at the centre.[3]

Model solution

$\dfrac{\theta}{360}\times 2\pi r = r$

Divide by $r$: $\dfrac{\theta}{360}\times 2\pi = 1$

$\theta = \dfrac{360}{2\pi} = 57.3^\circ$ (3 s.f.).

Answer: 57.3°

Marks

✔1Sets arc length $=r$ and cancels $r$

✔1Rearranges for θ

✔157.3° (3 s.f.)

2A sector has an angle of $120^\circ$ at the centre. The perimeter of the sector is $40$ cm. Work out the radius.[4]

Model solution

Perimeter $= \dfrac{120}{360}\times 2\pi r + 2r = \dfrac{2\pi}{3}r + 2r$

So $r\left(\dfrac{2\pi}{3}+2\right) = 40$

$r = \dfrac{40}{\,2\pi/3 + 2\,} = 9.77$ cm (3 s.f.).

Answer: 9.77 cm

Marks

✔1Arc length $=\frac{120}{360}\times2\pi r$

✔1Perimeter equation $=40$

✔1Factorises out $r$

✔19.77 cm (3 s.f.)

3The ratio of the arc length of a sector to its perimeter is $2 : 5$. Work out the angle at the centre.[4]

Model solution

$\dfrac{\text{arc}}{\text{arc}+2r} = \dfrac{2}{5}$, so $5\,\text{arc} = 2\,\text{arc}+4r$

Arc $= \dfrac{4r}{3}$, and arc $= \dfrac{\theta}{360}\times 2\pi r$

$\dfrac{\theta}{360}\times 2\pi = \dfrac{4}{3}$, so $\theta = \dfrac{480}{2\pi} = 76.4^\circ$ (3 s.f.).

Answer: 76.4°

Marks

✔1Ratio → $5\,\text{arc}=2\,\text{arc}+4r$

✔1Arc $=\frac{4r}{3}$

✔1Angle equation

✔176.4° (3 s.f.)

4A sector of radius $9$ cm has the same area as a semicircle of radius $6$ cm. Work out the angle of the sector.[3]

Model solution

Semicircle area $= \dfrac{1}{2}\pi(6)^2 = 18\pi$

Sector area $= \dfrac{\theta}{360}\pi(9)^2 = 18\pi$

$\dfrac{\theta}{360} = \dfrac{18}{81} = \dfrac{2}{9}$, so $\theta = 80^\circ$.

Answer: 80°

Marks

✔1Semicircle area $=18\pi$

✔1Sets sector area $=18\pi$

✔180°

5A sector has a perimeter of $20$ cm and an area of $16$ cm$^2$. Work out the two possible values of the radius.[5]

Model solution

Perimeter: arc $= 20 - 2r$. Area $= \dfrac{1}{2}\times\text{arc}\times r = 16$

$\dfrac{1}{2}(20-2r)r = 16 \Rightarrow 20r - 2r^2 = 32$

$r^2 - 10r + 16 = 0 \Rightarrow (r-2)(r-8)=0$

$r = 2$ cm or $r = 8$ cm.

Answer: r = 2 cm or r = 8 cm

Marks

✔1Uses area $=\frac12\,\text{arc}\times r$

✔1Arc $=20-2r$

✔1Forms quadratic

✔1Solves

✔1Both values 2 and 8

★Extension

Original area $A = \dfrac{\theta}{360}\pi r^2$.

New area $= \dfrac{2\theta}{360}\pi\left(\dfrac{r}{2}\right)^2 = \dfrac{2\theta}{360}\pi\cdot\dfrac{r^2}{4} = \dfrac{1}{2}\cdot\dfrac{\theta}{360}\pi r^2 = \dfrac{1}{2}A.$