smc-978-10-Arc Length/Perimeter

Trigonometry, 2ADV T1 2023 HSC 16

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place. (4 marks)

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`23.8\ text{m}`

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`text{Arc}\ PQ`	`=110/360 xx pi xx 2 xx 2.1`
	`=4.03171… \ text{m}`

`text{Consider}\ ΔOPQ:`

`sin 55^@`	`=x/2.1`
`x`	`=2.1 xx sin 55^@`
	`=1.7202…`

`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}`	`=8+(2xx3.6)+4.031+(8-3.440)`
	`=23.79…\ text{m}`
	`=23.8\ text{m (to 1 d.p.)}`

Trigonometry, 2ADV T1 2019 HSC 13b

The diagram shows a circle with centre `O` and radius 20 cm.

The points `A` and `B` lie on the circle such that `∠AOB = 70^@`.

Find the perimeter of the shaded segment, giving your answer correct to one decimal place. (3 marks)

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`47.4\ text{cm}`

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`text(Arc)\ AB`	`= 70/360 xx 2pi xx 20`
	`= 24.43…`

`text(Using cosine rule,)`

`AB^2`	`= 20^2 xx 20^2 – 2 ⋅ 20 ⋅ 20 xx cos 70`
	`= 526.383…`
`AB`	`= 22.94…`

`:.\ text(Perimeter)`	`= 24.43 + 22.94…`
	`= 47.37`
	`= 47.4\ text{cm (1 d.p.)}`

Trigonometry, 2ADV T1 2005 HSC 4a

A pendulum is 90 cm long and swings through an angle of 0.6 radians. The extreme positions of the pendulum are indicated by the points `A` and `B` in the diagram.

Find the length of the arc `AB`. (1 mark)

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Find the straight-line distance between the extreme positions of the pendulum. (2 marks)

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Find the area of the sector swept out by the pendulum. (1 mark)

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`text(54 cm)`
`text{53.2 cm (to 1 d.p.)}`
`text(2430 cm²)`

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i.	`text(Arc)\ AB`	`= theta/(2 pi) xx 2 pi r`
		`= rtheta`
		`= 90 xx 0.6`
		`= 54\ text(cm)`

ii.

`text(Using the cosine rule)`

`text(Distance)\ AB\ text(in straight line)`

`AB^2`	`= 90^2 xx 90^2 − 2 xx 90 xx 90 xx cos\ 0.6`
	`= 2829.563…`
`:.AB`	`= 53.193…`
	`= 53.2\ text{cm (to 1 d.p.)}`

iii. `text(Area of Sector)`

`= 0.6/(2pi) xx pir^2`

`= 0.3 xx 90^2`

`= 2430\ text(cm²)`

Trigonometry, 2ADV T1 2008 HSC 7b

The diagram shows a sector with radius `r` and angle `theta` where `0 < theta <= 2pi`.

The arc length is `(10pi)/3`.

Show that `r >= 5/3`. (2 marks)

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Calculate the area of the sector when `r = 4`. (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`(20pi)/3\ text(u²)`

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i.	`text(Show)\ r >= 5/3`

`text(Arc length)\ `	`= r theta\ \ text(where)\ \ 0 < theta <= 2pi`
`r theta`	`= (10pi)/3`
`:.theta`	`= (10pi)/(3r)`

`text(Using)\ \ \ 0 <= theta <= 2 pi`

`0 <= (10pi)/(3r)`	`<= 2pi`
`(10pi)/3`	`<= 2 pi r`
`5/3`	`<= r`

`:.\ r >= 5/3\ \ \ text(… as required.)`

ii.	`text(Area)`	`= 1/2 r^2 theta`
		`= 1/2 xx 4^2 xx (10pi)/(3 xx 4)`
		`= (20pi)/3\ text(u²)`

Trigonometry, 2ADV T1 2014 HSC 11g

The angle of a sector in a circle of radius 8 cm is `pi/7` radians, as shown in the diagram.

Find the exact value of the perimeter of the sector. (2 marks)

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`(8pi)/7 + 16\ text(cm)`

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`text(Arc length)`	`= theta/(2pi) xx 2 pi r`
	`= pi/7 xx 8`
	`= (8pi)/7\ text(cm)`

`text(S)text(ector perimeter)`	`= text(arc) + 2 xx text(radius)`
	`= (8pi)/7 + 16\ text(cm)`

Trigonometry, 2ADV T1 2009 HSC 5c

The diagram shows a circle with centre `O` and radius 2 centimetres. The points `A` and `B` lie on the circumference of the circle and `/_AOB = theta`.

There are two possible values of `theta` for which the area of `Delta AOB` is `sqrt 3` square centimetres. One value is `pi/3`.

Find the other value. (2 marks)

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Suppose that `theta = pi/3`.

(1) Find the area of sector `AOB` (1 mark)

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(2) Find the exact length of the perimeter of the minor segment bounded by the chord `AB` and the arc `AB`. (2 marks)

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`(2pi)/3`
(1) `(2pi)/3\ \ text(cm²)`
(2) `(2 + (2pi)/3)\ text(cm)`

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i.	`text(Area)\ Delta AOB`	`= 1/2 ab sin theta`
		`= 1/2 xx 2 xx 2 xx sin theta`
		`= 2 sin theta`

`2 sin theta`	`= sqrt 3\ \ \ text{(given)}`
`sin theta`	`= sqrt3/2`
`:. theta`	`= pi/3,\ pi\ – pi/3`
	`= pi/3,\ (2pi)/3`

`:.\ text(The other value of)\ theta\ text(is)\ \ (2pi)/3\ \ text(radians)`

ii. (1)	`text(Area of sector)\ AOB`	`= pi r^2 xx theta/(2pi)`
		`= 1/2 r^2 theta`
		`= 1/2 xx 2^2 xx pi/3`
		`= (2pi)/3\ text(cm²)`

ii. (2)

`text(Using the cosine rule:)`

`AB^2`	`= OA^2 + OB^2\ – 2 xx OA xx OB xx cos theta`
	`= 2^2 + 2^2\ – 2 xx 2 xx 2 xx cos (pi/3)`
	`= 4 + 4\ – 4`
	`= 4`
`:.\ AB`	`= 2`

`text(Arc)\ AB`	`= 2 pi r xx theta/(2pi)`
	`= r theta`
	`= (2pi)/3\ text(cm)`

`:.\ text(Perimeter) = (2 + (2pi)/3)\ text(cm)`

Trigonometry, 2ADV T1 2012 HSC 11f

The area of the sector of a circle with a radius of 6 cm is 50 cm².

Find the length of the arc of the sector. (2 marks)

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`50/3\ text(cm)`

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TIP: Many students find it easier to think of the area of a sector by calculating `theta/(2 pi)` multiplied by the area of a circle rather than remembering a formula.

`text(Area of sector, radius 6 cm = 50 cm²)`

`theta/(2 pi) xx pi r^2`	`= 50`
`1/2 r^2 theta`	`=50`
`1/2 xx 6^2 xx theta`	`=50`
`theta`	`=50/18=25/9\ text(radians)`

`:.\ text(Length of Arc)`	`= theta/(2pi) xx 2pi r`
	`= theta xx r`
	`= 25/9xx6`
	`= 50/3 \ text(cm)`