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Area, SM-Bank 139

The diagram shows a sector with an angle of 120° cut from a circle with radius 10 m.

What is the area of the sector? Write your answer correct to 1 decimal place.  (2 marks)

NOTE:  \(\text{Sector area}=\dfrac{\theta}{360}\times \pi r^2\)

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\(104.7\ \text{m}^2\ (1\text{ d.p.})\)

Show Worked Solution
\(\text{Sector area}\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{120}{360}\times \pi\times 10^2\)
  \(=104.719\dots\)
  \(\approx 104.7\ \text{m}^2\ (1\text{ d.p.})\)

Area, SM-Bank 137

The sector shown has a radius of 13 cm and an angle of 230°. 
 

 

 What is the area of the sector to the nearest square centimetre?    (2 marks) 

NOTE:  \(\text{Sector area}=\dfrac{\theta}{360}\times \pi r^2\)

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\(339\ \text{cm}^2\ (\text{nearest cm}^2)\)

Show Worked Solution
\(\text{Sector area}\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{230}{360}\times \pi\times 13^2\)
  \(=339.204\dots\)
  \(\approx 339\ \text{cm}^2\ (\text{nearest cm}^2)\)

Area, SM-Bank 128

Calculate the area of the following sector, giving your answer as an exact value in terms of \(\pi\).   (2 marks)

NOTE:  \(\text{Sector area}=\dfrac{\theta}{360}\times \pi r^2\)

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\(27\pi\ \text{mm}^2\)

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\(\theta=30^{\circ} \ \ r=18\ \text{mm}\)

\(A\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{30}{360}\times \pi \times 18^2\)
  \(=\dfrac{1}{12}\times 324\pi\)
  \(=27\pi\ \text{mm}^2\)

Area, SM-Bank 127

Calculate the area of the following sector, giving your answer as an exact value in terms of \(\pi\).   (2 marks)

NOTE:  \(\text{Sector area}=\dfrac{\theta}{360}\times \pi r^2\)

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\(\dfrac{100\pi}{3}\ \text{m}^2\)

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\(\theta=120^{\circ} \ \ r=10\ \text{m}\)

\(A\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{120}{360}\times \pi\times 10^2\)
  \(=\dfrac{1}{3}\times 100\pi\)
  \(=\dfrac{100\pi}{3}\ \text{m}^2\)

Area, SM-Bank 121

Milan cuts a sector from a circle so that  \(\dfrac{3}{8}\)  of the area of the circle remains.
 


 

If the circle's radius is 4 cm, what is the area of the shape, to the nearest square centimetre?  (2 marks)

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\(19\ \text{cm}^2\ (\text{nearest cm}^2)\)

Show Worked Solution
\(\text{Area}\) \(=\dfrac{3}{8}\times \pi r^2\)
  \(=\dfrac{3}{8}\times \pi\times 4^2\)
  \(=18.849\dots\)
  \(=19\ \text{cm}^2\ (\text{nearest cm}^2)\)

Area, SM-Bank 115

Calculate the area of the following shape, giving your answer correct to 1 decimal place.  (2 marks)
 

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\(855.3\ \text{m}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area quadrant}\) \(=\dfrac{1}{4}\times\pi r^2\)
  \(=\dfrac{1}{4}\times\pi\times 33^2\)
  \(=855.2985\dots\)
  \(\approx 855.3\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 114

Calculate the area of the following shape, giving your answer correct to 1 decimal place.  (2 marks)

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\(84.8\ \text{m}^2\ (1\ \text{d.p.})\)

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\(\text{Diameter}=12\ \text{m}\)

\(\therefore\ \text{Radius}=6\ \text{m}\)

\(\text{Area}\) \(=\dfrac{3}{4}\times\pi r^2\)
  \(=\dfrac{3}{4}\times\pi\times 6^2\)
  \(=84.8230\dots\)
  \(\approx 84.8\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 113

Calculate the area of the following shape, giving your answer as an exact value in terms of \(\pi\).  (2 marks)

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\(12\pi \ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(=\dfrac{3}{4}\times\pi r^2\)
  \(=\dfrac{3}{4}\times\pi\times 4^2\)
  \(=12\pi \ \text{m}^2\)

Area, SM-Bank 112

Calculate the area of the following quadrant, giving your answer as an exact value in terms of \(\pi\).  (2 marks)

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\(\dfrac{225\pi}{4}\ \text{cm}^2\)

Show Worked Solution
\(\text{Area quadrant}\) \(=\dfrac{1}{4}\times\pi r^2\)
  \(=\dfrac{1}{4}\times\pi\times 15^2\)
  \(=\dfrac{225}{4}\pi\)
  \(=\dfrac{225\pi}{4}\ \text{cm}^2\)

Area, SM-Bank 041

A golf course has a sprinkler system that waters the grass in the shape of a sector, as shown in the diagram below. 
 

A sprinkler is positioned at point \(L\) and can turn through an angle of 100°.

The section of grass that is watered is 4.5 m wide at all points.

Water can reach a maximum of 12 m from the sprinkler at \(L\).

What is the area of grass that this sprinkler will water?

Round your answer to the nearest square metre.  (2 marks)

NOTE:  \(\text{Sector Area}=\dfrac{\theta}{360}\times \pi r^2\)

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`199\ text{m²  (nearest m²)}`

Show Worked Solution
\(\text{Area}\) \(=\dfrac{360-\theta}{360}\times \pi\times R^2-\dfrac{360-\theta}{360}\times \pi\times r^2\)
  \(=\dfrac{260}{360}\times \pi\times 12^2-\dfrac{260}{360}\times \pi\times 7.5^2\)
  \(= 199.09\dots\)
  \(= 199\ \text{m²  (nearest m}^2)\)

 

Area, SM-Bank 037

The pie chart below displays the results of a survey.
 


 

Eighty per cent of the people surveyed selected ‘agree’.

Twenty per cent of the people surveyed selected ‘disagree’.

The radius of the pie chart is 16 mm.

Calculate the area of the "agree" sector, correct to the nearest square millimetre.  (2 marks)

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\(643\ \text{mm}^2\)

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\(\text{Agree represents }80\text{% of the circle}\)

\(\text{Area}\) \(=80\text{%}\times \pi r^2\)
  \(=0.8\times \pi\times 16^2\)
  \(=643.39\dots\)
  \(\approx 643\ \text{mm}^2\)

 

Area, SM-Bank 036

A windscreen wiper blade can clean a large area of windscreen glass, as shown by the shaded area in the diagram below.
 

 

The windscreen wiper blade is \(30\) cm long and it is attached to a \(9\) cm long arm.

The arm and blade move back and forth in a circular arc with an angle of \(110^\circ\) at the centre.

Calculate the area cleaned by this blade, in square centimetres, correct to one decimal place.  (2 marks)

NOTE:  \(\text{Sector area}=\dfrac{\theta}{360}\times \pi r^2\)

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\(1382.3\ \text{cm}^2\)

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\(\theta=110^\circ,\ \text{Radius large sector}=39,\ \text{Radius small sector}=9\)

\(\text{Area cleaned}\) \(\ =\ \text{large sector − small sector}\)
  \(=\dfrac{110}{360}\times \pi\times 39^2-\dfrac{110}{360}\times \pi\times 9^2\)
  \(=1382.300\dots\)
  \(\approx 1382.3\ \text{cm}^2\)

Measurement, STD2 M1 2009 HSC 11 MC

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  

  1. 34 cm²
  2. 42 cm²
  3. 50 cm²
  4. 193 cm²
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`B`

Show Worked Solution
`text(Area)` `=\ text(Area of Sector – Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx bh)`
  `= (90/360 xx pi xx 8^2)-(1/2 xx 4 xx 4)`
  `= 50.2654…-8`
  `= 42.265…\ text(cm²)`

`=>  B`

Measurement, STD2 M1 2013 HSC 16 MC

The shaded region shows a quadrant with a rectangle removed.
  

What is the area of the shaded region, to the nearest cm2?

  1. 38 cm²
  2. 52 cm²
  3. 61 cm²
  4. 70 cm²
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`B`

Show Worked Solution
`text(Shaded area)` `=\ text(Area of segment – Area of rectangle)`
  `=1/4 pi r^2-(6xx2)`
  `=1/4 pi xx9^2-12`
  `=51.617…\ text(cm²)`

`=>\ B`