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Area, SM-Bank 157

An Aussie Rules football team has booked half of the SCG for a training session. The field available to them under this booking covers 8257 square metres.

Assuming the SCG is perfectly round, determine its diameter, giving your answer in metres to 1 decimal place.  (2 marks)

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\(145.0\ \text{m}\)

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\(\text{Area of full SCG}\ =2 \times 8257 = 16\ 514 \ \text{m}^{2}\)

\(A\) \(=\pi r^{2} \)
\(16\ 514\) \(= \pi r^2\)
\(r^{2}\) \(=\dfrac{16\ 514}{\pi} \)
\(r\) \(=\sqrt{5256.569…}\)
  \(=72.5022…\ \text{m} \)

 
\(\therefore\ \text{Diameter}\ = 2 \times 72.502 = 145.0\ \text{m (1 d.p.)} \)

Area, SM-Bank 156

The semi-circle, pictured below, has an area of 32 square centimetres.
 

Calculate the diameter of the semi-circle, giving your answer to 2 decimal places.  (2 marks)

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\(9.03\ \text{cm}\)

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\(\text{Area of full circle}\ =2 \times 32 = 64 \ \text{cm}^{2}\)

\(A\) \(=\pi r^{2} \)
\(64\) \(= \pi r^2\)
\(r^{2}\) \(=\dfrac{64}{\pi} \)
\(r\) \(=\sqrt{20.3718…}\)
  \(=4.5135…\ \text{cm} \)

 
\(\therefore\ \text{Diameter}\ = 2 \times 4.513 = 9.03\ \text{cm (2 d.p.)} \)

Area, SM-Bank 155

A semi-circle has an area of 470 square centimetres.

Calculate the diameter of the semi-circle, giving your answer to 1 decimal place.  (2 marks)

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\(34.6\ \text{cm}\)

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\(\text{Area of full circle}\ =2 \times 470 = 940\ \text{cm}^{2}\)

\(A\) \(=\pi r^{2} \)
\(940\) \(= \pi r^2\)
\(r^{2}\) \(= \dfrac{940}{\pi} \)
\(r\) \(=\sqrt{299.211…}\)
  \(=17.30\ \text{cm}\)

 
\(\therefore\ \text{Diameter}\ = 2 \times 17.30 = 34.6\ \text{cm (1 d.p.)}\)

Area, SM-Bank 150

The cross-section of the circular road tunnel, pictured below, has an area of \(46.24\pi \) square metres.
 

Calculate the radius of the tunnel.  (2 marks)

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\(6.8\ \text{m}\)

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\(A\) \(=\pi r^{2} \)
\(46.24 \pi\) \(= \pi r^2\)
\(r^{2}\) \(=46.24\)
\(r\) \(=\sqrt{46.24}\)
  \(=6.8\ \text{m}\)

Area, SM-Bank 154

The entrance to a tunnel, pictured below, is a semi-circle with an area of \(28.88\pi \) square metres.
 

Calculate the diameter of the tunnel.  (2 marks)

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\(7.6\ \text{m}\)

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\(\text{Area of full circle}\ =2 \times 28.88\pi = 57.76\pi \ \text{m}^{2}\)

\(A\) \(=\pi r^{2} \)
\(57.76 \pi\) \(= \pi r^2\)
\(r^{2}\) \(=57.76\)
\(r\) \(=\sqrt{57.76}\)
  \(=7.6\ \text{m}\)

Area, SM-Bank 151

The circle pictured below has an area of \(144\pi \) square centimetres.
 

Calculate the radius of the circle.  (2 marks)

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\(12\ \text{cm}\)

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\(A\) \(=\pi r^{2} \)
\(144 \pi\) \(= \pi r^2\)
\(r^{2}\) \(=144\)
\(r\) \(=\sqrt{144}\)
  \(=12\ \text{cm}\)

Area, SM-Bank 153

A circular cricket ground has an area of 11 028 square metres.

Determine the radius of the cricket ground, in metres, giving your answer to 1 decimal place.   (2 marks)

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\(59.2\ \text{m}\)

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\(A\) \(=\pi r^{2} \)
\(11\ 028\) \(= \pi r^2\)
\(r^{2}\) \(=\dfrac{11\ 028}{\pi}\)
\(r\) \(=\sqrt{3510.321…}\)
  \(=59.247…\)
  \(=59.2\ \text{m (1 d.p.)}\)

Area, SM-Bank 152

If a circle has an area of 100 square millimetres, find the radius of the circle, giving your answer to 2 decimal places.  (2 marks)

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\(5.64\ \text{mm}\)

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\(A\) \(=\pi r^{2} \)
\(100\) \(= \pi r^2\)
\(r^{2}\) \(=\dfrac{100}{\pi}\)
\(r\) \(=\sqrt{31.8309}\)
  \(=5.641…\)
  \(=5.64\ \text{mm (2 d.p.)}\)

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

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

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

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

Area, SM-Bank 136

The diagram shows an annulus.
 

 
Calculate the area of the shaded region (annulus), correct to 2 decimal places.  (2 marks)

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\(\approx 50.27\ \text{cm}^2\ (2\text{ d.p.})\)

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\(\text{Method 1:}\)

\(\text{Radius small circle}\ (r)=3\)

\(\text{Radius large circle}\ (R)=\dfrac{10}{2}=5\)

\(\text{Shaded region}\) \(=\text{Area large circle}-\text{Area small circle}\)
  \(=\pi\times 5^2-\pi \times 3^2\)
  \(=25\pi-9\pi\)
  \(=16\pi\)
  \(=50.27\ \text{cm}^2\ (2\text{ d.p.})\)

 
\(\text{Method 2:}\)

\(\text{Area of annulus}\)

\(=\pi(R^2 − r^2)\)

\(=\pi(5^2 − 3^2)\)

\(=\pi(25 − 9)\)

\(=16\pi\)

\(=50.27\ \text{cm}^2\ (2\text{ d.p.})\)

Area, SM-Bank 135

A shape consisting of a quadrant and a right-angled triangle is shown.
 

  1. Use Pythagoras' Theorem to calculate the radius of the quadrant.  (2 marks)

  2. What is the area of this shape, correct to one decimal place?  (2 marks)
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a.    \(8\ \text{cm}\)

b.    \(74.3\ \text{cm}^2\ (1\text{d.p.})\)

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a.    \(\text{Using Pythagoras to find radius}\ (r):\)

\(a^2+b^2\) \(=c^2\)
\(r^2+6^2\) \(=10^2\)
\(r^2\) \(=10^2-6^2\)
\(r\) \(=\sqrt{64}\)
  \(=8\ \text{cm}\)

 

b.    \(\text{Total area}\) \(=\text{Area of triangle}+\text{Area of quadrant}\)
    \(=\dfrac{1}{2}\times 8\times 6+\dfrac{1}{4}\times \pi\times 8^2\)
    \(=24+50.265\dots\)
    \(=74.265\dots\)
    \(\approx 74.3\ \text{cm}^2\ (1\text{d.p.})\)

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 126

Calculate the area of the shaded region in the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

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\(2789.0\ \text{m}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter semi-cirle}=114\ \text{cm}\)

\(\text{Radius semi-cirle}(r)=57\ \text{cm}\)

\(\text{Total area}=\text{Area square}-\text{Area semi-cirle}\)

\(A\) \(=s^2-\pi r^2\)
  \(=114^2-\pi\times 57^2\)
  \(=12\ 996-10\ 207.034\dots\)
  \(=2788.965\dots\approx 2789.0\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 125

Calculate the area of the shaded region in the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

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\(22.0\ \text{m}^2\ (1 \text{ d.p.})\)

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\(\text{Radius small cirle}(r)=3\ \text{m}\)

\(\text{Radius large cirle}(R)=4\ \text{m}\)

\(\text{Total area}=\text{Area large cirle}-\text{Area small cirle}\)

\(A\) \(=\pi R^2-\pi r^2\)
  \(=\pi\times 4^2-\pi\times 3^2\)
  \(=50.265\dots-28.274\dots\)
  \(=21.991\dots\approx 22.0\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 124

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

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\(321.0\ \text{cm}^2\ (1 \text{ d.p.})\)

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\(\text{Radius small semi-cirle}(r)=4.5\ \text{mm}\)

\(\text{Radius large semi-cirle}(R)=9\ \text{mm}\)

\(\text{Total area}=\text{Area small semi-cirle}+\text{Area large semi-cirle}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{2}\times \pi r^2+\dfrac{1}{2}\times \pi R^2+lb\)
  \(=\dfrac{1}{2}\times \pi\times 4.5^2+\dfrac{1}{2}\times \pi\times 9^2+18\times 9\)
  \(=31.808\dots+127.234\dots+162\)
  \(=321.043\dots\approx 321.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 123

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

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\(178.3\ \text{cm}^2\ (1 \text{ d.p.})\)

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\(\text{Radius}=8\ \text{cm}\)

\(\text{Rectangle length}=24-8=16\ \text{cm}\)

\(\text{Total area}=\text{Area Quadrant}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{4}\times \pi r^2+lb\)
  \(=\dfrac{1}{4}\times \pi\times 8^2+16\times 8\)
  \(=50.265\dots+128\)
  \(=178.265\dots\approx 178.3\ \text{cm}^2\ (1 \text{ d.p.})\)

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

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

Area, SM-Bank 120

A one-on-one basketball court is a composite shape made up of a rectangle and a semicircle, as shown below.
 

Calculate the area of the court, giving your answer correct to one decimal place.   (2 marks)

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\(104.5\ \text{m}^2\ (1 \text{ d.p.})\)

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

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

\(\text{Total area}=\text{Area semi-circle}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{2}\times \pi r^2+lb\)
  \(=\dfrac{1}{2}\times \pi\times 6^2+12\times 4\)
  \(=104.548\dots\)
  \(=104.5\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 119

Calculate the area of a semi-circle with a diameter of 60 centimetres. Give your answer correct to one decimal place.   (2 marks)

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\(1413.7\ \text{cm}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter}=60\ \text{cm}\)

\(\therefore\ \text{Radius}=30\ \text{cm}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times \pi r^2\)
  \(=\dfrac{1}{2}\times \pi\times 30^2\)
  \(=1413.716\dots\)
  \(=1413.7\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 118

Calculate the area of a circle with a diameter of 37.4 millimetres. Give your answer correct to one decimal place.   (2 marks)

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\(1098.6\ \text{mm}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter}=37.4\ \text{mm}\)

\(\therefore\ \text{Radius}=18.7\ \text{mm}\)

\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 18.7^2\)
  \(=1098.583\dots\)
  \(=1098.6\ \text{mm}^2\ (\text{1 d.p.})\)

Area, SM-Bank 117

Calculate the area of a circle with a radius of 72.3 centimetres. Give your answer correct to one decimal place.   (2 marks)

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\(16\ 422.0\ \text{cm}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 72.3^2\)
  \(=16\ 422.015\dots\)
  \(=16\ 422.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 116

Calculate the area of a circle with a radius of 20 metres. Give your answer as an exact value in term of \(\pi\).   (2 marks)

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\(400\pi\ \text{m}^2\)

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\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 20^2\)
  \(=400\pi\ \text{m}^2\)

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

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

Area, SM-Bank 111

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

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\(1250\pi\ \text{m}^2\)

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\(\text{Diameter}=100\ \text{m}\)

\(\therefore\ \text{Radius}=50\ \text{m}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times\pi r^2\)
  \(=\dfrac{1}{2}\times\pi\times 50^2\)
  \(=1250\pi\ \text{m}^2\)

Area, SM-Bank 110

Calculate the area of the following semi-circle, giving your answer to 2 decimal places.   (2 marks)
 

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\(0.57\ \text{mm}^2\ (2\ \text{d.p.})\)

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\(\text{Diameter}=1.2\ \text{mm}\)

\(\therefore\ \text{Radius}=0.6\ \text{mm}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times\pi r^2\)
  \(=\dfrac{1}{2}\times\pi\times 0.6^2\)
  \(=0.5654\dots\)
  \(\approx 0.57\ \text{mm}^2\ (2\ \text{d.p.})\)

Area, SM-Bank 109

Calculate the area of the following semi-circle, giving your answer to 2 decimal places.   (2 marks)
 

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\(173.18\ \text{m}^2\ (\text{2 d.p.})\)

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\(\text{Diameter}=21\ \text{m}\)

\(\therefore\ \text{Radius}=10.5\ \text{m}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times\pi r^2\)
  \(=\dfrac{1}{2}\times\pi\times 10.5^2\)
  \(=173.1802\dots\)
  \(\approx 173.18\ \text{m}^2\ (2\ \text{d.p.})\)

Area, SM-Bank 108

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

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\(25\pi\ \text{m}^2\)

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\(\text{Diameter}=10\ \text{m}\)

\(\therefore\ \text{Radius}=5\ \text{m}\)

\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 5^2\)
  \(=25\pi\ \text{m}^2\)

Area, SM-Bank 107

Calculate the area of the following circle, correct to one decimal place.  (2 marks)
 

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\(422.7\ \text{cm}^2\ (1\ \text{d.p.})\)

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\(\text{Diameter}=23.2\ \text{cm}\)

\(\therefore\ \text{Radius}=11.6\ \text{cm}\)

\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 11.6^2\)
  \(=422.7327\dots\)
  \(\approx 422.7\ \text{cm}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 106

Calculate the area of the following circle, correct to one decimal place.   (2 marks)
 

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\(50.3\ \text{m}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 4^2\)
  \(=50.2654\dots\)
  \(\approx 50.3\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 105

Calculate the area of the following circle, correct to one decimal place.  (2 marks)
 

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\(514.7\ \text{cm}^2\ (1\ \text{d.p.})\)

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\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 12.8^2\)
  \(=514.7185\dots\)
  \(\approx 514.7\ \text{cm}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 064 MC

A circular pool is located in a square lawn, as shown below.
 

The sides of the square lawn are 10 m in length.

The pool has a radius of 3 m.

The area of the lawn surrounding the pool, in square metres, is closest to

  1. \(59\)
  2. \(72\)
  3. \(81\)
  4. \(128\)
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\(B\)

Show Worked Solution
\(\text{Area of square}\) \(=\text{side}^2\)
  \(=10^2\)
  \(=100\ \text{m}^2\)

   

\(\text{Area of pool}\) \(=\pi r^2\)
  \(=\pi \times 3^2\)
  \(= 28.27\dots\ \text{m}^2\)

 

\(\therefore\ \text{Area of lawn}\) \(=100-28.27\dots\)
  \(=71.72\dots\ \text{m}^2\)

 
\(\Rightarrow B\)

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

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

Area, SM-Bank 034

The radius of a circle is 6.5 centimetres.

A square has the same area as this circle.

Calculate the side length of the square, in centimetres correct to one decimal place.   (3 marks)

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\(11.5\ \text{cm}\)

Show Worked Solution
\(\text{Area of circle}\) \(=\pi r^2\)
  \(=\pi\times 6.5^2\)
  \(= 132.73\dots\ \text{cm}^2\)

 
\(\text{Area of square}\ =s^2=\ \text{Area of circle}\)

\(s^2\) \(= 132.73\dots\)
\(s\) \(=\sqrt{132.73\dots}\)
  \(= 11.5\ \text{cm (1 d.p.)}\)

Measurement, STD1 M1 2019 HSC 15

The diagram shows a shape made up of a square of side length 8 cm and a semicircle.
  


 

Find the area of the shape to the nearest square centimetre.  (3 marks)

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`89\ text(cm²  (nearest cm²))`

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♦♦ Mean mark 26%.

`text(Area)` `=\ text(Area of square + Area of semicircle)`
  `= 8 xx 8 + 1/2 xx pi xx 4^2`
  `= 89.13…`
  `= 89\ text(cm²  (nearest cm²))`

Measurement, STD2 M1 2008 HSC 11 MC

The diagram shows the floor of a shower. The drain in the floor is a circle with a diameter of 10 cm.

What is the area of the shower floor, excluding the drain?
 

 
 

  1. 9686 cm²
  2. 9921 cm²
  3. 9969 cm²
  4. 10 000 cm²
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`B`

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COMMENT: Students should see that answers are all in cm², and therefore use cm as the base unit for their calculations. 
`text(Area)` `=\ text(Square – Circle)`
  `= (100 xx 100)-(pi xx 5^2)`
  `= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm²)`

 
`=>  B`

Measurement, STD2 M1 2014 HSC 12 MC

A path 1.5  metres wide surrounds a circular lawn of radius 3 metres. 

What is the approximate area of the path?

  1. 7.1 m²
  2. 21.2 m²
  3. 35.3 m²
  4. 56.5 m²
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`C`

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`text(Area of annulus)`

`= pi (R^2-r^2)`

`= pi (4.5^2-3^2)`

`= pi (11.25)`

`=35.3\ text{m²  (1 d.p.)}`
 

`=>  C`

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`

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`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`

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`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`