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