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Properties of Geometric Figures, SM-Bank 007

In the diagram \(AB\) is a straight line.

Calculate the size of the angle marked \(x^{\circ}\), giving reasons for your answer.    (3 marks)

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\(\text{Equilateral triangle}\ \ \Rightarrow\ \ \text{all angles}\ = 60^{\circ}\)

\(\text{Vertically opposite angles of 60° are equal (see diagram)}\)

\(x^{\circ} = 180-(90+60) = 30^{\circ}\ \ \text{(180° in straight line)}\)

Show Worked Solution

\(\text{Equilateral triangle}\ \ \Rightarrow\ \ \text{all angles}\ = 60^{\circ}\)

\(\text{Vertically opposite angles of 60° are equal (see diagram)}\)

\(x^{\circ} = 180-(90+60) = 30^{\circ}\ \ \text{(180° in straight line)}\)

Properties of Geometric Figures, SM-Bank 006

Pablo creates a design that is made up of 3 rectangles and 2 straight lines, as shown below.
 

What is the size of angle \(x^{\circ}\)?   (3 marks)

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\(\text{135 degrees}\)

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\(\text{Isosceles triangle}\ \ \Rightarrow\ \ \text{angles opposite equal sides are equal} \)

\(\text{Since there is 180° in a  straight line:}\)

\(x + 45\) \(= 180\)
\(x^{\circ}\) \(= 135^{\circ}\)

Solving Problems, SM-Bank 029

What is the value of \(x^{\circ}\) in this diagram?   (2 marks)

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\(54^{\circ}\)

Show Worked Solution

\(\text{Adjacent angle to 144°}\ = 180-144=36^{\circ}\ \ \text{(180° in straight line)}\)

\(x^{\circ}= 180-(90 + 36)=54^{\circ}\ \ \text{(180° in straight line)}\)

Properties of Geometrical Figures, SM-Bank 004 MC

A triangle is divided into 2 parts by a straight line.

The angles are then labelled.
 

Which statement is true about the sum of angles?

  1. `b + c + d = 180^@`
  2. `c + d + e = 360^@`
  3. `a + b + f + g = 360^@`
  4. `d + e + f + g = 180^@`
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`C`

Show Worked Solution

`text(Consider each option:)`

`text(Option A:)\ \ b + c + d != 180\ \ => \ b+c = 180^@`

`text(Option B:)\ \ c + d + e != 360^@\ \ => \ c + d + e = 180^@\ \ text{(angle sum of triangle)}`

`text(Option C:)\ \ a + b + f + g = 360^@`

  `=>\ text(Correct since the angle sum of a quadrilateral = 360°)`

`text(Option D:)\ \ d + e + f + g != 180\ \ => \ e + f  = 180^@`

`=> C`

Properties of Geometrical Figures, SM-Bank 003 MC

Tom drew this shape on grid paper.

Which one of the shapes below when joined to Tom's shape without an overlap, will not make isosceles triangle?

A.  
 B.  
 C.  
 D.  
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\(C\)

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\(\text{An isosceles triangle has two sides of the same length.}\)

\(\text{Option C will form a scalene triangle (all sides different lengths).}\)

\(\Rightarrow C\)

Unit Conversion, SM-Bank 023

  1. By considering the dimensions of a square with an area of 1 square kilometre, complete the unit conversion equation below:

1 km² = ________ m  ×  ________ m = ______________ km²   (1 mark)

  1. The Barangaroo precinct in Sydney has a total floor space of 681 000 square metres.
  2. Find the area of this floor space in square kilometres.  (1 mark)
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a.    \(1\ \text{km}^{2} = 1000\ \text{m}\ \times\ 1000\ \text{m}\ = 1\ 000\ 000\ \text{km}^{2} \)

b.    \(0.681\ \text{km}^{2}\)

Show Worked Solution

a.    \(1\ \text{km}^{2} = 1000\ \text{m}\ \times\ 1000\ \text{m}\ = 1\ 000\ 000\ \text{km}^{2} \)
 

b.    \(\text{Area}\) \(= \dfrac{681\ 000}{1\ 000\ 000}\)  
  \(=0.681\ \text{km}^{2}\)  

Unit Conversion, SM-Bank 022

  1. By considering the dimensions of a square with an area of 1 square kilometre, complete the unit conversion equation below:

1 km² = _______ m  ×  _______ m = ____________ km²   (1 mark)

  1. A rural property is in the shape of a rectangle with sides of length 1.20 kilometres and 2.36 kilometres.
  2. Find the area of the property in square kilometres and using the conversion equation, express the area in square metres.  (2 marks)
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a.    \(1\ \text{km}^{2} = 1000\ \text{m}\ \times\ 1000\ \text{m}\ = 1\ 000\ 000\ \text{km}^{2} \)

b.    \(2\ 832\ 000\ \text{m}^{2}\)

Show Worked Solution

a.    \(1\ \text{km}^{2} = 1000\ \text{m}\ \times\ 1000\ \text{m}\ = 1\ 000\ 000\ \text{km}^{2} \)
 

b.    \(\text{Area}\) \(= 1.20 \times 2.36\)  
  \(=2.832\ \text{km}^{2}\)  
  \(=2.832 \times 1\ 000\ 000\  \text{m}^{2}\)  
  \(=2\ 832\ 000\ \text{m}^{2}\)  

Unit Conversion, SM-Bank 021

The square below has an area of 1 square metre.
 

  1. Complete the unit conversion equation below:

1 m² = _____ cm  ×  _____ cm = __________ cm²   (1 mark)

  1. A circular poster has an area of 0.09\(\pi\) square metres.
  2. Express this area in square centimetres, giving your answer in exact form.  (1 mark)
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a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

b.    \(900\pi\ \text{cm}^{2}\)

Show Worked Solution

a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)
 

b.    \(\text{0.09}\pi\ \text{m}^{2}\) \(= 0.09\pi \times 10\ 000 \)  
  \(=900\pi\ \text{cm}^2 \)

Unit Conversion, SM-Bank 020

The square below has an area of 1 square metre.
 

  1. Complete the unit conversion equation below:

1 m² = _____ cm  ×  _____ cm = __________ cm²   (1 mark)

  1. A circular stage has an area of 265 000 square centimetres.
  2. Express this area in square metres.  (1 mark)
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a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

b.    \(26\ \text{m}^{2}\)

Show Worked Solution

a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)
 

b.    \(\text{265 000 cm}^{2}\) \(= \dfrac{26\ 000}{10\ 000}\)  
  \(=26\ \text{m}^{2}\)  

Unit Conversion, SM-Bank 019

The square below has an area of 1 square metre.
 

  1. Complete the unit conversion equation below:

1 m² = _____ cm  ×  _____ cm = __________ cm²   (1 mark)

  1. A rectangular serviette has an area of 0.075 square metres.
  2. Express this area in square centimetres.  (1 mark)
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a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

b.    \(750\ \text{cm}^{2}\)

Show Worked Solution

a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)
 

b.    \(\text{0.075 m}^{2}\) \(=0.075 \times 10\ 000\)  
  \(=750\ \text{cm}^{2}\)  

Unit Conversion, SM-Bank 018

The square below has an area of 1 square metre.
 

  1. Complete the unit conversion equation below:

1 m² = _____ cm  ×  _____ cm = __________ cm²   (1 mark)

  1. A doubles tennis court has an area of 260 square metres.
  2. Express this area in square centimetres.  (1 mark)
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a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

b.    \(2\ 600\ 000\ \text{cm}^{2}\)

Show Worked Solution

a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)
 

b.    \(\text{260 m}^{2}\) \(=260 \times 10\ 000\)  
  \(=2\ 600\ 000\ \text{cm}^{2}\)  

Unit Conversion, SM-Bank 017

  1. By considering the dimensions of a square with an area of 1 square metre, complete the unit conversion equation below:

1 m² = _____ cm  ×  _____ cm = __________ cm²   (1 mark)

  1. A circular elevator has a diameter of 1.6 metres.
  2. Find the area of the elevator's floor in square metres and using the conversion equation, express the area to the nearest square centimetre.  (2 marks)
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a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

b.    \(20\ 106\ \text{cm}^{2}\)

Show Worked Solution

a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)
 

b.    \(\text{Radius}\ =\dfrac{1.6}{2} = 0.8\ \text{m} \)

\(\text{Area}\) \(=\pi \times 0.8^2\)  
  \(=2.01061…\ \text{m}^{2}\)  
  \(=2.01061… \times 10\ 000\ \text{cm}^{2}\)  
  \(=20\ 106\ \text{cm}^{2}\)  

Unit Conversion, SM-Bank 016

The square below has an area of 1 square metre.
 

  1. Complete the unit conversion equation below:

1 m² = _____ cm  ×  _____ cm = _______ cm²   (1 mark)

  1. A circular playground has an area of 2000 square metres.
  2. Using the conversion equation, or otherwise, express the area in square centimetres.  (1 mark)
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a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

b.    \(20\ 000\ 000\ \text{mm}^{2}\)

Show Worked Solution

a.    \(1\ \text{m}^{2} = 100\ \text{cm}\ \times\ 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)
 

b.    \(2000\ \text{m}^{2}\) \(= 2000 \times 10\ 000\ \text{cm}^{2} \)
  \(= 20\ 000\ 000\ \text{cm}^{2} \)

Unit Conversion, SM-Bank 015

The square below has an area of 1 square centimetre.
 

  1. Complete the unit conversion equation below:

1 cm² = _____ mm  ×  _____ mm = _______ mm²   (1 mark)

  1. A rectangle has an area of 1100 square millimetres.
  2. Using the conversion equation, or otherwise, express the area of the circle in square centimetres.  (1 mark)
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a.    \(1\ \text{cm}^{2} = 10\ \text{mm}\ \times\ 10\ \text{mm}\ = 100\ \text{mm}^{2} \)

b.    \(11\ \text{cm}^{2}\)

Show Worked Solution

a.    \(1\ \text{cm}^{2} = 10\ \text{mm}\ \times\ 10\ \text{mm}\ = 100\ \text{mm}^{2} \)
 

b.    \(1100\ \text{mm}^{2}\) \(= \dfrac{1100}{100} \)
  \(=11\ \text{cm}^{2} \)

Unit Conversion, SM-Bank 014

  1. By considering the dimensions of a square with an area of 1 square centimetre, complete the unit conversion equation below:

1 cm² = _____ mm  ×  _____ mm = _______ mm²   (1 mark)

  1. A circle has an area of 950 square centimetres.
  2. Using the conversion equation, or otherwise, express the area of the circle in square millimetres.  (1 mark)
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a.    \(1\ \text{cm}^{2} = 10\ \text{mm}\ \times\ 10\ \text{mm}\ = 100\ \text{mm}^{2} \)

b.    \(95\ 000\ \text{mm}^{2}\)

Show Worked Solution

a.    \(1\ \text{cm}^{2} = 10\ \text{mm}\ \times\ 10\ \text{mm}\ = 100\ \text{mm}^{2} \)
 

b.    \(900\ \text{cm}^{2}\) \(= 900 \times 100\ \text{mm}^{2} \)
  \(= 90\ 000\ \text{mm}^{2} \)

Solving Problems, SM-Bank 022

In the diagram below, \(DG\) is parallel to \(BC\), and \(\angle ABC = 115^{\circ} \).
  

Find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)

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\(\angle CBE = 180-115=65^{\circ}\ \ \text{(180° in a straight line)}\)

\(x^{\circ} = 65^{\circ}\ \ \text{(alternate angles)}\)

Show Worked Solution

\(\angle CBE = 180-115=65^{\circ}\ \ \text{(180° in a straight line)}\)

\(x^{\circ} = 65^{\circ}\ \ \text{(alternate angles)}\)

Solving Problems, SM-Bank 021

In the diagram below, \(BE\) is parallel to \(CD\), and \(\angle ABE = 160^{\circ} \).
  

Find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)

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\(\angle DBE = 180-160=20^{\circ}\ \ \text{(180° in a straight line)}\)

\(180^{\circ}\) \(=x+20+110\ \ \text{(cointerior angles)} \)  
\(x^{\circ}\) \(=180-130\)  
  \(=50^{\circ}\)  
Show Worked Solution

\(\angle DBE = 180-160=20^{\circ}\ \ \text{(180° in a straight line)}\)

\(180^{\circ}\) \(=x+20+110\ \ \text{(cointerior angles)} \)  
\(x^{\circ}\) \(=180-130\)  
  \(=50^{\circ}\)  

Solving Problems, SM-Bank 020

In the diagram below, \(QR\) is parallel to \(SU\).
  

Find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)

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\(\angle STP = 38^{\circ}\ \ \text{(corresponding angles)}\)

\((x+30)^{\circ}\) \(=180-38\ \ \text{(180° in straight line)} \)  
\(x^{\circ}\) \(=142-30\)  
  \(=112^{\circ}\)  
Show Worked Solution

\(\angle STP = 38^{\circ}\ \ \text{(corresponding angles)}\)

\((x+30)^{\circ}\) \(=180-38\ \ \text{(180° in straight line)} \)  
\(x^{\circ}\) \(=142-30\)  
  \(=112^{\circ}\)  

Solving Problems, SM-Bank 019

In the diagram below, \(PR\) is parallel to \(TU\) and reflex \(\angle QST = 255^{\circ}\)
  

Find the value of \(x^{\circ}\), giving reasons for your answer.   (3 marks)

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\(\angle QST = 360-255 = 105^{\circ}\ \ \text{(360° about a point)}\)

\(\angle VSQ =70^{\circ} \ \ \text{(alternate angles)} \)

\(\angle VST\ =x^{\circ} \ \ \text{(alternate angles)} \)

\(x^{\circ}\) \(=105-70\)  
  \(=35^{\circ}\)  
Show Worked Solution

\(\text{Add middle parallel line:}\)
 

\(\angle QST = 360-255 = 105^{\circ}\ \ \text{(360° about a point)}\)

\(\angle VSQ =70^{\circ} \ \ \text{(alternate angles)} \)

\(\angle VST\ =x^{\circ} \ \ \text{(alternate angles)} \)

\(x^{\circ}\) \(=105-70 \)  
  \(=35^{\circ}\)  

Solving Problems, SM-Bank 018

In the diagram below, find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)
  

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\(y^{\circ}\ =70^{\circ} \ \ \text{(alternate angles)} \)

\(x^{\circ}\ =z^{\circ} \ \ \text{(alternate angles)} \)

\((z+y)^{\circ}\) \(=110^{\circ}\ \)  
\((x+y)^{\circ}\) \(=110^{\circ}\ \)  
\(x^{\circ}\) \(=110-70\)  
  \(=40^{\circ}\)  
Show Worked Solution

\(\text{Extend middle parallel line:}\)
 

\(y^{\circ}\ =70^{\circ} \ \ \text{(alternate angles)} \)

\(x^{\circ}\ =z^{\circ} \ \ \text{(alternate angles)} \)

\((z+y)^{\circ}\) \(=110^{\circ}\ \)  
\((x+y)^{\circ}\) \(=110^{\circ}\ \)  
\(x^{\circ}\) \(=110-70\)  
  \(=40^{\circ}\)  

Solving Problems, SM-Bank 017

In the diagram below, find the value of \(x^{\circ}\), giving reasons for your answer.   (3 marks)
  

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\(\text{Full interior angle}\ = 360-275=85^{\circ} \ \ \text{(360° about a point)} \)
 

\(\text{Since cointerior angles sum to 180°,}\)

\(\Rightarrow \text{interior angle (1)}\ = 180-125=55^{\circ} \)

\(\text{Since angles about a point sum to 360°,}\)

\(\Rightarrow \text{interior angle (2)}\ = 85-55=30^{\circ} \)
 

\(x^{\circ}\) \(=180-30\ \ \text{(cointerior angles)} \)  
  \(=150^{\circ}\)  
Show Worked Solution

\(\text{Add parallel line:}\)
 

\(\text{Full interior angle}\ = 360-275=85^{\circ} \ \ \text{(360° about a point)} \)
 

\(\text{Since cointerior angles sum to 180°,}\)

\(\Rightarrow \text{interior angle (1)}\ = 180-125=55^{\circ} \)

\(\text{Since angles about a point sum to 360°,}\)

\(\Rightarrow \text{interior angle (2)}\ = 85-55=30^{\circ} \)
 

\(x^{\circ}\) \(=180-30\ \ \text{(cointerior angles)} \)  
  \(=150^{\circ}\)  

Solving Problems, SM-Bank 016

In the diagram below, find the value of \(a^{\circ}\), giving reasons for your answer.   (2 marks)
  

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\(46^{\circ}\)

Show Worked Solution

\(\text{Since cointerior angles sum to 180°:}\)

\(180^{\circ}\) \(=a+60+74\)  
\(a^{\circ}\) \(=180-134\)  
  \(=46^{\circ}\)  

Solving Problems, SM-Bank 015

In the diagram below, find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)
  

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\(45^{\circ}\)

Show Worked Solution

\(\text{Since cointerior angles sum to 180°:}\)

\(180^{\circ}\) \(=x+70+65\)  
\(x^{\circ}\) \(=180-135\)  
  \(=45^{\circ}\)  

Solving Problems, SM-Bank 014

In the diagram below, find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)
  

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\(\angle y^{\circ} = 40^{\circ}\ \ \text{(alternate angles)} \)

\(\angle x^{\circ}=360=40 = 320^{\circ}\ \ \text{(360° about a point)}\)

Show Worked Solution

\(\angle y^{\circ} = 40^{\circ}\ \ \text{(alternate angles)} \)

\(\angle x^{\circ}=360=40 = 320^{\circ}\ \ \text{(360° about a point)}\)

Solving Problems, SM-Bank 013

In the diagram below, find the value of \(a^{\circ}\), giving reasons for your answer.   (2 marks)
  

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\(\angle b^{\circ} = 360-325 = 35^{\circ}\ \ \text{(360° about a point)} \)

\(\angle a^{\circ}=35^{\circ}\ \ \text{(alternate angles)}\)

Show Worked Solution

\(\angle b^{\circ} = 360-325 = 35^{\circ}\ \ \text{(360° about a point)} \)

\(\angle a^{\circ}=35^{\circ}\ \ \text{(alternate angles)}\)

Solving Problems, SM-Bank 012

In the diagram below, find the value of \(x^{\circ}\), giving reasons for your answer.   (2 marks)
  

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\(\angle y^{\circ} = 180-(52+90) = 38^{\circ}\ \ \text{(180° in straight line)} \)

\(\angle x^{\circ}=38^{\circ}\ \ \text{(alternate angles)}\)

Show Worked Solution

\(\angle y^{\circ} = 180-(52+90) = 38^{\circ}\ \ \text{(180° in straight line)} \)

\(\angle x^{\circ}=38^{\circ}\ \ \text{(alternate angles)}\)

Solving Problems, SM-Bank 011

In the diagram below, \(QR\) is parallel to lines \(SU\) and \(VW\).
 

Find the value of \(x^{\circ}\), giving reasons for your answer.   (3 marks)

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\(\angle UTQ = 125^{\circ}\ \ \text{(corresponding angles)} \)

\(\angle STX=125^{\circ}\ \ \text{(vertically opposite angles)}\)

\(x^{\circ} = 180-125=55^{\circ} \ \ \text{(cointerior angles)}\)

Show Worked Solution

\(\angle UTQ = 125^{\circ}\ \ \text{(corresponding angles)} \)

\(\angle STX=125^{\circ}\ \ \text{(vertically opposite angles)}\)

\(x^{\circ} = 180-125=55^{\circ} \ \ \text{(cointerior angles)}\)

Solving Problems, SM-Bank 010

In the diagram below, \(BC\) is parallel to \(DE\) and \(\angle ACB\) is a right-angle.
 

Find the value of \(x^{\circ}\), giving reasons for your answer.   (3 marks)

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\(\text{Extend line}\ BC: \)
 

\(\angle GCF=180-120=60^{\circ}\ \ \text{(180° in a straight line)}\)

\(x^{\circ} = 60^{\circ} \ \ \text{(corresponding angles)}\)

Show Worked Solution

\(\text{Extend line}\ BC: \)
 

\(\angle GCF=180-120=60^{\circ}\ \ \text{(180° in a straight line)}\)

\(x^{\circ} = 60^{\circ} \ \ \text{(corresponding angles)}\)

Solving Problems, SM-Bank 009

In the diagram below, two parallel lines \(OB\) and \(DC\) cut the horizontal transversal \(OE\), and \(OA\) is perpendicular to \(OE\).
 

Find the value of \(a^{\circ}\), giving reasons for your answer.   (2 marks)

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\(\angle BOE=90-20=70^{\circ}\ \ \text{(complementary angles)}\)

\(a^{\circ} = 70^{\circ} \ \ \text{(corresponding angles)}\)

Show Worked Solution

\(\angle BOE=90-20=70^{\circ}\ \ \text{(complementary angles)}\)

\(a^{\circ} = 70^{\circ} \ \ \text{(corresponding angles)}\)

Solving Problems, SM-Bank 007

Determine if two lines in the diagram below are parallel, giving reasons for your answer.   (2 marks)
 

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\(\angle \text{unknown} = 360-310=50^{\circ}\ \ \text{(360° about a point)}\)

\(\text{Since cointerior angles sum to 180°:}\)

\(140 + 50= 190^{\circ} \neq 180^{\circ}\)

\(\therefore \ \text{Lines are not parallel.}\)

Show Worked Solution

\(\angle \text{unknown} = 360-310=50^{\circ}\ \ \text{(360° about a point)}\)

\(\text{Since cointerior angles sum to 180°:}\)

\(140 + 50 = 190^{\circ} \neq 180^{\circ}\)

\(\therefore \ \text{Lines are not parallel.}\)

Solving Problems, SM-Bank 008

Find the value of \(x^{\circ}\) in the diagram below, giving reasons for your answer.   (2 marks)
 

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\(40^{\circ}\)

Show Worked Solution

\(\text{Since cointerior angles sum to 180°:}\)

\(180\) \(=x+65+75\)  
\(180\) \(=x+140\)  
\(x^{\circ}\) \(=180-40\)  
  \(=40^{\circ}\)  

Volume, SM-Bank 070

Convert 485 millilitres to litres.  (1 mark)

Show Answers Only

\(0.485\ \text{ L}\)

Show Worked Solution
\(1000\ \text{mL}\) \(=1\ \text{L}\)
\(\therefore\ 485\ \text{mL}\) \(=\Bigg(\dfrac{485}{1000}\Bigg)\text{ L}\)
  \(=0.485\ \text{ L}\)

Volume, SM-Bank 069

Convert 7350 millilitres to litres.  (1 mark)

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\(7.350\ \text{ L}\)

Show Worked Solution
\(1000\ \text{mL}\) \(=1\ \text{L}\)
\(\therefore\ 7350\ \text{mL}\) \(=\Bigg(\dfrac{7350}{1000}\Bigg)\text{ L}\)
  \(=7.350\ \text{ L}\)

Volume, SM-Bank 068

Convert 60 000 millilitres to litres.  (1 mark)

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\(60\ \text{L}\)

Show Worked Solution
\(1000\ \text{mL}\) \(=1\ \text{L}\)
\(\therefore\ 60\ 000\ \text{mL}\) \(=\Bigg(\dfrac{60\ 000}{1000}\Bigg)\text{ L}\)
  \(=60\ \text{ L}\)

Volume, SM-Bank 067

A triangular prism has a volume of 680 cubic centimetres and a height of 5 centimetres. What is the cross-sectional area?  (2 marks)

Show Answers Only

\(136\ \text{m}^2\)

Show Worked Solution
\(V\) \(=Ah\)
\(680\) \(=A\times 5\)
\(A\) \(=\dfrac{680}{5}\)
  \(=136\)

 
\(\therefore\ \text{The cross-sectional area is }136\ \text{m}^2.\)

Volume, SM-Bank 066

Gavin is going camping in the summer holidays and purchased the two-person tent shown below.

  1. Given the triangular face of the tent is isosceles, use Pythagoras' Theorem to calculate the perpendicular height of the tent.  (2 marks)
  2. Using your answer from (a), calculate the volume of the tent in cubic metres.  (2 marks)
  3. What is the capacity of the tent in litres?  (1 mark)
Show Answers Only

a.    \(2\ \text{m}\)

b.    \(12 \text{m}^3\)

c.    \(12\ 000\ \text{L}\)

Show Worked Solution
a.    \(a^2+b^2\) \(=c^2\)
  \(x^2+1.5^2\) \(=2.5^2\)
  \(x^2\) \(=2.5^2-1.5^2\)
  \(x^2\) \(=4\)
  \(x\) \(=\sqrt{4}=2\)

 
\(\therefore\ \text{The perpendicular height of the tent is }2\ \text{metres.}\)
 

b.    \(V\) \(=Ah\)
    \(=\Big(\dfrac{1}{2}\times 3\times 2\Big)\times 4\)
    \(=3\times 4\)
    \(=12\ \text{m}^3\)

 

c.    \(1\ \text{m}^3\) \(=1000\ \text{L}\)
  \(\therefore\ 12\ \text{m}^3\) \(=12\ 000\ \text{L}\)

  
\(\therefore\ \text{The capacity of the tent is }12\ 000\ \text{litres.}\)

Volume, SM-Bank 065

  1. Given the triangular face of the prism above is isosceles, use Pythagoras' Theorem to calculate its perpendicular height. Give your answer correct to the nearest whole centimetre.  (2 marks)
  2. Using your answer from (a), calculate the volume of the prism in cubic centimetres.  (2 marks)
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a.    \(7\ \text{cm}\)

b.    \(168\ \text{cm}^3\)

Show Worked Solution
a.    \(a^2+b^2\) \(=c^2\)
  \(a^2+3^2\) \(=7.6^2\)
  \(a^2\) \(=7.6^2-3^2\)
  \(a^2\) \(=48.76\)
  \(a\) \(=\sqrt{48.76}=6.982\dots\)
  \(a\) \(\approx 7\)

 
\(\therefore\ \text{The perpendicular height of the triangle is }7\ \text{cm, (nearest whole centimetre).}\)
 

b.    \(V\) \(=Ah\)
    \(=\Big(\dfrac{1}{2}\times 6\times 7\Big)\times 8\)
    \(=21\times 8\)
    \(=168\ \text{cm}^3\)

Volume, SM-Bank 064

  1. For the triangular prism above, use Pythagoras' Theorem to calculate the perpendicular height, \(x\), of the triangular face.  (2 marks)
  2. Using your answer from (a), calculate the volume of the prism in cubic millimetres.  (2 marks)
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a.    \(5\ \text{mm}\)

b.    \(480\ \text{mm}^3\)

Show Worked Solution
a.    \(a^2+b^2\) \(=c^2\)
  \(x^2+12^2\) \(=13^2\)
  \(x^2\) \(=13^2-12^2\)
  \(x^2\) \(=25\)
  \(x\) \(=\sqrt{25}=5\)

 
\(\therefore\ \text{The perpendicular height of the triangle is }5\ \text{mm}\)
 

b.    \(V\) \(=Ah\)
    \(=\Big(\dfrac{1}{2}\times 12\times 5\Big)\times 16\)
    \(=30\times 16\)
    \(=480\ \text{mm}^3\)

Volume, SM-Bank 063

Calculate the volume of the triangular prism below in cubic metres.  (2 marks)

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\(648\ \text{m}^3\)

Show Worked Solution
\(V\) \(=Ah\)
  \(=\Big(\dfrac{1}{2}\times 8\times 9\Big)\times 18\)
  \(=36\times 18\)
  \(=648\ \text{m}^3\)

Volume, SM-Bank 062

Calculate the volume of the triangular prism below in cubic centimetres.   (2 marks) 
 

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\(102.362\ \text{cm}^3\)

Show Worked Solution
\(V\) \(=Ah\)
  \(=\Big(\dfrac{1}{2}\times 3.1\times 5.2\Big)\times 12.7\)
  \(=8.06\times 12.7\)
  \(=102.362\ \text{cm}^3\)

Volume, SM-Bank 061

Calculate the volume of the triangular prism below in cubic metres.  (2 marks)

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\(105\ \text{m}^3\)

Show Worked Solution
\(V\) \(=Ah\)
  \(=\Big(\dfrac{1}{2}\times 6\times 5\Big)\times 7\)
  \(=15\times 7\)
  \(=105\ \text{m}^3\)

Volume, SM-Bank 060

A children's rectangular swimming pool measures 175 cm × 180 cm × 30 cm.

  1. Find the volume of the swimming pool in cubic centimetres.  (2 marks)

  2. What is the capacity of the swimming pool in litres?  (1 mark)

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a.    \(945\ 000\ \text{cm}^3\)

b.    \(945\ \text{L}\)

Show Worked Solution
a.     \(V\) \(=Ah\)  
    \(=(175\times 180)\times 30\)
    \(=945\ 000\ \text{cm}^3\)

 
b.    \(1\ \text{L}=1000\ \text{cm}^3\)

\(945\ 000\ \text{cm}^3\) \(=\Bigg(\dfrac{945\ 000}{1000}\Bigg)\ \text{L}\)
  \(=945\ \text{L}\)

 
\(\therefore\ \text{The swimming pool will hold 945 litres of water.}\)

Volume, SM-Bank 059

During the construction of a new house a concrete slab in the shape of a rectangular prism is to be poured.

The slab measures 20 m × 15 m × 0.15 m.

  1. Find the volume of the concrete required for the slab in cubic metres.  (2 marks)

  2. Calculate the cost of the concrete if it costs $350 per cubic metre.  (2 marks)

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a.    \(45\ \text{m}^3\)

b.    \($15\ 750\)

Show Worked Solution
a.     \(V\) \(=Ah\)  
  \(V\) \(=(20\times 15)\times 0.15\)
    \(=45\ \text{m}^3\)

 

b.    \(\text{Cost}\) \(=\text{Price}\times \text{Concrete}\)
    \(=$350\times 45\)
    \(=$15\ 750\)

Volume, SM-Bank 058

A rectangular sand pit measures 150 cm × 200 cm × 45 cm.

  1. Find the volume of the sand pit in cubic centimetres.  (2 marks)

  2. How many cubic metres of sand will the sand pit hold?  (1 mark)

Show Answers Only

a.    \(1\ 350\ 000\ \text{cm}^3\)

b.    \(1.35\ \text{m}^3\)

Show Worked Solution
a.     \(V\) \(=Ah\)  
  \(V\) \(=(150\times 200)\times 45\)
    \(=1\ 350\ 000\ \text{cm}^3\)

 
b.    \(1\ \text{m}=100\ \text{cm}\)

\(\therefore\ 1\ \text{m}^3=(100\times 100\times 100)\ \text{cm}^3\)

\(\therefore\ 1\ 350\ 000\ \text{cm}^3\) \(=\Bigg(\dfrac{1\ 350\ 000}{100\times 100\times 100}\Bigg)\ \text{m}^3\)
  \(=1.35\ \text{m}^3\)

 
\(\therefore\ \text{The sandpit will hold }1.35\ \text{cubic metres of sand.}\)

Volume, SM-Bank 057

Calculate the volume of the rectangular prism below in cubic millimetres.   (2 marks)
 

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\(104\ 832\ \text{mm}^3\)

Show Worked Solution
\(V\) \(=Ah\)
\(V\) \(=(39\times 42)\times 64\)
  \(=104\ 832\ \text{mm}^3\)

Volume, SM-Bank 056

Calculate the volume of the rectangular prism below in cubic centimetres.  (2 marks)
 

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\(402.57\ \text{cm}^3\)

Show Worked Solution
\(V\) \(=Ah\)
\(V\) \(=(4.2\times 13.5)\times 7.1\)
  \(=402.57\ \text{cm}^3\)

Volume, SM-Bank 055

Calculate the volume of the rectangular prism below in cubic metres.  (2 marks)
 

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\(360\ \text{m}^3\)

Show Worked Solution
\(V\) \(=Ah\)
\(V\) \(=(5\times 9)\times 8\)
  \(=360\ \text{m}^3\)

Volume, SM-Bank 054

The prism below has a height of 100 metres.

Given its volume is 485 cubic metres, calculate the cross-sectional area, \((A)\), of the prism.  (2 marks)
 

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\(4.85\ \text{m}^2\)

Show Worked Solution

\(V=485\ \text{m}^3\ \text{and }h=100\ \text{m}\)

\(V\) \(=Ah\)
\(485\) \(=100A\)
\(A\) \(=\dfrac{485}{100}\)
\(A\) \(=4.85\ \text{m}^2\)

Volume, SM-Bank 053

The prism below has a height of 23 centimetres.

Given its volume is 391 cubic centimetres, calculate the cross-sectional area, \((A)\), of the prism.  (2 marks)
 

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\(17\ \text{cm}^2\)

Show Worked Solution

\(V=391\ \text{cm}^3\ \text{and }h=23\ \text{cm}\)

\(V\) \(=Ah\)
\(391\) \(=23A\)
\(A\) \(=\dfrac{391}{23}\)
\(A\) \(=17\ \text{cm}^2\)

Volume, SM-Bank 052

The prism below has a cross-sectional area of 15 square millimetres.

Given its volume is 184.5 cubic millimetres, calculate the height, \((h)\), of the prism.  (2 marks)
 

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\(12.3\ \text{mm}\)

Show Worked Solution

\(V=184.5\ \text{mm}^3\ \text{and }A=15\ \text{mm}^2\)

\(V\) \(=Ah\)
\(184.5\) \(=15h\)
\(h\) \(=\dfrac{184.5}{15}\)
\(h\) \(=12.3\ \text{mm}\)

Volume, SM-Bank 051

The prism below has a cross-sectional area of 50 square centimetres.

Its volume is 425 cubic centimetres.
 

Calculate the height, \((h)\), of the prism.  (2 marks)

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\(8.5\ \text{cm}\)

Show Worked Solution

\(V=425\ \text{cm}^3\ \text{and }A=50\ \text{cm}^2\)

\(V\) \(=Ah\)
\(425\) \(=50h\)
\(h\) \(=\dfrac{425}{50}\)
\(h\) \(=8.5\ \text{cm}\)