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Area, SMB-009

Find the surface area of the rectangular prism pictured below, correct to 1 decimal place. All measurements are in metres.  (2 marks)
 

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\(68.0\ \text{m}^2\)

Show Worked Solution
\(\text{S.A.}\) \(=2 \times (2.8 \times 1.7) + 2 \times (2.8 \times 6.5) + 2 \times (1.7 \times 6.5) \)  
  \(= 2(4.76 + 18.2 + 11.05) \)  
  \(=68.02\)  
  \(=68.0\ \text{m}^2\ \ \text{(1 d.p.)}\)  

Area, SMB-008

Find the surface area of the rectangular prism pictured below. All measurements are in metres.  (2 marks)
 

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\(368\ \text{m}^2\)

Show Worked Solution
\(\text{S.A.}\) \(=2 \times (6 \times 14) + 2 \times (6 \times 5) + 2 \times (5 \times 14) \)  
  \(=2(84+30+70)\)  
  \(=368\ \text{m}^2\)  

Area, SMB-007

The diagram shows the shape of Carmel’s garden bed. All measurements are in metres.
  

Show that the area of the garden bed is 57 square metres.  (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
`text(Area of)\ Delta ABC` `= 1/2 xx b xx h`
  `= 1/2 xx 10 xx 5.1`
  `= 25.5\ text(m²)`
`text(Area of)\ Delta ACD` `= 1/2 xx 10 xx 6.3`
  `= 31.5\ text(m²)`

 

`:.\ text(Total Area)` `= 25.5 + 31.5`
  `= 57\ text(m² … as required)`

Area, SMB-001

Find the area of the shaded part in the diagram below, giving your answer in square metres to 1 decimal place.  (3 marks)
 

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\(62.8\ \text{m}^2 \)

Show Worked Solution
\(\text{Shaded area}\) \(=\ \text{Area of large sector – Area of small sector}\)  
  \(= \dfrac{30}{360} \times \pi R^2-\dfrac{30}{360} \times \pi r^2 \)  
  \(= \dfrac{1}{12} \pi (16^2-4^2) \)  
  \(= 62.83… \)  
  \(=62.8\ \text{m}^2\ \ \text{(1 d.p.)} \)  

Area, SMB-008 MC

Four identical circles of radius `r` are drawn inside a square, as shown in the diagram below

The region enclosed by the circles has been shaded in the diagram.
 

 
The shaded area can be found using

  1. `4 r^2-2 pi r`
  2. `4 r^2-pi r^2`
  3. `4 r-pi r^2`
  4. `2 r^2-pi r^2`
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`B`

Show Worked Solution

`text(Area of square) = 4r xx 4r = 16r^2`

`text(Area of circles) = 4 pi r^2`

`text(Shaded Area)`

`= 1/4 xx (16r^2-4 pi r^2)`

`= 4 r^2-pi r^2`
 

`=>  B`

Area, SMB-006

A child's toy has the following design.
 

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

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`27\ text(cm²)`

Show Worked Solution

`text{Circle radius = 3 cm}`

`text(Consider the rectangle starting from the middle of the left circle.)`

`text{Shaded Area}` `= text{Area rectangle}-3.5 xx text{Area circle}`
  `= (21 xx 6)-3.5 xx pi xx 3^2`
  `= 27.03 …`
  `=27\ text{cm²  (nearest whole)}`

Area, SMB-005

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

Find the area of the path, correct to the nearest square metre.   (2 marks)

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`35\ text{m²}`

Show Worked Solution

`text(Area of annulus)`

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

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

`= pi (11.25)`

`=35.3…`

`=35\ text{m²  (nearest m²)}`

Area, SMB-004

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

Find the area of the shower floor, excluding the drain, to the nearest square centimetre.   (3 marks)
 

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`9921\ text(cm²)`

Show Worked Solution
`text(Area)` `=\ text(Square – Circle)`
  `= (100 xx 100)-(pi xx 5^2)`
  `= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm²)`
  `= 9921\ text{cm² (nearest cm²)}`

Area, SMB-003

The shaded region shows a quadrant with a rectangle removed.
  

Find the area of the shaded region, to the nearest cm2.   (2 marks)

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`52\ text(cm²)`

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…`
  `=52\ text(cm²)`

Area, SMB-002

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  (3 marks)

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`42\ text{cm²}`

Show Worked Solution
`text(Area)` `=\ text(Area of Sector – Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx l xx h)`
  `= (90/360 xx pi xx 8^2)-(1/2 xx 4 xx 4)`
  `= 50.2654…-8`
  `= 42.265…\ text(cm²)`
  `= 42\ text{cm²  (nearest cm²)}`

Trigonometry, SMB-031

  1. Express the ratio of `tan theta`  as a fraction.  (2 marks)

  2. Hence or otherwise, find the value of `theta`, correct to two decimal places.  (1 mark)

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  1. `tan theta=15/8`
  2. `61.93^@`
Show Worked Solution

i.    `text{By Pythagoras,}`

`x^2+8^2` `=17^2`  
`x^2` `=289-64`  
  `=225`  
`x` `=15`  

 
`:.tan theta = 15/8`

 

ii.   `theta` `=tan^{-1}(15/8)`  
  `=61.927…^@`  
  `=61.93^@\ text{(to 2 d.p.)}`  

Trigonometry, SMB-030

Find the value of `theta`, correct to the nearest minute.  (2 marks)

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`54^@ 28^{′}`

Show Worked Solution
`tan theta` `=7/5`  
`theta` `=tan^{-1}(7/5)`  
  `=54.462…`  
  `=54^@ 27^{′}44^{″}`  
  `=54^@ 28^{′}`  
NOTE: Seconds > 30 are rounded up to the higher minute.

Trigonometry, SMB-029

Find the value of `theta`, correct to the nearest minute.  (2 marks)

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`61^@ 56^{′}`

Show Worked Solution
`tan theta` `=15/8`  
`theta` `=tan^{-1}(15/8)`  
  `=61.927…`  
  `=61^@55^{′}39^{″}`  
  `=61^@ 56^{′}`  
NOTE: Seconds > 30 round up to the higher minute.

Trigonometry, SMB-024

Express `tan theta` as a fraction in its simplest form.  (3 marks)

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`tan theta = (2sqrt5)/5`

Show Worked Solution

`text{Let unknown side =}\ x`

`text{By Pythagoras:}`

`6^2` `=x^2 + 4^2`  
`x^2` `=36-16`  
`x` `=sqrt20\ \ (x>0)`  

 

`:.tan theta` `=4/sqrt20`  
  `= 4/(2sqrt5) xx sqrt5/sqrt5`  
  `= (2sqrt5)/5`  

Trigonometry, SMB-023

Using Pythagoras and showing your working, express `tan 30^{@}` as a fraction.  (2 marks)

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`tan 30^{@} = 1/sqrt3`

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`text{Let}\ \ x =\ text{unknown side}`

`text{By Pythagoras:}`

`2^2` `=x^2 + 1^2`  
`x^2` `=4-1`  
`x` `=sqrt3\ \ (x>0)`  

 
`tan 30^{@} = text{opp}/text{adj} = 1/sqrt3`

Trigonometry, SMB-022

Find the value of `theta`, correct to the nearest minute.  (3 marks)

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`66^@ 25^{′}`

Show Worked Solution

`cos theta` `=2/5`  
`theta` `=cos^{-1}(2/5)`  
  `=66.421…^@`  
  `=66^@ 25^{′} text{(to 1 d.p.)}`  

Trigonometry, SMB-021

  1. Express the ratio of `cos theta`  in its simplest form.  (2 marks)

  2. Hence, find the value of `theta`, correct to two decimal places.  (1 mark)

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  1. `cos theta = 4/5`
  2. `theta = 36.87^@`
Show Worked Solution

i.    `text{By Pythagoras,}`

`x^2+8^2` `=10^2`  
`x^2` `=100-36`  
  `=64`  
`x` `=8`  

 
`:.cos theta = 8/10=4/5`

 

ii.   `theta` `=cos^{-1}(4/5)`  
  `=36.869…^@`  
  `=36.87^@\ text{(to 2 d.p.)}`  

Trigonometry, SMB-020


  

Find the value of `theta`, correct to the nearest minute.  (2 marks)

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`70^@ 32^{′}`

Show Worked Solution
`cos theta` `=4/12`  
`theta` `=cos^{-1}(1/3)`  
  `=70.528…`  
  `=70^@ 31^{′}43^{″}`  
  `=70^@ 32^{′}`  
NOTE: Seconds > 30 are rounded up to the higher minute.

Trigonometry, SMB-014

Express `cos theta` as a fraction in its simplest form.  (3 marks)

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`cos theta = sqrt3/2`

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`text{Let unknown side =}\ x`

`text{By Pythagoras:}`

`4^2` `=x^2 + 2^2`  
`x^2` `=16-4`  
`x` `=sqrt12\ \ (x>0)`  

 

`:.cos theta` `=sqrt12/4`  
  `= (2sqrt3)/4`  
  `= sqrt3/2`  

Trigonometry, SMB-013

Using Pythagoras and showing your working, express `cos 30^{@}` as a fraction.  (2 marks)

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`cos 30^{@} = sqrt3/2`

Show Worked Solution

`text{Let}\ \ x =\ text{unknown side}`

`text{By Pythagoras:}`

`2^2` `=x^2 + 1^2`  
`x^2` `=4-1`  
`x` `=sqrt3\ \ (x>0)`  

 
`cos 30^{@} = text{adj}/text{hyp} = sqrt3/2`

Trigonometry, SMB-004

Without the use of a calculator, express the following as a fraction

  1. `sin 30^{@}`   (1 mark)

  2. `sin 60^{@}`   (2 marks)

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  1. `sin 30^{@} = 1/2`
  2. `sin 60^{@} = sqrt3/2`
Show Worked Solution

i.    `sin 30^{@} = text{opp}/text{hyp} = 1/2`

ii.   `text{Let}\ \ x =\ text{unknown side}`

`text{By Pythagoras:}`

`2^2` `=x^2 + 1^2`  
`x^2` `=4-1`  
`x` `=sqrt3\ \ (x>0)`  

 
`sin 60^{@} = text{opp}/text{hyp} = sqrt3/2`

Trigonometry, SMB-003

Express the following as a fraction

  1. `sin theta`  (1 mark)

  2. `tan theta`  (1 mark)

Show Answers Only
  1. `sin theta = 7/25`
  2. `tan theta = 7/24`
Show Worked Solution

i.    `sin theta = text{opp}/text{hyp} = 7/25`

ii.   `tan theta = text{opp}/text{adj} = 7/24`

Trigonometry, SMB-002

Express the following as a simplified fraction

  1. `cos theta`  (1 mark)

  2. `tan theta`  (1 mark)

Show Answers Only
  1. `cos theta = 3/5`
  2. `tan theta = 4/3`
Show Worked Solution

i.    `cos theta = text{adj}/text{hyp} = 6/10 = 3/5`

ii.   `tan theta = text{opp}/text{adj} = 8/6 = 4/3`

Trigonometry, SMB-001

Express the following as a fraction

  1. `cos theta`  (1 mark)

  2. `sin theta`  (1 mark)

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  1. `cos theta = 12/13`
  2. `sin theta = 5/13`
Show Worked Solution

i.    `cos theta = text{adj}/text{hyp} = 12/13`

ii.   `sin theta = text{opp}/text{hyp} = 5/13`

Numbers of Any Magnitude, SMB-024 MC

The first three days of the Brisbane cricket test had the following attendances:

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Day 1}\rule[-1ex]{0pt}{0pt} & 20\ 156\\
\hline
\rule{0pt}{2.5ex}\text{Day 2}\rule[-1ex]{0pt}{0pt} & 18\ 397\\
\hline
\rule{0pt}{2.5ex}\text{Day 3}\rule[-1ex]{0pt}{0pt} & 29\ 981\\
\hline
\end{array}

What was the total crowd over the first 3 days, to the nearest 1000?

  1. `68\ 000`
  2. `69\ 000`
  3. `70\ 000`
  4. `71\ 000`
Show Answers Only

`B`

Show Worked Solution

`20\ 156 + 18\ 397 + 29\ 981`

`= 68\ 534`

`= 69\ 000\ text{(nearest 1000)}`
 

`=>B`

Numbers of Any Magnitude, SMB-023

A red blood cell and a white blood cell have a combined total mass of `4.2 xx 10^-11` grams.

If the red blood cell has a mass of `2 xx 10^-12` grams, what is the mass of the white blood cell?  (2 marks)

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`4.0 xx 10^-11\ text(grams)`

Show Worked Solution
`text{Mass}` `=4.2 xx 10^-11-2 xx 10^-12`  
  `= 4.0 xx 10^-11\ text{(by calculator)}`  

Numbers of Any Magnitude, SMB-021

An insect expert estimates that an anthill 30 cm high contains `10^8` ants.

He also estimates that a 25 cm high anthill contains `10^7` ants.

If his estimations are correct, how many more ants live in the 30 cm high anthill than the 25 cm high one?  (2 marks)

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`text(90 million)`

Show Worked Solution

`text(The extra ants in the 30 cm high anthill)`

`= 100\ 000\ 000-10\ 000\ 000`

`=90\ 000\ 000`

`=90\ text(million)`

Numbers of Any Magnitude, SMB-015

The width of red blood cell is 8 nanometres or `8 xx 10^{-9}` metres.

How many red blood cells, lined up side by side in straight line, would it take to form a 1 cm distance.  (2 marks)

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`1.25 xx 10^6`

Show Worked Solution

`text{Convert cm to metres:}`

`text{100 cm = 1 m}\ => \ text{1 cm = 0.01 m}`

`text{Number of cells}` `=0.01/(8xx10^{-9})`  
  `=1\ 250\ 000`  
  `=1.25 xx 10^6`  

Measurement, STD2 M1 2009 HSC 25b

The mass of a sample of microbes is 50 mg. There are approximately  `2.5 × 10^6` microbes in the sample.

In scientific notation, what is the approximate mass in grams of one microbe?   (2 marks)

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`2 xx 10^-8\ text(grams)`

Show Worked Solution
♦♦♦ Mean mark 20%.
IMPORTANT: Can you solve: 8 apples weigh 1kg, what does 1 apple weigh? This is exactly the same concept.
`text(We need to convert 50 mg into grams)`
`50\ text(mg) = 50/1000 = 0.05\ text(g) = 5 xx 10^-2\ text(grams)`

 

`:.\ text(Mass of 1 microbe)` `= text(mass of sample)/text(# microbes)`
  `= (5 xx 10^-2)/(2.5 xx 10^6)`
  `= 2 xx 10^-8\ text(grams)`