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Measurement, STD1 M1 2025 HSC 13

A trapezium is shown.
 

  1. Write an expression for the area of this trapezium.   (1 mark)

  2. Find the value of \(a\), given that the area of the trapezium is 330 cm².   (2 marks)
Show Answers Only

a.   \(\text{A}=6(a+30)\ \text{or}\ A=6a+180\)

b.   \(a=25\)

Show Worked Solution
a.     \(A\) \(=\dfrac{h}{2}\Big(a+b\Big)\)
    \(=\dfrac{12}{2}\Big(a+30\Big)\)
    \(=6a+180\)

♦♦♦ Mean mark (a) 24%.

b.    \(\text {When} \ A=330:\)

\(6a+180\) \(=330\)
\(6a\) \(=330-180\)
\(6a\) \(=150\)
\(a\) \(=25\)

♦♦♦ Mean mark (b) 25%.

Measurement, STD1 M1 2025 HSC 11

Consider the composite shape shown.
 

What is the area of the shape in cm² ?   (3 marks)

Show Answers Only

\(21\ \text{cm}^2\)

Show Worked Solution

\(\text{Height of triangle}=3+2=5\ \text{cm}\)

\(\text{Area}\) \(=\text{Area of triangle}+\text{Area of rectangle}\)
  \(=\dfrac{1}{2}\times 6\times 5+3\times 2\)
  \(=15+6\)
  \(=21\ \text{cm}^2\)
♦♦ Mean mark 30%.

Measurement, STD1 M5 2024 HSC 29

A floor plan for a living area is shown. All measurements are in millimetres.
 

  1. What is the length and width of the cupboard, in metres?   (1 mark)

  2. The floor of the living area is to be tiled. Tiles will NOT be placed under the cupboard.
  3. Each tile is 0.2 m × 0.5 m. The tiles are supplied in boxes of 15 at a cost of $100 per box. Only full boxes can be purchased.
  4. What is the cost of the tiles for the living area?   (4 marks)
Show Answers Only

a.    \(4\ \text{m}\times 0.5\ \text{m}\)

b.    \($1100\)

Show Worked Solution

a.    \(\text{1 metre = 1000 mm}\)

\(\text{Cupboard}\ = 4\ \text{m}\times 0.5\ \text{m}\)

Mean mark (a) 53%.

b.    \(\text{Method 1}\)

\(\text{Total area to be tiled}\ =6\times 3-4\times 0.5=16\ \text{m}^2\)

\(\text{Area of each tile}\ =0.2\times0.5=0.1\ \text{m}^2\)

\(\text{Tiles needed}\ =\dfrac{16}{0.1}=160\ \text{tiles}\)

\(\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}\)

\(\therefore\ \text{Total cost}\ =100\times 11=$1100\)

   
\(\text{Method 2}\)

\(\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 \)

\(\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 \)
 

\(\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160\)

\(\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}\)

\(\therefore\ \text{Total cost}\ =100\times 11=$1100\)

♦♦ Mean mark (b) 35%.

Measurement, STD1 M5 2023 HSC 12

A floor plan is shown.
 

  1. What are the dimensions, in metres, of the living room?  (2 marks)

  2. The shaded area of the kitchen floor is to be tiled. Each tile is 40 cm by 40 cm. How many tiles are needed to cover the kitchen floor?  (2 marks)

  3. The tiles are supplied in boxes of 10 . Only full boxes can be purchased. How many boxes need to be purchased to tile the kitchen floor?  (1 mark)

Show Answers Only

a.    \(5.2\times 5.94\)

b.    \(72\text{ tiles}\)

c.    \(8\text{ boxes}\)

Show Worked Solution

a.    \(\text{Dimensions}=5.2\times (2.15+0.16+3.6)=5.2\times5.94\)


♦ Mean mark (a) 51%.

b.    \(\text{Method 1}\)

\(\text{Kitchen Area}\) \(=3.2\times 3.6\)  
  \(=11.52\text{ m}^2\)  
\(\text{Tile area}\) \(=0.4\times 0.4\)  
  \(=0.16\text{ m}^2\)  
\(\text{Tiles needed}\) \(=\dfrac{11.52}{0.16}\)  
  \(=72\text{ tiles}\)  

  
\(\text{Method 2}\)

\(\text{Tiles to fit width}\) \(=\dfrac{3.6}{0.4}\)
  \(=9\)
\(\text{Tiles to fit length}\) \(=\dfrac{3.2}{0.4}\)
  \(=8\)
\(\text{Tiles needed}\) \(=9\times 8\)
  \(=72\text{ tiles}\)

♦ Mean mark (b) 43%.

c.    \(\text{Boxes}=\dfrac{72}{10}=7.2\)

\(\therefore\ \text{Boxes }=8\)

Measurement, STD2 M7 2023 HSC 26

Kim is building a path around a garden at the back of a house, as shown. The path is 0.5 m wide.

  1. Find the area of the path.   (2 marks)

  2. Kim is mixing some concrete for the path. The concrete mix is made up of crushed rock, sand and cement in the ratio of 4 : 2 : 1 by weight.
  3. Kim needs 2.1 tonnes of concrete in the correct ratio.
  4. Calculate how many 15 kg bags of cement Kim needs to buy.   (3 marks)

Show Answers Only
  1. `6.5\ text{m}^2`
  2. `20\ text{bags}`
Show Worked Solution

a.    `text{Area outer rectangle}\ = 3xx8=24\ text{m}^2`

`text{Area garden}\ = 2.5xx7=17.5\ text{m}^2`

`A_text{path}` `=24-17.5`  
  `=6.5\ text{m}^2`  

 
b.    `text{7 parts = 2.1 tonnes}`

`text{1 part}\ = 2.1/7=0.3\ text{tonnes}\ =300\ text{kgs}`

`text{Rock}:text{Sand}:text{Cement} = 4:2:1 = 1200:600:300`

`=>\ text{300 kgs of cement are required}`

`:.\ text{Bags of cement}` `=300/15`  
  `=20\ text{bags}`  

Measurement, STD1 M1 2019 HSC 36

A path 1.8 m wide is being built around a rectangular garden. The garden is 8.4 m long and 5.4 m wide. The path is shaded in the diagram.
 

 
 

The path is to be covered with triangular pavers with side lengths of 15 cm and 20 cm as shown.
 


 

The pavers are to be laid to cover the path with no gaps or overlaps.

How many pavers are needed?  (4 marks)

Show Answers Only

`4176`

Show Worked Solution
`text(Shaded Area)` `=\ text(Large rectangle − garden area)`
  `=(1.8+8.4+1.8) xx (1.8+5.4+1.8) – (8.4 xx 5.4)`
  `= 12 xx 9 – 8.4 xx 5.4`
  `= 62.64\ text(m²)`

♦♦ Mean mark 18%.
COMMENT: Convert all measurements to the same units to minimise errors (either m² or cm²).

`text(Area of 1 paver (in m²))` `= 1/2 xx 0.15 xx 0.20`
  `= 0.015\ text(m²)`

 
`:.\ text(Number of pavers needed)`

`= 62.64/0.015`

`= 4176`

Measurement, STD2 M1 2004 HSC 23a

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

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

  2. Carmel decides to add a 5 cm layer of straw to the garden bed.

     

    Calculate the volume of straw required. Give your answer in cubic metres.   (2 marks)

  3. Each bag holds 0.25 cubic metres of straw.

     

    How many bags does she need to buy?   (2 marks)

  4. A straight fence is to be constructed joining point A to point B.

     

    Find the length of this fence to the nearest metre.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(2.85 m³)`
  3. `text(She needs to buy 12 bags)`
  4. `8\ text{m  (nearest metre)}`
Show Worked Solution
a.    `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)`

 

b.    `V` `= Ah`
  `= 57 xx 0.05`
  `= 2.85\ text(m³)`

 

c.    `text(Bags to buy)` `= 2.85/0.25`
  `= 11.4`

 
`:.\ text(She needs to buy 12 bags.)`

 

d.   `text(Using Pythagoras,)`

`AB^2` `= 6.0^2 + 5.1^2`
  `= 36 + 26.01`
  `= 62.01`
`AB` `= 7.874…`
  `=8\ text{m  (nearest metre)}`

Measurement, STD2 M1 2009 HSC 23c

The diagram shows the shape and dimensions of a terrace which is to be tiled.
 

  1. Find the area of the terrace.   (2 marks)

  2. Tiles are sold in boxes. Each box holds one square metre of tiles and costs $55. When buying the tiles, 10% more tiles are needed, due to cutting and wastage.

     

    Find the total cost of the boxes of tiles required for the terrace.   (2 marks)

Show Answers Only

a.    `13.77\ text(m²)`

b.   `$880`

Show Worked Solution
a.
`text(Area)` `=\ text(Area of big square – Area of 2 cut-out squares`
  `= (2.7 + 1.8) xx (2.7 + 1.8)\-2 xx (1.8 xx 1.8)`
  `= 20.25\-6.48`
  `= 13.77\ text(m²)`

 

b. `text(Tiles required)` `= (13.77 +10 text{%}) xx 13.77`
    `= 15.147\ text(m²)`

 

 `=>\ text(16 boxes are needed)`

`:.\ text(Total cost of boxes)` `=16 xx $55`
  `= $880`