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Measurement, STD2 EQ-Bank 1

2UG-2005-25b

Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle.   (1 mark)

Show Answers Only

`ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2:`

`a^2 + b^2= 5^2 + 12^2= 169= 13^2= c^2`

Show Worked Solution

`ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2:`

`a^2 + b^2= 5^2 + 12^2= 169= 13^2= c^2`

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 M1 2024 HSC 18

A garden is made up of a right-angled triangle and a semicircle as shown.
 

What is the area of the garden, correct to the nearest square metre?   (3 marks)

Show Answers Only

\(35\ \text{m}^2\)

Show Worked Solution

\(\text{Base of triangle }=2\times\ \text{radius} = 6\ \text{metres}\)

\(\text{Total Area }\) \(=\text{ Area of triangle + area of semicircle}\)
  \(=\dfrac{1}{2}\times 6\times 7 + \dfrac{1}{2}\times \pi\times 3^2\)
  \(=21+14.137\dots\)
  \(=35.137\dots\)
  \(=35\ \text{m}^2\ (\text{nearest m}^{2})\)
♦♦ Mean mark 33%.

Measurement, STD1 M1 2024 HSC 9 MC

An equilateral triangle and an isosceles triangle are shown.

The triangles have the same perimeter.
 

What is the value of \(x\)?

  1. \(8\)
  2. \(9\)
  3. \(11\)
  4. \(12\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Perimeter triangle 1 }=3\times 11=33\ \text{m}\)

\(\text{Perimeter triangle 2 }=2\times x+9=(2x+9)\ \text{m}\)

\(\therefore\ 2x+9\) \(=33\)
\(2x\) \(=24\)
\(x\) \(=12\)

  
\(\Rightarrow D\)

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, STD2 M6 2023 HSC 12 MC

A cylindrical pipe with a radius of 12.5 cm is filled with water to a depth, `d` cm, as shown.

The surface of the water has a width of 20 cm.
 

What is the depth of water in the pipe?

  1. 2.5 cm
  2. 5.0 cm
  3. 7.5 cm
  4. 12.5 cm
Show Answers Only

`B`

Show Worked Solution

`text{Triangle base}\ = 20/2=10\ text{cm}`

`text{Let}\ x =\ text{⊥ distance from centre to water}`

`text{Using Pythagoras:}`

`12.5^2` `=x^2+10^2`  
`x^2` `=12.5^2-10^2=56.25`  
`x` `=\sqrt{56.25}=7.5\ text{cm}`  

 
`d=12.5-7.5=5.0\ text{cm}`

`=>B`

♦ Mean mark 42%.

Measurement, STD1 M1 2019 HSC 25

The diagram shows a sector with an angle of 120° cut from a circle with radius 10 m.

What is the perimeter of the sector? Write your answer correct to 1 decimal place.  (3 marks)

Show Answers Only

`40.9\ \ (text(1 d. p.))`

Show Worked Solution
`text(Arc length)` `= 120/360 xx 2 xx pi xx 10`
  `= 20.94`

 

`:.\ text(Perimeter)` `= 20.94 + 2 xx 10`
  `= 40.94`
  `= 40.9\ \ (text(1 d. p.))`

Measurement, STD1 M1 2022 HSC 4 MC

The area `(A)` of a circle is given by the formula `A=\pi r^2`, where `r` is the radius.

What is the value of `A`, correct to three significant figures, if `r=3.55` ?

  1. 39.5
  2. 39.6
  3. 39.591
  4. 39.592
Show Answers Only

`B`

Show Worked Solution
`A` `=pi\r^2`  
  `=pi xx3.55^2`  
  `=39.591…`  
  `=39.6\ \ text{(3 sig fig)}`  

  
`=>B`


♦♦ Mean mark 30%.
COMMENT: Significant figures warrant attention.

Measurement, STD2 M1 2021 FUR1 6

A child's toy has the following design.
 

Find the area of the shaded region to the nearest square centimetre.  (2 marks)

Show Answers Only

`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 …\ text{cm}^2`
  `=27\ text{cm²  (nearest whole)}`

Measurement, STD2 M1 2021 HSC 1 MC

Which of the following shapes has the largest perimeter?
 

Show Answers Only

`A`

Show Worked Solution

`\text{Consider each option:}`

`\text{Option A:} \ 4 \times 8 = 32 \ \text{cm}`

`\text{Option B:} \ 2 \times (3 + 11) = 28 \ \text{cm}`

`\text{Option C:} \ 3 \times 10 = 30 \ \text{cm}`

`\text{Option D:} \ 4 \times 2 + 3 + 9 = 20 \ \text{cm}`
 

`=> A`

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, 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)

Show Answers Only

`89\ text(cm²  (nearest cm²))`

Show Worked Solution

♦♦ 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 2018 HSC 22 MC

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

 
What is the perimeter of this shape, correct to one decimal place?

  1. 28.6 cm
  2. 36.6 cm
  3. 66.3 cm
  4. 74.3 cm
Show Answers Only

`text(B)`

Show Worked Solution

`text(Using Pythagoras to find radius)\ (r):`

`r` `= sqrt(10^2 – 6^2)`
  `= sqrt64`
  `= 8\ text(cm)`

 

`text(Arc length)` `= 1/4 xx 2 pi r`
  `= 1/4 xx 2 xx pi xx 8`
  `= 12.56…\ text(cm)`

 

`:.\ text(Perimeter)` `= 8 + 6 + 10 + 12.56…`
  `= 36.57…`

`=>\ text(B)`

Measurement, STD2 M1 2017 HSC 25 MC

In the circle, centre `O`, the area of the quadrant is 100 cm².
 


 

What is the arc length `l`, correct to one decimal place?

  1. 8.9 cm
  2. 11.3 cm
  3. 17.7 cm
  4. 25.1 cm
Show Answers Only

`C`

Show Worked Solution

`text(Find)\ r:`

♦ Mean mark 44%.
`text(Area)` `= 1/4 pir^2`
`100` `= 1/4 pir^2`
`r^2` `= 400/pi`
`:. r` `= 11.283…\ text(cm)`

 

`text(Arc length)` `= theta/360 xx 2pir`
  `= 90/360 xx 2pi xx 11.283…`
  `= 17.724…`
  `= 17.7\ text(cm)`

 
`=> C`

Measurement, STD2 M1 2015 HSC 28a

The diagram shows an annulus.
 

2015 28a

 
Calculate the area of the annulus.  (1 mark)

Show Answers Only

`≈ 50.26…\ text(cm²)`

Show Worked Solution

`text(Area of annulus)`

`= pi(R^2 − r^2)`

`= pi(5^2 − 3^2)`

`= pi(25 − 9)`

`= 16pi\ text(cm²)`

`≈ 50.26…\ text(cm²)`

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 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
Show Answers Only

`B`

Show Worked Solution
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
Show Answers Only

`C`

Show Worked Solution

`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 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`

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
Show Answers Only

`B`

Show Worked Solution
`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 2012 HSC 27b

The sector shown has a radius of 13 cm and an angle of 230°. 
 

 2012 27b
 

 What is the perimeter of the sector to the nearest centimetre?    (2 marks) 

Show Answers Only

`text(78 cm)\ \ \ \  text((nearest cm))`

Show Worked Solution
♦ Mean mark 39%
MARKER’S COMMENT: The formula of the length of an arc is given in the formula sheet.
`text(Perimeter)` `= 2 xx text(radius) + text(arc length)`
`text(Arc length)` `= theta/360 xx 2 xx pi xx r`
  `= 230/360 xx 2 xx pi xx 13`
  `= 52.1853…`
   
`:.\ text(Perimeter)` `= 2 xx 13 + 52.1853…`
  `= 78.1853…`
  `=text(78 cm)\ \ \ \ text((nearest cm))`

Measurement, STD2 M6 2011 HSC 9 MC

Two trees on level ground, 12 metres apart, are joined by a cable. It is attached 2 metres above the ground to one tree and 11 metres above the ground to the other.

What is the length of the cable between the two trees, correct to the nearest metre? 

  1.  `9\ text(m)`
  2. `12\ text(m)`
  3. `15\ text(m)`
  4. `16\ text(m)`
Show Answers Only

`C`

Show Worked Solution

`text(Using Pythagoras)`

`c^2` `=12^2+9^2`
  `=144+81`
  `=225`
`c` `=\sqrt{225}=15,\ \ c>0`

 
`=>C`

Measurement, STD2 M6 2010 HSC 3 MC

A field diagram has been drawn from an offset survey.
 

What is the distance from `G` to `H` correct to the nearest metre?

  1. \(11\)
  2. \(13\)
  3. \(16\)
  4. \(20\)
Show Answers Only

`B`

Show Worked Solution

`text(Using Pythagoras:)`

`GH^2` `=12^2+(16-11)^2`
  `=144+25`
  `=169`
`GH` `=sqrt169=13\ text(m)`

 
` =>  B`

Measurement, STD2 M1 2013 HSC 19 MC

A logo is designed using half of an annulus.
  

What is the area of the logo, to the nearest cm²?

  1. `25\ text(cm²)`
  2. `33\ text(cm²)`
  3. `132\ text(cm²)`
  4. `143\ text(cm²)`
Show Answers Only

`B`

Show Worked Solution
`text(Area)` `=1/2xxpi(R^2-r^2)`
  `=1/2xxpi(5^2-2^2)`
  `=21/2pi`
  `=33\ text(cm²)\ \ \ text{(nearest 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
Show Answers Only

`B`

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…\ text(cm²)`

`=>\ B`