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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 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, STD2 M1 2022 HSC 32

The diagram shows a park consisting of a rectangle and a semicircle. The semicircle has a radius of 100 m. The dimensions of the rectangle are 200 m and 250 m.

A lake occupies a section of the park as shown. The rest of the park is a grassed section. Some measurements from the end of the grassed section to the edge of the lake are also shown.
 

  1. Using two applications of the trapezoidal rule, calculate the approximate area of the grassed section.  (2 marks)

  2. Hence calculate the approximate area of the lake, to the nearest square metre.   (2 marks)

Show Answers Only
  1. `35\ 500\ text{m}^2`
  2. `30\ 208\ text{m}^2`
Show Worked Solution

a.   `h=100\ text{m}`

`A` `~~h/2(x_1+x_2)+h/2(x_2+x_3)`  
  `~~100/2(160+150)+100/2(150+250)`  
  `~~50xx310+50xx400`  
  `~~35\ 500\ text{m}^2`  

 

b.    `text{Total Area}` `=\ text{Area of Rectangle + Area of Semi-Circle}`
    `= (250 xx 200) + 1/2 xx pi xx 100^2`
    `=65\ 707.96\ text{m}^2`

 

`text{Area of Lake}` `=67\ 708-35\ 500`  
  `=30\ 208\ text{m}^2`  

♦ Mean mark (b) 44%.

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, STD2 M1 SM-Bank 2 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`
Show Answers Only

`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`

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 2018 HSC 24 MC

The coordinates of city A are (39°N, 75°W). City B lies on the same longitude and is 5700 km south of city A.

What is the latitude of city B, given the earth's radius is 6400 km?

  1. 51°N
  2. 51°S
  3. 12°N
  4. 12°S
Show Answers Only

`text(D)`

Show Worked Solution

`text(Circumference) = 2 xx pi xx 6400`

♦ Mean mark 45%.
COMMENT: If the earth’s radius is stated, this past question is arguably within the scope of the new syllabus.

`theta/360` `= 5700/(2pi xx 6400)`
`theta` `= (5700 xx 360)/(2pi xx 6400)`
  `~~ 51°`

   
`:. text(City)\ B\ text(is 12° South.)`

`=>\ text(D)`

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 2016 HSC 30c

A school playground consists of part of a circle, with centre `O`, and a rectangle as shown in the diagram. The radius `OB` of the circle is 45 m, the width `BC` of the rectangle is 20 m and `AOB` is 100°.
 

What is the area of the whole playground, correct to the nearest square metre?  (5 marks)

Show Answers Only

`6971\ text{m²  (nearest m²)}`

Show Worked Solution

`text(In)\ DeltaOEB,`

`sin50^@` `= (EB)/45`
`EB` `= 45 xx sin50^@`
  `= 34.47…`
`:. AB` `= 2 xx 34.47…`
  `= 68.944\ \ (text(3 d.p.))`

 

`cos50^@` `= (OE)/45`
`:. OE` `= 45 xx cos50^@`
  `= 28.925\ \ (text(3 d.p.))`

 

`text(Area of)\ DeltaOAB`

`= 1/2 xx AB xx OE`

`= 1/2 xx 68.944 xx 28.925`

`= 997.12\ text(m²)`

 

`text(Area)\ ABCD` `= 20 xx 68.944`
  `= 1378.88\ text(m²)`

 

`text(Area of major sector)\ OAB`

`= pi xx 45^2 xx 260/360`

`= 4594.58\ text(m²)`

 

`:.\ text(Area of playground)`

`= 997.12 + 1378.88 + 4594.58`

`= 6970.58`

`= 6971\ text{m²  (nearest m²)}`

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)}`

Algebra, STD2 A1 2007 HSC 28b

This shape is made up of a right-angled triangle and a regular hexagon.
 

 

The area of a regular hexagon can be estimated using the formula  `A = 2.598H^2`  where  `H`  is the side-length.

Calculate the total area of the shape using this formula.   (3 marks)

Show Answers Only

`22.784\ text(cm²)`

Show Worked Solution

`text(Area) = 2.598H^2`

`text(Using Pythagoras)`

`H^2` `= 2^2 + 2^2`
  `= 8`
`H` `= sqrt 8`
`:.\ text(Area of hexagon)` `= 2.598 xx (sqrt 8)^2`
  `= 20.784\ text(cm²)`
`text(Area of triangle)` `= 1/2 bh`
  `= 1/2 xx 2 xx 2`
  `= 2\ text(cm²)`

 

`:.\ text(Total Area)` `= 20.784 + 2`
  `= 22.784\ text(cm²)`

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 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`