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v1 Measurement, STD2 M1 2022 HSC 34

A composite solid is shown. The top section is a hemisphere with a diameter of 6 cm. The bottom section is a cylinder with a height of 3 cm and a diameter of 4 cm
 

Find the total volume of the composite solid in cm³, correct to 1 decimal place.  (4 marks)

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`94.2 \ text{cm}^3`

Show Worked Solution
`text{Volume of Hemisphere}` `=2/3 pi r^3`  
  `=2/3 pi xx 3^3`  
  `=56.54\ text{cm}^3`  

 

`text{Volume of Cylinder}` `=pi r^2 h`  
  `=pi (2^2) xx 3`  
  `=37.69\ text{cm}^3`  

 

`text{Total Volume}` `=56.54+37.69`
  `=94.23`
  `=94.2 \ text{cm}^3`

v1 Measurement, STD2 M1 2014 HSC 27c

A swimming pool is in the shape of a rectangle with a semicircle at each end, as shown.

The pool is 7000 mm long, 4000 mm wide, and has a depth of 2100 mm.  
  

How much water is needed to fill the pool, to the nearest litre?   (4 marks) 

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`51 \ 589 \ text(L)`

Show Worked Solution

`V = Ah` 

♦ Mean mark 41%
STRATEGY: Adjusting measurements to metres makes the final conversion to litres simple.

`text(Finding Area of base)`

`text(Semi-circles have radius 2000 mm) = 2 \ text(m)`

`:.\ text(Area of 2 semicircles)`

`=2 xx 1/2 xx pi r^2`

`= pi xx 2^2`

`= 12.56 \ text(m)^2`
 

`text(Area of rectangle)`

`= l xx b`

`= (7-2 xx 2) xx 4`

`= 12\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= (12.56… + 12) xx 2.1`
  `= 51.589…\ text(m)^3`
  `= 51 \ 589 \ text(L)\ \ text{(using 1m³} = 1000\ text{L)}`
   

v1 Measurement, STD2 M1 2013 HSC 12 MC

A hemisphere sits perfectly on top of a cylinder to form a solid. 

What is the volume of the solid?

  1. 1750 cm³
  2. 1950 cm³
  3. 2150 cm³
  4. 2350 cm³
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`C`

Show Worked Solution
`text(Volume )` `=text{Vol (cylinder)} +text{Vol (hemisphere)}`
  `= pi r^2h+2/3pi r^3`
  `= pi xx 6^2 xx 15 + 2/3pi xx 6^3`
  `=2149.84\ text(cm)^3`

 
`=>\ C`

v1 Measurement, STD2 M1 2010 HSC 17 MC

During a heavy storm, 2.4 hectares of farmland received rainfall to a depth of 12 cm.

How many kilolitres of rainwater fell on the farmland? (1 hectare = 10 000 m²)

  1. 2.88 kL
  2. 2880 kL
  3. 288 000 kL
  4. 2 880 000 kL
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`B`

Show Worked Solution
♦♦ Mean mark 36%
NOTE: The unit conversion 1 m³ = 1000 L is contained in the Formulae and Data sheet given out in the exam.
`text(Area of farmland)` `=2.4xx10\ 000`
  `=24\ 000\ text(m²)`
`text(Volume)` `=Ah`
  `=24\ 000xx0.12`
  `=2880\ text(m³)`

 

`text{1 m³}` `=1000\ text(L)=1\ text(kL)`
`:.\ text(Volume)` `=2880\ text(kL)`

`=>B`

v1 Measurement, STD2 M1 SM-Bank 26 MC

A rectangular swimming pool measures 15 m by 8 m and has an average depth of 1.8 m. The pool is being filled at a rate of 250 litres per minute.

What is the volume of water needed to fill the pool completely?

(1 m³ = 1000 L)

  1. 216 L
  2. 2 160 L
  3. 21 600 L
  4. 216 000 L
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`=>\ text(D)`

Show Worked Solution

`text(Volume of pool in m³:)`

`text(Vol)` `= l xx w xx h`
  `= 15 xx 8 xx 1.8`
  `= 216 \ text(m³)`
  `= 216 xx 1000 \ text(L)`
  `= 216\ 000 \ text(L)`

`=>\ text(D)`

v1 Measurement, STD2 M1 2017 HSC 22 MC

A concrete storm drain is constructed in the shape of a rectangular prism with a cylindrical tunnel running through its center. The dimensions are shown in the diagram.
 
 

What is the approximate volume of concrete needed to construct the storm drain?

  1. `text(4.5 m)³`
  2. `text(4.8 m)³`
  3. `text(5.1 m)³`
  4. `text(5.4 m)³`
Show Answers Only

`A`

Show Worked Solution
`text(Volume of concrete)` `= text(Volume of rectangular prism)-text(Volume of cylinder)`
  `= l xx w xx h-pi r^2 h`
  `= 2.5 xx 1.8 xx 1.2-pi xx (0.4)^2 xx 1.8`
  `= 5.4-pi xx 0.16 xx 1.8`
  `= 5.4-0.9047…`
  `= 4.4952…`
  `= 4.5\ text(m)³`

`=>A`

v1 Measurement, STD2 M1 2017 HSC 18 MC

A trapezoidal prism is illustrated below, with the following dimensions:
 

What is the volume of the trapezoidal prism?

  1. `24.38\ text(m)^3`
  2. `32.75\ text(m)^3`
  3. `39.38\ text(m)^3`
  4. `47.25\ text(m)^3`
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`C`

Show Worked Solution
`text(Area of trapezoid)` `= 1/2h (a + b)`
  `= 1/2 xx 3 xx (6 + 4.5)`
  `= 15.75\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= 15.75 xx 2.5`
  `= 39.38\ text(m)^3`

 
`=>C`

v1 Measurement, STD2 M1 2009 HSC 19 MC

A tennis ball canister holds 3 balls inside, as illustrated below.

The diameter of each ball is 6.5 cm.

What is the volume of the tennis ball canister, to the nearest cubic centimetre?

  1. `527\ text(cm)^3`
  2. `597\ text(cm)^3`
  3. `637\ text(cm)^3`
  4. `647\ text(cm)^3`
Show Answers Only

`D`

Show Worked Solution

`text(S)text(ince diameter sphere = 12 cm) `

`=>\ text(Radius of cylinder = 6 cm)`

`text(Height of cylinder)` `= 3 xx text(diameter of sphere)`
  `= 3 xx 6.5`
  `= 19.5\ text(cm)`
   
`:.\ text(Volume cylinder)` `= pi r^2 h`
  `= pi xx 3.25^2 xx 19.5`
  `= 647.07…\ text(cm³)`

 
`=>  D`

v1 Measurement, STD2 M1 2018 HSC 30a

A cylindrical oil tank has a height of 7 metres and a capacity of 1.5 megalitres.
 

What is the diameter of the oil tank? Give your answer in metres, correct to two decimal places.  (3 marks)

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`16.50\ text{m}`

Show Worked Solution

`text{Converting megalitres to m³  (using 1 m³ = 1000 L):}`

♦ Mean mark 48%.

`1.5\ text(ML)` `= (1.5 xx 10^6)/(10^3)`
  `= 1.5 xx 10^3\ text(m)^3`
  `= 1500\ text(m)^3`

 

`V` `= pir^2h`
`1500` `= pi xx r^2 xx 7`
`r^2` `= 1500/(pi xx 7)`
`sqrt(r^2)` `= sqrt(68.21)`
  `= 8.26\ text{m}`

 

`text{Diameter}` `=2r`  
  `=2xx8.26 \ text{m}`  
  `=16.52\ text{m}`  

 

v1 Measurement, STD2 M1 2009 HSC 11 MC

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  

  1. 68 m²
  2. 73 m²
  3. 95 m²
  4. 193 m²
Show Answers Only

`C`

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 12^2)-(1/2 xx 6 xx 6)`
  `= 113.097…-18`
  `= 95.097…\ text(m²)`

`=> C`

v1 Measurement, STD2 M1 2016 HSC 30a

The area of a school roof is 45 m². All rain that falls on the roof flows into a storage tank.

How many litres of water are collected in the tank when 25 mm of rain falls?   (2 marks)

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`1125\ \text{L}`

Show Worked Solution
`text{Volume}` `= A × h`
  `= 45 × (25 ÷ 1000)`
  `= 1.125\ \text{m}^3`
  `= 1125\ \text{L}`

v1 Measurement, STD2 M7 2023 HSC 26

Jo is constructing a concrete border around a rectangular vegetable patch, as shown. The border is 0.6 m wide.
 

  1. Find the area of the border.   (2 marks)
  2. Jo is preparing concrete using a mix of gravel, sand, and cement in the ratio 5 : 3 : 2 by weight.
    Jo needs 2 tonnes of concrete in the correct ratio.
    Calculate how many 20 kg bags of cement Jo needs to buy.   (3 marks)

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  1. `11.4\ \text{m}^2`
  2. `20\ \text{bags}`
Show Worked Solution

a.    `text{Area of outer rectangle} = 6.5 × 4.2 = 27.3\ \text{m}^2`

`text{Area of garden} = 5.3 × 3.0 = 15.9 \ text{m}^2`

`text{Area of border}` `= 27.3-15.9`  
  `=11.4\ text{m}^2`  

 

b.   `text{Ratio parts:} \ 5 + 3 + 2 = 10` parts

`text{Total concrete} = 2\ \text{tonnes} = 2000\ \text{kg}`

`text{Each part} = 2000 ÷ 10 = 200\ \text{kg}`

`text{Cement} = 2 \ text{parts} = 2 × 200 = 400\ \text{kg}`

`text{Bags of cement} = 400 ÷ 20 = 20`

`⇒ \ text{Jo needs 20 bags of cement.}`

v1 Measurement, STD2 M1 2021 HSC 16

The surface area, `A`, of a sphere is given by the formula

`A = 4 pi r^2,`

where `r` is the radius of the sphere.

A satellite dish resembles the inner surface of the lower half of a sphere with a radius of 1.5 meters.

 

Find the surface area of the satellite dish in square metres, correct to one decimal place.   (2 marks)

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`14.1\ text{m}^2`

Show Worked Solution
`A` `= frac{1}{2} times 4 pi r^2`
  `= 2 pi r^2`
  `= 2 pi times (1.5)^2`
  `= 2 pi times 2.25`
  `= 14.137…`
  `= 14.1\ text{m}^2\ \text{(1 d.p.)}`

v1 Measurement, STD2 M1 2019 HSC 16

A decorative light fixture is in the shape of a hollow hemisphere with a diameter of 24 cm.
 

The inside of the fixture is to be coated with reflective paint.

What is the area to be painted on the inside surface? Give your answer correct to the nearest square centimetre.   (2 marks)

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`905\ \text{cm}^2`

Show Worked Solution
`A` `= 2 pi r^2`
  `= 2 × pi × 12^2`
  `= 2 × pi × 144 = 905.0…`
  `≈ 905\ \text{cm}^2`

v1 Measurement, STD1 M1 2019 HSC 25

The diagram illustrates a sector formed by a central angle of 105°, taken from a circle with a radius of 15 metres.

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

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

Show Worked Solution
`text(Arc length)` `= 105/360 xx 2 xx pi xx 15`
  `= 27.49`

 

`:.\ text(Perimeter)` `= 27.49 + 2 xx 15`
  `= 57.49`
  `= 57.5\ \ (text(1 d. p.))`

v1 Measurement, STD2 M1 2016 HSC 30c

A landscape artist was commissioned to design a garden consisting of part of a circle, with centre `O`, and a rectangle, as shown in the diagram. The radius `OC` of the circle is 20 m, the width `BC` of the rectangle is 10 m, and `DOC` is 100°.
 

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

Show Answers Only

`6281\ text{m²  (nearest m²)}`

Show Worked Solution

`text(In)\ \triangle ODC,`

`sin50^@` `= (ED)/20`
`ED` `= 20 \times \sin50^@`
  `= 34.472`
`:. DC` `= 2 \times 34.472 = 68.944\ \text{m}`

 

`cos50^@` `= (OE)/20`
`:. OE` `= 20 \times \cos50^@ = 28.925`

 

`text(Area of)\ \triangle ODC`

`= \frac{1}{2} \times 68.944 \times 28.925 = 997.12\ \text{m}^2`

 

`text(Area of rectangle ABCD)` `= 10 \times 68.944 = 689.44\ \text{m}^2`

 

`text(Area of major sector DOAC)`

`= \pi \times 20^2 \times \frac{260}{360} = 4594.58\ \text{m}^2`

 

`:.\ \text{Area of garden}`

`= 997.12 + 689.44 + 4594.58 = 6281.14`

`= 6281\ \text{m² (nearest m²)}`

v1 Measurement, STD2 M1 2018 HSC 27c

A farmer is designing a chicken coop with a roof shaped like half a cylinder, open at both ends. The structure has a diameter of 4 metres and a length of 12 metres.
 

 
The curved roof is to be made of aluminum sheets.

What area of aluminum sheets is required, to the nearest m²?  (2 marks)

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`75\ text(m²  (nearest m²))`

Show Worked Solution

`text(Flatten out the half cylinder,)`

`text(Width)` `= 1/2 xx text(circumference)`
  `= 1/2 xx pi xx 4`
  `= 6.283…`

 

`:.\ text(Sheeting required)` `= 12 xx 6.283…`
  `= 75.39…`
  `= 75\ text(m²  (nearest m²))`

v1 Measurement, STD2 M1 2015 HSC 28a

The diagram shows a circular garden bed with a circular path around it.
 

The radius of the entire structure (garden + path) is 6 m, and the radius of the inner garden is 4 m.

Calculate the area of the path. (1 mark)

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`≈ 62.83…\ text(m²)`

Show Worked Solution

`text(Area of path)`

`= pi(R^2 − r^2)`

`= pi(6^2 − 4^2)`

`= pi(36 − 16)`

`= 20pi\ \text(m²)`

`≈ 62.83…\ \text(m²)`

v1 Measurement, STD2 M1 2009 HSC 23c

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

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

  2. Tiles are sold in boxes. Each box holds one square metre of tiles and costs $60. 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 floor.   (2 marks)

Show Answers Only

i.    `17\ text(m²)`

ii.   `$1140`

Show Worked Solution
i.
`text(Area)` `=\ text(Area of big square – Area of 2 cut-out squares`
  `= (3 + 2) xx (3 + 2)\-2 xx (2 xx 2)`
  `= 25\-8`
  `= 17\ text(m²)`

 

ii. `text(Tiles required)` `= (17 +10 text{%}) xx 17`
    `= 18.7\ text(m²)`

 

 `=>\ text(19 boxes are needed)`

`:.\ text(Total cost of boxes)` `=19 xx $60`
  `= $1140`