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Measurement, STD2 M1 2024 HSC 34

A container for soccer balls is made using two half spheres joined to each end of a cylindrical body.
 

Three soccer balls fit exactly inside the container. Each ball has a diameter of 23 cm.

The hemispherical ends of the container just touch the surface of the soccer balls.

What is the total surface area of the container? Give your answer in square metres correct to one decimal place.   (4 marks)

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\(\text{0.5 m}^{2}\)

Show Worked Solution

\(\text{Ball diameter = 23 cm}\ \Rightarrow \ r=11.5\ \text{cm}\)

\(\text{S.A. (2 half spheres)}\) \(=2 \times \dfrac{1}{2} \times 4 \pi \times 11.5^2\)  
  \(=1662\ \text{cm}^{2}\)  

 
\(\text{Cylinder width}\ =\ \text{ball circumference}= 2 \times \pi \times 11.5\)

\(\text{Cylinder height}\ = 2 \times 23 = 46\ \text{cm}\)

\(\text{S.A. (cylinder)}\ = 2\pi rh = 2 \times \pi \times 11.5 \times 46 = 3324\ \text{cm}^{2}\)
 

\(\text{Note:}\ 1\ \text{m}^{2} = 100\ \text{cm} \times 100\ \text{cm}\ = 10\,000\ \text{cm}^{2}\)

\(\text{S.A. (container)}\) \(=1662 + 3324\)  
  \(=4986\ \text{cm}^{2}\)  
  \(=\dfrac{4986}{10\,000}\ \text{m}^{2}\)  
  \(=0.5\ \text{m}^{2}\ \text{(1 d.p.)}\)  

Measurement, STD2 M1 2022 HSC 34

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


 

Find the total surface area of the composite solid in cm², correct to 1 decimal place.  (4 marks)

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`122.5\ text{cm}^2`

Show Worked Solution
`text{S.A. of Cylinder}` `=pir^2+2pirh`  
  `=pi(2^2)+2pi(2)(3)`  
  `=16pi\ text{cm}^2`  

 

`text{S.A. of Hemisphere}` `=1/2 xx 4pir^2`  
  `=2pi(3^2)`  
  `=18pi\ text{cm}^2`  

 

`text{Area of Annulus}` `=piR^2-pir^2`  
  `=pi(3^2)-pi(2^2)`  
  `=5pi\ text{cm}^2`  

 

`text{Total S.A.}` `=16pi+18pi+5pi`  
  `=39pi`  
  `=122.522…`  
  `=122.5\ text{cm}^2\ \ text{(to 1 d.p.)}`  

♦ Mean mark 50%.

Measurement, STD2 M1 2020 HSC 25

A composite solid consists of a triangular prism which fits exactly on top of a cube, as shown.
 

Find the surface area of the composite solid.   (3 marks)

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`424 \ text{cm}^2`

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`text{S.A. of 1 face of cube} = 8 xx 8 = 64 \ text{cm}^2`

`text{Height of triangle} = 11 – 8 = 3 \ text{cm}`

`therefore \ text{S.A. (triangular prism)}` `= 2 xx ( frac{1}{2} xx 8 xx 3 ) + 2 xx (5 xx 8)`
  `= 24 + 80`
  `= 104 \ text{cm}^2`

 

`therefore \ text{Total S.A.}` `= 5 xx 64 + 104`
  `= 424 \ text{cm}^2`

Measurement, STD2 M1 2012 HSC 25 MC

The solid shown is made of a cylinder with a hemisphere (half a sphere) on top.
 

What is the total surface area of the solid, to the nearest square centimetre?

  1. 628 cm²
  2. 679 cm²
  3. 729 cm²
  4. 829 cm²
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`B`

Show Worked Solution

`text(Total surface area)`

`= pir^2 + 2pirh + 1/2 xx 4pir^2`

`= pi xx 4^2 + 2pi xx 4 xx 21 + 1/2 xx 4pi xx 4^2`

`= 678.58…\ text(cm²)`

`=> B`

Measurement, STD2 M1 2018 HSC 27c

A shade shelter is to be constructed in the shape of half a cylinder with open ends. The diameter is 3.8 m and the length is 10 m.
 

 
The curved roof is to be made of plastic sheeting.

What area of plastic sheeting is required, to the nearest m²?  (2 marks)

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`60\ text(m²  (nearest m²))`

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`text(Flatten out the half cylinder,)`

`text(Width)` `= 1/2 xx text(circumference)`
  `= 1/2 xx pi xx 3.8`
  `= 5.969…`

 

`:.\ text(Sheeting required)` `= 10 xx 5.969…`
  `= 59.69…`
  `= 60\ text(m²  (nearest m²))`

Measurement, STD2 M1 2016 HSC 26a

Calculate the surface area of a sphere with a radius of 5 cm, correct to the nearest whole number.  (1 mark)
 

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`314\ text{cm²}`

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`SA` `= 4pir^2`
  `= 4 xx pi xx 5^2`
  `= 314.15…`
  `= 314\ text{cm²  (nearest cm²)}`

Measurement, STD2 M1 2015 HSC 26f

Approximately 71% of Earth’s surface is covered by water. Assume Earth is a sphere with a radius of 6400 km.

Calculate the number of square kilometres covered by water.  (2 marks)

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`3.7 xx 10^8\ text(km²)\ \ text{(nearest km²)}`

Show Worked Solution

`text(Surface area of Earth)`

`= 4pir^2`

`= 4pi xx 6400^2`

 

`:.\ text(Surface covered by water)`

`= text(71%) xx 4pi xx 6400^2`

`= 365\ 450\ 163.7…`

`= 3.7 xx 10^8\ text(km²)\ \ text{(nearest km²)}`

Measurement, STD2 M1 2014 HSC 25 MC

A grain silo is made up of a cylinder with a hemisphere (half a sphere) on top. The outside of the silo is to be painted.
  

 What is the area to be painted?

  1. `8143\ text(m²)`
  2. `11\ 762\ text(m²)`
  3. `12\ 667\ text(m²)`
  4. `23\ 524\ text(m²)`
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`A`

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`text(Total Area) = text(Area of cylinder) + text(½ sphere)`

♦ Mean mark 40%
`text(Area of cylinder)` `= 2 pi rh`
  `= 2pi xx 24 xx 30`
  `= 4523.9`
`text(Area of ½ sphere)` `= 1/2 xx 4 pi r^2`
  `= 1/2 xx 4 pi xx 24^2`
  `= 3619.1`
`:.\ text(Total area)` `= 4523.9 + 3619.1`
  `= 8143\ text(m²)`

`=>  A`

Measurement, STD2 M1 2010 HSC 28b

Moivre’s manufacturing company produces cans of Magic Beans. The can has a diameter of  10 cm and a height of  10 cm.

2010 28b1

  1. Cans are packed in boxes that are rectangular prisms with dimensions  30 cm × 40 cm × 60 cm.

     

    What is the maximum number of cans that can be packed into one of these boxes?   (1 mark)

  2. The shaded label on the can shown wraps all the way around the can with no overlap. What area of paper is needed to make the labels for all the cans in this box when the box is full?   (2 marks)

  3. The company is considering producing larger cans. Monica says if you double the diameter of the can this will double the volume.

     

    Is Monica correct? Justify your answer with suitable calculations.   (2 marks)

      
    The company wants to produce a can with a volume of 1570 cm³, using the least amount of metal. Monica is given the job of determining the dimensions of the can to be produced. She considers the following graphs.
     
    2010 28b2

  4. What radius and height should Monica recommend that the company use to minimise the amount of metal required to produce these cans? Justify your choice of dimensions with reference to the graphs and/or suitable calculations.   (2 marks)

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  1. `72\ text(cans)`
  2. `20\ 358\ text{cm²  (nearest cm²)}`
  3. `text(Monica is incorrect because the volume)`

     

    `text(doesn’t double.  It increases by a factor of 4.)`

  4. `text(Radius 6.3 cm and height 12.6 cm.)`
Show Worked Solution
♦♦ Mean mark 27%
i.    `text(Maximum # Cans)` `= 4 xx 3 xx 6`
    `= 72\ text(cans)`

 

♦ Mean mark 38%
MARKER’S COMMENT: Many students didn’t account for the clearance of 0.5 cm at the top and bottom of each can.
ii.    `text(Label Area)\ text{(1 can)}` `= 2 pi rh`
    `= 2 xx pi xx 5 xx 9`
    `= 90 pi`
    `= 282.7433…\ text(cm²)`

 

`:.\ text(Label Area)\ text{(72 cans)}`

`= 72 xx 282.7433…`

`= 20\ 357.52…`

`= 20\ 358\ text(cm²)`  `text{(nearest cm²)}`

 

♦ Mean mark 44%
MARKER’S COMMENT: Many students performed calculations in this part without concluding if Monica is correct or not. Read the question carefully.
iii.   `text(Original volume)` `= pi r^2 h`
    `= pi xx 5^2 xx 10`
    `= 785.398…\ text(cm³)`
`text(If the diameter doubles, radius) = 10\ text(cm)`
`text(New volume)` `= pi xx 10^2 xx 10`
  `= 3141.592…\ text(cm³)`

 

`:.\ text(Monica is incorrect because the volume)`
`text(doesn’t double.  It increases by a factor of 4.)`

 

♦♦ Mean mark 26%
iv.
`text(Minimum metal used when surface area is a minimum.)`
`text(From graph, minimum surface area when)\ r = 6.3\  text(cm)`
`text(When)\ r = 6.3\  text(cm,)\ h = 12.6\  text(cm)\ \ \ text{(from graph)}`
`:.\ text(She should recommend radius 6.3 cm and height 12.6 cm)`