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Measurement, STD2 M6 2025 HSC 37

The diagram shows a park consisting of two equilateral triangles. The shaded triangle is a grassed section. All measurements on the diagram are in metres.
 

How long will it take to mow the grassed section if it takes 5 minutes to mow 20 m² ? Give your answer to the nearest minute.   (4 marks)

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\(7 \ \text{minutes}\)

Show Worked Solution

\(\text{Large triangle is equilateral (all sides = 12 m)}\)

\(\text{Area of large} \ \triangle\) \(=\dfrac{1}{2}ab \, \sin C\)
  \(=\dfrac{1}{2} \times 12 \times 12 \times \sin 60^{\circ}\)
  \(=36 \sqrt{3}\)

 

\(\text{Area of} \ \ \triangle_1=\dfrac{1}{2} \times 3 \times 9 \times \sin 60^{\circ}=\dfrac{27 \sqrt{3}}{4}\)

\(\text{Grassed area}=36 \sqrt{3}-3 \times \dfrac{27 \sqrt{3}}{4}=27.2798 \ldots \ \text{m}^2\)

\(\text{Time to mow}\) \(=\dfrac{27.2798 \ldots}{20} \times 5\)
  \(=6.81 \ldots\)
  \(=7 \ \text{minutes (nearest min)}\)

Filed Under: Non-Right Angled Trig (Std2) Tagged With: Band 6, smc-804-30-Sine Rule (Area), smc-804-60-X-topic with PAV

Measurement, STD2 M6 2024 HSC 32

A regular pentagon \(ABCDE\) is drawn inside a circle with a radius 30 cm.

\(O\) is the centre of the circle.

What is the area of the shaded region of the circle. Answer correct to 2 significant figures.   (4 marks)

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\(\text{690 cm}^{2}\)

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\(\text{Method 1}\)

\(\text{Area}\ \Delta ODC\) \(= \dfrac{1}{2} ab \sin C\)  
  \(=\dfrac{1}{2} \times 30 \times 30 \times  \sin 72^{\circ} \)  
  \(=427.98\ \text{cm}^{2}\)  

 

\(\text{Shaded Area}\) \(=\ \text{Area of circle}-5 \times\ \text{Area}\ \Delta ODC\)  
  \(=(\pi \times 30^2)-(5 \times 427.98)\)  
  \(=687.5\)  
  \(=690\ \text{cm}^{2}\ \text{(2 sig.fig)}\)  

  

♦ Mean mark 48%.
\(\text{Method 2}\)

\(\text{Consider}\ \Delta ODC:\)

\(\text{Area}=427.98\ \text{cm}^{2}\ \ \text{(see above)}\)

 \(\text{Area of sector}\ ODC = \dfrac{72}{360} \times \pi \times 30^{2} = 565.49\ \text{cm}^{2}\)

\(\text{Shaded area (total)}\) \(=(565.49-427.98) \times 5\)  
  \(=690\ \text{cm}^{2}\ \text{(2 sig.fig)}\)  

Filed Under: Non-Right Angled Trig (Std2) Tagged With: Band 5, smc-804-30-Sine Rule (Area), smc-804-60-X-topic with PAV

Measurement, STD2 M6 2023 HSC 33

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

 

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place.  (4 marks)

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`23.8\ text{m}`

Show Worked Solution
`text{Arc}\ PQ` `=110/360 xx pi xx 2.1^2`  
  `=4.03171… \ text{m}`  

 
`text{Consider}\ ΔOPQ:`
 

♦ Mean mark 42%.
`sin 55^@` `=x/2.1`  
`x` `=2.1 xx sin 55^@`  
  `=1.7202…`  

 
`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}` `=8+(2xx3.6)+4.031+(8-3.440)`  
  `=23.79…`  
  `=23.8\ text{m  (to 1 d.p.)}`  

Filed Under: Non-Right Angled Trig (Std2) Tagged With: 2adv-std2-common, Band 5, common-content, smc-804-60-X-topic with PAV

Measurement, STD2 M6 2021 HSC 32

A right-angled triangle  `XYZ`  is cut out from a semicircle with centre `O`. The length of the diameter  `XZ`  is 16 cm and  `angle YXZ`  = 30°, as shown on the diagram.
 


 

  1. Find the length of  `XY`  in centimetres, correct to two decimal places.  (2 marks)

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  2. Hence, find the area of the shaded region in square centimetres, correct to one decimal place.  (3 marks)

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  1. `13.86 \ text{cm}`
  2. `45.1 \ text{cm}^2`
Show Worked Solution

 

a.    `cos 30^@` `=(XY)/16`
  `XY` `= 16 \ cos 30^@`
    `= 13.8564`
    `= 13.86 \ text{cm (2 d.p.)}`

 

b.    `text{Area of semi-circle}` `= 1/2 times pi r^2`
    `= 1/2 pi times 8^2`
    `= 100.531 \ text{cm}^2`

♦ Mean mark part (b) 36%.
`text{Area of} \ Δ XYZ` `= 1/2 ab\ sin C`  
  `= 1/2 xx 16 xx 13.856 xx sin 30^@`  
  `= 55.42 \ text{cm}^2`  

 

`:. \ text{Shaded Area}` `= 100.531-55.42`  
  `= 45.111`  
  `= 45.1 \ text{cm}^2 \ \ text{(1 d.p.)}`  

Filed Under: Non Right-Angled Trig, Non-Right Angled Trig (Std2) Tagged With: 2adv-std2-common, Band 4, Band 5, num-title-ct-pathc, num-title-qs-hsc, smc-4553-30-Sine Rule (Area), smc-804-30-Sine Rule (Area), smc-804-60-X-topic with PAV

Measurement, STD2 M6 2008 HSC 25c

Pieces of cheese are cut from cylindrical blocks with dimensions as shown.

 

Twelve pieces are packed in a rectangular box. There are three rows with four pieces of cheese in each row. The curved surface is face down with the pieces touching as shown.
  

  1. What are the dimensions of the rectangular box?  (4 marks)

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    To save packing space, the curved section is removed.
     
             
     

  2. What is the volume of the remaining triangular prism of cheese? Answer to the nearest cubic centimetre.    (2 marks)

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  1. `41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`
  2. `506\ text(cm)³\ text{(nearest whole)}`
Show Worked Solution

i.  `text(Box height) = 15\ text(cm)`

♦ Mean mark 45%.

`text{(radius of the arc)}`

`text(Box width)` `= 3 xx 7`
  `= 21\ text(cm)`
`text(Box length)` `= 4x`

`text(Using cosine rule)`

`c^2` `= a^2 + b^2 – 2ab cos C`
`x^2` `= 15^2 + 15^2 – 2 xx 15 xx 15 xx cos 40^@`
  `= 450 – 344.7199…`
  `= 105.2800…`
`x` `= 10.2606…`

 

`text(Box length)` `= 4 xx 10.2606…`
  `= 41.04…`

 
`:.\ text(Dimensions are)\ \ 41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`

 

ii.  `text(Volume) = Ah`

♦♦♦ Mean mark 22%.

`h = 7\ text(cm)`

`text(Find)\ A:`

`A` `= 1/2 ab sin C`
  `= 1/2 xx 15 xx 15 xx sin 40^@`
  `= 72.3136…`

 

`:. V` `= 72.3136… xx 7`
  `= 506.195…`
  `= 506\ text(cm³)\ \ text{(nearest whole)}`

Filed Under: Areas and Volumes (Harder), Non-Right Angled Trig, Non-Right Angled Trig (Std2), Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027) Tagged With: Band 5, Band 6, smc-6304-50-Volume (Circular Measure), smc-798-50-Volume (Circular Measure), smc-804-10-Cosine Rule, smc-804-30-Sine Rule (Area), smc-804-60-X-topic with PAV

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