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Measurement, STD2 M1 2025 HSC 26

A toy has a curved surface on the top which has been shaded as shown. The toy has a uniform cross-section and a rectangular base.
 

  1. Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy.   (2 marks)

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  2. The total surface area of the plastic toy is 1300 cm².
  3. What is the approximate area of the curved surface?   (2 marks)

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  4. The measurements shown on the diagram are given to the nearest millimetre.
  5. What is the percentage error of the measurement of 10.2 cm? Give your answer correct to 3 significant figures.   (2 marks)

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Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

c.   \(0.490 \%\)

Show Worked Solution

a.    \(\begin{array}{|c|c|c|c|}
\hline\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 5.1 \quad & \quad 10.2 \quad\\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 6 & 3.8 & 0 \\
\hline
\end{array}\)

\(A\) \(\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)\)
  \(\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)\)
  \(\approx 34.68 \ \text{cm}^2\)

 

b.    \(\text{Toy has 5 sides.}\)

\(\text{Area of base}=10.2 \times 40=408 \ \text{cm}^2\)

\(\text{Area of rectangle}=6.0 \times 40=240 \ \text{cm}^2\)

\(\text{Approximated areas}=2 \times 34.68=69.36 \ \text{cm}^2\)

\(\therefore \ \text{Area of curved surface}\) \(=1300-(408+240+69.36)\)
  \(=582.64 \ \text{cm}^2\)

 

c.   \(\text{Convert cm to mm:}\)

\(10.2 \ \text{cm}\times 10=102 \ \text{mm}\)

\(\text{Absolute error}=\dfrac{1}{2} \times \text{precision}=0.5 \ \text{mm}\)

\(\% \ \text{error}=\dfrac{0.5}{102} \times 100=0.490 \%\)

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027), Units and Measurement Error (Std 2) Tagged With: Band 3, Band 4, Band 5, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

v1 Measurement, STD2 M1 2020 HSC 27

The shaded region on the diagram represents a lot. Each grid represents 6 m × 6 m.
 


 

  1. Use two applications of the trapezoidal rule to calculate the approximate area of the lot.   (3 marks)

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  2. Should the answer to part (a) be more than, equal to or less than the actual area of the lot? Referring to the diagram above, briefly explain your answer.   (2 marks)

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Show Answers Only
  1. `1080 \ text{m}^2`
  2. `text{The estimate will be more than actual area}`
Show Worked Solution

a.     

`h = 4 xx 6 = 24 \ text{m}`

♦ Mean mark 47%.
`text{Area}` `= frac{h}{2} (x_1 + x_2) + frac{h}{2} (x_2 + x_3)`
   `= frac{24}{2} (30 + 18) + frac{24}{2} (18 + 24}`
  `= 576 + 504`
  `= 1080 \ text{m}^2`

 

♦♦ Mean mark 23%.

b.     

`text{The trapezoidal rule captures the shaded area plus the}`

`text{the extra area highlighted above.}`

`therefore \ text{The estimate will be more than actual area}`

Filed Under: Trapezoidal Rule (Std2-X) Tagged With: Band 5, smc-941-10-1-3 Approximations

v1 Measurement, STD2 M1 2023 HSC 24

The diagram shows the cross-section of a wall across a creek. 
 


 
  1. Use two applications of the trapezoidal rule to estimate the area of the cross-section of the wall.  (2 marks)

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  2. The wall has a uniform thickness of 0.9 m. The weight of 1 m³ of concrete is 3.52 tonnes.  
  3. How many tonnes of concrete are in the wall? Give the answer to two significant figures.  (3 marks)

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Show Answers Only
  1. `21.25\ text{m}^2`
  2. `text{67 tonnes}`
Show Worked Solution

a.    `h=10.0/2=5`

`A` `~~h/2[2+1.5+2(2.5)]`  
  `~~5/2(8.5)`  
  `~~21.25\ text{m}^2`  

 
b.
   `V_text{wall}=21.25 xx 0.9=19.13\ text{m}^3`

`text{Mass of concrete}` `=19.13 xx 3.52`  
  `=67.33`  
  `=67\ text{tonnes (2 sig.fig.)}`  
♦ Mean mark (b) 46%.
 

Filed Under: Trapezoidal Rule (Std2-X) Tagged With: Band 4, Band 5, smc-799-30-Mass, smc-941-10-1-3 Approximations

v1 Measurement, STD2 M1 2021 HSC 12 MC

A block of land is represented by the shaded region on the number plane. All measurements are in kilometres. 
 

What is the approximate area of the land, in square kilometres, using two applications of the trapezoidal rule?

  1. 11.25
  2. 14.85
  3. 16.75
  4. 22.55
Show Answers Only

`C`

Show Worked Solution

Using two applications of the trapezoidal rule:

`\text{Area}` `≈ \frac{5}{2} (1.2 + 2 × 2 + 1.5)`
  `= 2.5 (1.2 + 4 + 1.5)`
  `= 2.5 × 6.7`
  `= 16.75 \ \text{km}^2`

`⇒ C`

Filed Under: Trapezoidal Rule (Std2-X) Tagged With: Band 4, smc-941-10-1-3 Approximations

v1 Measurement, STD2 M1 2013 HSC 15a*

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.4 m apart. The height of each outer pole is 1.6 m, and the height of the middle pole is 2 m. The roof hangs between the poles.

The front of the tent has area `A\ text(m²)`. 

  1. Use the trapezoidal rule to estimate `A`.    (2 marks)

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  2. Explain whether the trapezoidal rule give a greater or smaller estimate of  `A`?  (1 mark)

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Show Answers Only
  1. `5.04\ text(m²)`
  2. `text(The trapezoidal rule assumes a straight line between)`

     

    `text(all points and therefore would estimate a greater)`

     

    `text(area than the actual area of the tent front.)`

Show Worked Solution
i.    `A` `~~ h/2 [y_0 + 2y_1 + y_2]`
    `~~ 1.4/2 [1.6 + (2 xx 2) + 1.6]`
    `~~ 0.7 [7.2]`
    `~~ 5.04\ text(m²)`

 

ii.  `text(The trapezoidal rule assumes a straight line between)`

`text(all points and therefore would estimate a greater)`

`text(area than the actual area of the tent front.)`

Filed Under: Trapezoidal Rule (Std2-X) Tagged With: Band 4, Band 5, smc-941-10-1-3 Approximations

Measurement, STD2 M1 2023 HSC 24

The diagram shows the cross-section of a wall across a creek. 
 


 
  1. Use two applications of the trapezoidal rule to estimate the area of the cross-section of the wall.  (2 marks)

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  2. The wall has a uniform thickness of 0.80 m. The weight of 1 m³ of concrete is 3.52 tonnes.  
  3. How many tonnes of concrete are in the wall? Give the answer to two significant figures.  (3 marks)

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Show Answers Only
  1. `18\ text{m}^2`
  2. `text{51 tonnes}`
Show Worked Solution

a.    `h=8.0/2=4`

`A` `~~h/2[1.9+1.7+2(2.7)]`  
  `~~4/2(9)`  
  `~~18\ text{m}^2`  

 
b.
   `V_text{wall}=18 xx 0.8=14.4\ text{m}^3`

`text{Mass of concrete}` `=14.4 xx 3.52`  
  `=50.688`  
  `=51\ text{tonnes (2 sig.fig.)}`  
♦ Mean mark (b) 46%.
 

Filed Under: Energy and Mass (Std 2), Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 4, Band 5, smc-6328-10-1-3 Approximations, smc-799-30-Mass, smc-941-10-1-3 Approximations

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)

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

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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%.

Filed Under: Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027), Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 4, Band 5, smc-6304-20-Perimeter and Area (Circular Measure), smc-6328-10-1-3 Approximations, smc-798-20-Perimeter and Area (Circular Measure), smc-941-10-1-3 Approximations

Measurement, STD2 M1 2021 HSC 12 MC

A block of land is represented by the shaded region on the number plane. All measurements are in kilometres. 
 

Which of the following is the approximation for the area of this block of land in square kilometres, using two applications of the trapezoidal rule?

  1. 99
  2. 19.8
  3. 39.6
  4. 72
Show Answers Only

`B`

Show Worked Solution

`\text{Area}` `≈ \frac{6}{2} (1.2 + 2 \times 2 + 1.4)`
  `≈ 3 (6.6)`
  `≈ 19.8 \ \text{km}^2`

 
`=> B`

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 4, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

Measurement, STD2 M1 2020 HSC 27

The shaded region on the diagram represents a garden. Each grid represents 5 m × 5 m.
 


 

  1. Use two applications of the trapezoidal rule to calculate the approximate area of the garden.   (3 marks)

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  2. Should the answer to part (a) be more than, equal to or less than the actual area of the garden? Referring to the diagram above, briefly explain your answer.   (2 marks)

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Show Answers Only
  1. `850 \ text{m}^2`
  2. `text{The estimate will be more than actual area}`
Show Worked Solution

a.     

`h = 4 xx 5 = 20 \ text{m}`

♦ Mean mark 47%.
`text{Area}` `= frac{h}{2} (x_1 + x_2) + frac{h}{2} (x_2 + x_3)`
   `= frac{20}{2} (25 + 20) + frac{20}{2} (20 + 20}`
  `= 450 + 400`
  `= 850 \ text{m}^2`

 

♦♦ Mean mark 23%.

b.     

`text{The trapezoidal rule captures the shaded area plus the}`

`text{the extra area highlighted above.}`

`therefore \ text{The estimate will be more than actual area}`

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 5, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

Measurement, STD2 M7 2019 HSC 41

A map is drawn to scale, on 1-cm paper, showing the position of a supermarket and a cinema. A reservoir is also shown.
 


 

  1. It takes 10 minutes to walk in a straight line from the cinema to the supermarket at a constant speed of 3 km/h. Show that the scale of the map is 1 cm = 100 m.  (3 marks)

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  2. The reservoir is initially empty. During a storm 20 mm of rain falls on the reservoir.

     

    With the aid of one application of the trapezoidal rule, estimate the amount of water in the reservoir immediately after the storm. Assume that all rain which falls over the reservoir is stored. Give your answer in cubic metres.  (3 marks)

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Show Answers Only
  1. `text(See Worked Solutions)`
  2. `1600\ text(m)^3`
Show Worked Solution

a.   `text(3 km/h = 3000 metres per 60 minutes)`

♦ Mean mark part (a) 49%.

`text(In 10 minutes:)`

`text(Actual distance) = 3000 xx 10/60 = 500\ text(metres)`

`text(Distance on map = 5 cm)`

`:.\ text(Scale   5 cm)` `: 500\ text(metres)`
`text(1 cm)` `: 100\ text(metres)`

 

b.   

 

`A` `~~ h/2(a + b)`
  `~~ 400/2(100 + 300)`
  `~~ 80\ 000\ text(m²)`

 
`text(Converting mm to metres:)`

♦♦ Mean mark part (b) 28%.

`text(20 mm) = 20/1000 text(m = 0.02 metres)`

 
`:.\ text(Volume of water)`

`= A xx h`

`= 80\ 000 xx 0.02`

`= 1600\ text(m)^3`

Filed Under: Ratio and Scale (Std2), Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 5, smc-1187-40-Maps and Scale Drawings, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

Measurement, STD2 M1 2009 HSC 25c*

There is a lake inside the rectangular grass picnic area  `ABCD`, as shown in the diagram.
 

2UG-2009-25c
  

  1. Use Trapezoidal’s Rule to find the approximate area of the lake’s surface.   (3 marks)

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The lake is 60 cm deep. Bozo the clown thinks he can empty the lake using a four-litre bucket.

  1. How many times would he have to fill his bucket from the lake in order to empty the lake? (Note that 1 m³ = 1000 L).    (2 marks)

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Show Answers Only
  1. 426 m²
  2. 63 900 times
Show Worked Solution
i.     `text(Area of lake = Area of rectangle)\ – text(Area of grass)`
`text(Area of rectangle)` `= 24 xx 55`
  `= 1320\ text(m²)`

 
`text{Area of grass (two applications)}`

`~~ 12/2(20 + 5) + 12/2(5 + 10) + 12/2(35 + 22) + 12/2(22 + 30)`
`~~ 6(25 + 15 + 57 + 52)`
`~~ 894\ text(m²)`

 

`:.\ text(Area of lake)` `~~ 1320\ – 894`
  `~~ 426\ text(m²)`

 

♦ Mean mark 44%
STRATEGY: Most students who did calculations in cm² and cm³ made errors. Keeping calculations in metres is much easier here.
ii.    `V` `= Ah`
    `= 426 xx 0.6`
    `= 255.6\ text(m³)`
    `= 255\ 600\ text(L)\ \ \ text{(1 m³ = 1000  L)}`

 

`:.\ text(Times to fill bucket)` `= 255\ 600 -: 4`
  `= 63\ 900`

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 4, Band 5, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

Measurement, STD2 M1 SM-Bank 15 MC

The shaded region represents a block of land bounded on one side by a road.
 

2UG-2005-12MC

 
What is the approximate area of the block of land, using the Trapeziodal rule?

  1.  720 m²
  2.  880 m²
  3.  1140 m²
  4.  1440 m²
Show Answers Only

`A`

Show Worked Solution

`text(Area)` `≈ 20/2(23 + 15) + 20/2(15 + 19)`
  `≈ 10(38) + 10(34)`
  `≈ 720\ text(m²)`

 
`=> A`

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 3, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

Measurement, STD2 M1 SM-Bank 12

The scale diagram shows the aerial view of a block of land bounded on one side by a road. The length of the block, `AB`, is known to be 90 metres.
 


 

Calculate the approximate area of the block of land, using three applications of the Trapezoidal rule.  (3 marks)

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Show Answers Only

5175 m²

Show Worked Solution

`text(Solution 1)`

`text(6 cm → 90 metres)`

` text(1 cm → 15 metres)`
 

`text(Height) = 2 xx 15 = 30\ text(metres)`

`text(Area)` `~~ 30/2(75 + 60) + 30/2(60 + 45) + 30/2(45 + 60)`
  `~~ 15(135 + 105 + 105)`
  `~~ 5175\ text(m²)`

 

`text(Solution 2)`

`text(After converting from scale:)`

`text(Area)` `~~ 30/2(75 + 2 xx 60 + 2 xx 45 + 60)`
  `~~ 5175\ text(m²)`

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 4, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

Measurement, STD2 M1 2013 HSC 15a*

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.2 m apart. The height of each outer pole is 1.5 m, and the height of the middle pole is 1.8 m. The roof hangs between the poles.

2013 15a

The front of the tent has area `A\ text(m²)`. 

  1. Use the trapezoidal rule to estimate `A`.    (2 marks)

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  2. Explain whether the trapezoidal rule give a greater or smaller estimate of  `A`?  (1 mark)

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Show Answers Only
  1. `3.96\ text(m²)`
  2. `text(The trapezoidal rule assumes a straight line between)`

     

    `text(all points and therefore would estimate a greater)`

     

    `text(area than the actual area of the tent front.)`

Show Worked Solution
i.    `A` `~~ h/2 [y_0 + 2y_1 + y_2]`
    `~~ 1.2/2 [1.5 + (2 xx 1.8) + 1.5]`
    `~~ 0.6 [6.6]`
    `~~ 3.96\ text(m²)`

 

ii.  `text(The trapezoidal rule assumes a straight line between)`

`text(all points and therefore would estimate a greater)`

`text(area than the actual area of the tent front.)`

Filed Under: Trapezoidal Rule (Std 2), Trapezoidal Rule (Std2-2027) Tagged With: Band 4, Band 5, smc-6328-10-1-3 Approximations, smc-941-10-1-3 Approximations

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