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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)

  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)

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}`

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)

  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)

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%.
 

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`

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)

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

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.)`

v1 Measurement, STD2 M1 2015 HSC 28c*

Three equally spaced cross-sectional areas of a vase are shown.
 

 
Use the Trapezoidal rule to find the approximate capacity of the vase in litres.  (3 marks)

Show Answers Only

`3\ text(litres)`

Show Worked Solution

`text(Solution 1)`

`V` `≈ 15/2(35 + 170) + 15/2(170 + 25)`
  `≈ 15/2(205 + 195)`
  `≈ 3000\ text{mL   (1 cm³ = 1 mL)}`
  `~~3\ text(L)`

`text(Solution 2)`

`V` `≈ 15/2(35 + 2 xx 170 + 25)`
  `≈ 15/2(400)`
  `≈ 3000\ text{mL}`
  `~~3 \ text(L)`

v1 Measurement, STD2 M1 SM-Bank 16 MC

The diagram represents the shape of a paddock.

Estimate the area of the paddock using four applications of the Trapezoidal Rule.

  1. 123 m²
  2. 129 m²
  3. 135 m²
  4. 141 m²
Show Answers Only

`B`

Show Worked Solution

`text(Interval width:) \ h = 3 \ text(m)`

`text(Heights:) \ 6, 10, 14, 10, 12`

`text(Apply Trapezoidal Rule:)`

`text(Area)` `=3/2(6 + 2×10 + 2×14 + 2×10 + 12)`
  `= 3/2(6 + 20 + 28 + 20 + 12)`
  `=3/2(86)`
  `= 129 \ text(m)^2`