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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)

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

  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)

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.   \(\text{Convert cm to mm:}\)

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

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

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

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 0.80 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)

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%.
 

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)

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

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%.

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`

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)

  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)

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}`

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)

  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)

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`

Measurement, STD2 M1 2007 HSC 28c*

A piece of plaster has a uniform cross-section, which has been shaded, and has dimensions as shown.
 

 
 

  1. Use the Trapezoidal rule to approximate the area of the cross-section.    (3 marks)

  2. The total surface area of the piece of plaster is 7480.8 cm²
  3. Calculate the area of the curved surface as shown on the diagram. Give your answer to the nearest square centimetre   (2 marks)

Show Answers Only
  1. `48.96\ text(cm²)`
  2. `3502.88\ text(cm²)`
Show Worked Solution
i.   
`A` `~~ 3.6/2 [5 + 2(4.6 + 3.7 + 2.8) + 0]`
  `~~ 1.8(27.2)`
  `~~ 48.96\ text(cm²)`

 

ii.  `text(Total Area) = 7480.8\ text{cm²   (given)}`

`text(Area of Base)` `= 14.4 xx 200`
  `= 2880\ text(cm²)`
`text(Area of End)` `= 5 xx 200`
  `= 1000\ text(cm²)`
`text(Area of sides)` `= 2 xx 48.96`
  `= 97.92\ text(cm²)`

 

`:.\ text(Area of curved surface)`

`= 7480.8 – (2880 + 1000 + 97.92)`

`= 3502.88`

`=3503\ text{cm²  (nearest cm²)}`

Measurement, STD2 M1 2017 HSC 29a*

A new 200-metre long dam is to be built.

The plan for the new dam shows evenly spaced cross-sectional areas.
 
 


 

  1. Using the Trapezoidal rule, show that the volume of the dam is approximately 44 500 m³.  (2 marks)

  2. It is known that the catchment area for this dam is 2 km².

     

    Assuming no wastage, calculate how much rainfall is needed, to the nearest mm, to fill the dam.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution
i.   
`V` `~~ 50/2[360 + 2(300 + 270 + 140) + 0]`
  `~~ 25(1780)`
  `~~ 44\ 500\ text(m³)`

 

ii.   `text(Convert  2 km² → m²:)`

♦♦♦ Mean mark 11%.
`text(2 km²)` `= 2000\ text(m × 1000 m)`
  `= 2\ 000\ 000\ text(m²)`

 

`text(Using)\ \ V=Ah\ \ text(where)\ \ h= text(rainfall):`

`44\ 500` `= 2\ 000\ 000 xx h`
`:.h` `= (44\ 500)/(2\ 000\ 000)`
  `= 0.02225…\ text(m)`
  `= 22.25…\ text(mm)`
  `= 22\ text{mm  (nearest mm)}`

Measurement, STD2 M1 2008 HSC 28b*

A tunnel is excavated with a cross-section as shown.

 

  1. Find an expression for the area of the cross-section using the Trapezoidal rule.  (2 marks)

  2. The area of the cross-section must be 600 m2. The tunnel is 80 m wide. 

     

    If the value of `a` increases by 2 metres, by how much will `b` change?   (2 marks)

Show Answers Only
  1. `h(2a + b)`
  2. `b\ text(decreases by 4.)`
Show Worked Solution
i.   
`A` `~~ h/2[0 + 2(a + b + a) + 0]`
  `~~ h/2(4a + 2b)`
  `~~ h(2a + b)`

 

ii.   `A = 600\ text(m²)`

`text(If tunnel is 80 metres wide)`

`4h` `= 80`
`h` `= 20`

 
`text{Using part (i):}`

`600 = 20(2a + b)`

`2a + b` `= 30`
`b` `= 30 – 2a`

 
`:.\ text(If)\ a\ text(increases by 2,)\ b\ text(must decrease by 4.)`

Measurement, STD2 M1 2014 HSC 28d*

An aerial diagram of a swimming pool is shown. 

The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.

In the diagram of the swimming pool, the five widths are measured to be: 
 

`CD = 21.88\ text(m)`

`EF = 25.63\ text(m)`

`GH = 31.88\ text(m)`

`IJ = 36.25\ text(m)`

`KL = 21.88\ text(m)` 
 

  1. Use four applications of the Trapezoidal Rule to calculate the surface area of the pool.  (2 marks)

  2.  The average depth of the pool is 1.2 m

     

    Calculate the approximate volume of the swimming pool, in litres.  (1 mark)

Show Answers Only
  1. `1445.5\ text(m²)`
  2. `1\ 734\ 600\ text(L)`
Show Worked Solution
i.   

`text(Surface Area of pool)`

`~~ 12.5/2[21.88 + 2(25.63 + 31.88 + 36.25) + 21.88]`

`~~ 1445.5\ text(m²)`
 

♦ Mean mark 50%. Be careful not to give away easy marks! 
ii.    `V` `= Ah`
    `~~ 1445.5 xx 1.2`
    `~~ 1734.6\ text(m³)`
    `~~ 1\ 734\ 600\ text(L)`

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)

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)

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`

Measurement, STD2 M1 2015 HSC 28c*

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

2UG 2015 29c

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

Show Answers Only

`3.3\ text(litres)`

Show Worked Solution

`text(Solution 1)`

`V` `≈ 15/2(45 + 180) + 15/2(180 + 35)`
  `≈ 15/2(225 + 215)`
  `≈ 3300\ text{mL   (1 cm³ = 1 mL)}`
  `~~3.3\ text(L)`

 

`text(Solution 2)`

`V` `≈ 15/2(45 + 2 xx 180 + 35)`
  `≈ 15/2(440)`
  `≈ 3300\ text{mL}`
  `~~3.3\ text(L)`

Measurement, STD2 M1 SM-Bank 16 MC

The diagram represents a field.
 

What is the area of the field, using four applications of the Trapezoidal’s rule?

  1. 105 m²
  2. 136 m²
  3. 210 m²
  4. 420 m²
Show Answers Only

`A`

Show Worked Solution

`text(Solution 1)`

`text(Area)` `~~ 3/2(6 + 7) + 3/2(7 + 12) + 3/2(12 + 8) + 3/2(8 + 10)`
  `~~ 3/2(13 + 19 + 20 + 18)`
  `~~ 105\ text(m²)`

 

`text(Solution 2)`

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & 9 & 12 \\
\hline
\rule{0pt}{2.5ex} \text{height} \rule[-1ex]{0pt}{0pt} & \ \ \ 6\ \ \  & \ \ \ 7\ \ \  & \ \ 12\ \  & \ \ \ 8\ \ \  & \ \ 10\ \  \\
\hline
\rule{0pt}{2.5ex} \text{weight} \rule[-1ex]{0pt}{0pt} & 1 & 2 & 2 & 2 & 1 \\
\hline
\end{array}

`text(Area)` `~~ 3/2(6 + 2 xx 7 + 2 xx 12 + 2 xx 8 + 10)`
  `~~ 3/2(70)`
  `~~ 105\ text(m²)`

 
`=> A`

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`

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)

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)`

Measurement, STD2 M1 2018 HSC 28a

A field is bordered on one side by a straight road and on the other side by a river, as shown. Measurements are taken perpendicular to the road every 7.5 metres along the road.
 

 
Use four applications of the Trapeziodal rule to find an approximation to the area of the field. Answer to the nearest square metre.  (3 marks)

Show Answers Only

`242\ text{m²  (nearest m²)}`

Show Worked Solution

`text(Strategy 1)`

`A` `~~ 7.5/2(8.8 + 7.1) + 7.5/2(7.1 + 9.8) + 7.5/2(9.8 + 8.5) + 7.5/2(8.5 + 4.9)`
  `~~ 241.875`
  `~~ 242\ text{m²  (nearest m²)}`

 

`text(Strategy 2)`

`A` `~~ 7.5/2(8.8 + 2 xx 7.1 + 2 xx 9.8 + 2 xx 8.5 + 4.9)`
  `~~ 242\ text(m²)`

Measurement, STD2 M1 SM-Bank 2

A farmer wants to estimate the area of an irregular shaped paddock.
 

 
What is the estimated area of the land using the Trapezoidal Rule?  (2 marks)

Show Answers Only

`370\ text(m²)`

Show Worked Solution

`text(Solution 1)`

`text(Height) = 20 div 4 = 5\ text(m)`

`text(Area)` `~~ 5/2 (28 + 18) + 5/2 (18 + 17) + 5/2 (17 + 16) + 5/2 (16 + 18)`
  `~~ 370\ text(m²)`

 

`text(Solution 2)`

  `x` `0` `5` `10` `15` `20`
  `text(height)` `28` `18` `17` `16` `18`
  `text(weight)` `1` `2` `2` `2` `1`
`text(Area)` `~~ h/2 [28 + 2 (18 + 17 + 16) + 18]`
  `~~ 5/2 xx 148`
  `~~ 370\ text(m²)`

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)

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

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.)`