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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)

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  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(m)^2`

`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 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)

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  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution

a.    `\text{Solution 1}`

`A` `~~50/2(360+300)+50/2(300+270)+50/2(270+140)+50/2(140+0)`  
  `~~25(660+570+410+140)`  
  `~~45\ 500\ \text{m}^3`  

 
`\text{Solution 2}`

`V~~50/2[360 + 2(300 + 270 + 140) + 0]~~ 44\ 500\ text(m)^3`
 

b.   `text(Convert 2 km)^2\ \text{to m}^2:`

♦♦♦ Mean mark (b) 11%.

`text(2 km)^2= 2000\ text(m) xx 1000\ \text{m} = 2\ 000\ 000\ text(m)^2`

`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 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)

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  1. `1445.5\ text(m)^2`
  2. `1\ 734\ 600\ text(L)`
Show Worked Solution

a.    `\text{Strategy 1}`

`text(Surface Area of pool)`

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

`+ 12.5/2(36.25 + 21.88)`

`~~ 1445.5\ text(m)^2`
 

`\text{Strategy 2}`

`text(Surface Area of pool)`

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

`~~ 1445.5\ text(m)^2`
 

♦ Mean mark (b) 50%.
b.    `V` `= Ah`
    `~~ 1445.5 xx 1.2`
    `~~ 1734.6\ text(m)^3`
    `~~ 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)

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a.    `426\ \text{m}^2`

b.    `63\ 900\ \text{times}`

Show Worked Solution

a.    `text(Area of lake = Area of rectangle)\ – text(Area of grass)`

`text(Area of rectangle)\ = 24 xx 55= 1320\ text(m)^2`

 `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(m)^2`

 
`:.\ text(Area of lake)~~ 1320-894~~ 426\ text(m)^2`
 

♦ Mean mark 44%
STRATEGY: Most students who did calculations in cm² and cm³ made errors. Keeping calculations in metres is much easier here.
b.    `V` `= Ah`
    `= 426 xx 0.6`
    `= 255.6\ text(m)^3`
    `= 255\ 600\ text(L)\ \ \ text{(1 m}^3\ \text{= 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)

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`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)~~3.3\ text(L)`