smc-941-10-1-3 Approximations

Measurement, STD2 M1 2023 HSC 24

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

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

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

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`18\ text{m}^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.

Using two applications of the trapezoidal rule, calculate the approximate area of the grassed section. (2 marks)

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

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`35\ 500\ text{m}^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?

99
19.8
39.6
72

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`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.

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

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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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`850 \ text{m}^2`
`text{The estimate will be more than actual area}`

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`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%.

`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.

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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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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`text(See Worked Solutions)`
`1600\ text(m)^3`

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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)`

`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 2009 HSC 25c*

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

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.

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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426 m²
63 900 times

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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 SM-Bank 15 MC

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

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

720 m²
880 m²
1140 m²
1440 m²

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`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)

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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 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.

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

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

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

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`3.96\ text(m²)`
`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.)`