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