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)
--- 6 WORK AREA LINES (style=lined) ---
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)
--- 4 WORK AREA LINES (style=lined) ---
- 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)
--- 4 WORK AREA LINES (style=lined) ---
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 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)
--- 6 WORK AREA LINES (style=lined) ---
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)`