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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 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 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 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)`