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Measurement, STD2 M1 2025 HSC 26*

A toy has a curved surface on the top which has been shaded as shown. The toy has a uniform cross-section and a rectangular base.
 

  1. Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy.   (2 marks)

  2. The total surface area of the plastic toy is 1300 cm².
  3. What is the approximate area of the curved surface?   (2 marks)

Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

Show Worked Solution

a.   \(\text{Solution 1}\)

\(A\) \(=\dfrac{5.1}{2}(6.0+3.8) + \dfrac{5.1}{2}(3.8+0) \)  
  \(=34.68\ \text{cm}^2\)  

 
\(\text{Solution 2}\)

\(\begin{array}{|c|c|c|c|}
\hline\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 5.1 \quad & \quad 10.2 \quad\\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 6 & 3.8 & 0 \\
\hline
\end{array}\)

\(A\) \(\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)\)
  \(\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)\)
  \(\approx 34.68 \ \text{cm}^2\)

 

b.    \(\text{Toy has 5 sides.}\)

\(\text{Area of base}=10.2 \times 40=408 \ \text{cm}^2\)

\(\text{Area of rectangle}=6.0 \times 40=240 \ \text{cm}^2\)

\(\text{Approximated areas}=2 \times 34.68=69.36 \ \text{cm}^2\)

\(\therefore \ \text{Area of curved surface}\) \(=1300-(408+240+69.36)=582.64 \ \text{cm}^2\)
♦ Mean mark (b) 48%.

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

a.    `48.96\ text(cm)^2`

b.    `3502.88\ text(cm)^2`

Show Worked Solution

a.    `\text{Solution 1}`

`A` `~~3.6/2(5+4.6)+3.6/2(4.6+3.7)+3.6/2(3.7+2.8)+3.6/2(2.8+0)`  
  `~~1.8(9.6+8.3+6.5+2.8)`  
  `~~ 48.96\ text(cm)^2`  

 
`\text{Solution 2}`

`A~~ 3.6/2 [5 + 2(4.6 + 3.7 + 2.8) + 0]~~ 48.96\ text(cm)^2`
 

b.    `text(Total Area) = 7480.8\ text{cm}^2\ \text{(given)}`

`text(Area of Base)= 14.4 xx 200= 2880\ text(cm)^2`

`text(Area of End)= 5 xx 200= 1000\ text(cm)^2`

`text(Area of sides)= 2 xx 48.96= 97.92\ text(cm)^2`
 

`:.\ text(Area of curved surface)`

`= 7480.8-(2880 + 1000 + 97.92)`

`= 3502.88`

`=3503\ text{cm}^2\ \text{(nearest cm}^2\text{)}`