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.

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

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

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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%.