Part B (9 marks)
MS11-4 Performs calculations in relation to two-dimensional and three-dimensional figures.
Question 2
A wedding cake with three tiers rests on the table. Each tier is $6 cm$ high. The layers have radii of $20 cm, 15 cm$ and $10 cm$ respectively.
$2 π r \times h$
a) Calculate the area of the cake, that is iced, correct to the nearest ${cm}^{2}$ ?
(4 marks)
$\begin{aligned} A^{'} & = 2 r \times h \\ = 2 \times r \times 10 \times 6 \\ = 376.99 \\ A^{2} & = 2 π \times \times h \\ = 2 \times π \times 15 \times 6 \\ = 565.4 \end{aligned}$
$A^{3} = 2 π r \times h$
b) Find the total volume of the cake, correct to the nearest ${cm}^{3}$ ?
(2 marks)
6

Answer

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Answer

\boxed{\text{Total iced area} \approx 1696 \text{ cm}^2, \text{Total volume} \approx 13666 \text{ cm}^3}

Steps

Step 1 ： $A_{1} = 2 π r_{1} h = 2 π (20) (6) = 753.98 {cm}^{2}$

Step 2 ： $A_{2} = 2 π r_{2} h = 2 π (15) (6) = 565.49 {cm}^{2}$

Step 3 ： $A_{3} = 2 π r_{3} h = 2 π (10) (6) = 376.99 {cm}^{2}$

Step 4 ： $A_{t o t a l} = A_{1} + A_{2} + A_{3} = 753.98 + 565.49 + 376.99 = 1696.46 {cm}^{2}$

Step 5 ： $V_{1} = π r_{1}^{2} h = π (20)^{2} (6) = 7539.82 {cm}^{3}$

Step 6 ： $V_{2} = π r_{2}^{2} h = π (15)^{2} (6) = 4241.15 {cm}^{3}$

Step 7 ： $V_{3} = π r_{3}^{2} h = π (10)^{2} (6) = 1884.96 {cm}^{3}$

Step 8 ： $V_{t o t a l} = V_{1} + V_{2} + V_{3} = 7539.82 + 4241.15 + 1884.96 = 13665.93 {cm}^{3}$

Step 9 ：\boxed{\text{Total iced area} \approx 1696 \text{ cm}^2, \text{Total volume} \approx 13666 \text{ cm}^3}