Author Topic: Q5 + Q6 2008 FM Exam 1 (Read 595 times) Tweet Share

Einstein · « **on:** June 05, 2014, 08:08:26 pm »

Could someone please show me how these are done?

thanks

Conic · « **Reply #1 on:** June 05, 2014, 10:37:48 pm »

Q6: The total area is the area of the base (rectangle), the two ends (semi circles), and the curved surface (half a cylinder). So we have

$\mathrm{Area}_{\mathrm{total}}=(2.5\cdot1.5)+2\cdot\left ( \frac{1}{2}\cdot\pi \cdot\left (\frac{1.5}{2} \right )\right )^2+\left ( \frac{1}{2}\cdot2\pi \cdot\left (\frac{1.5}{2}\right )\cdot(2.5) \right )=11.4$ .

So D is correct. (We have 1.5/2 to convert the diameter to radius).

Q5: By the sine rule we have

$\frac{\sin(x^{\circ})}{4}=\frac{\sin(125.1^{\circ})}{8}\implies \sin(x^{\circ})=4\cdot\frac{\sin(125.1^{\circ})}{8}=\frac{\sin(125.1^{\circ})}{2}$ .

So A is correct.

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Author Topic: Q5 + Q6 2008 FM Exam 1 (Read 595 times) Tweet Share

Einstein

Q5 + Q6 2008 FM Exam 1

Conic

Re: Q5 + Q6 2008 FM Exam 1

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