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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: IntelxD on September 08, 2014, 08:33:10 pm

Title: Assistance with a Geo - Trig question
Post by: IntelxD on September 08, 2014, 08:33:10 pm: 'The combined total surface area of a hemisphere of radius 28mm mounted on top of a right circular cone with slant height 40mm is closest to:'

a) 8444.6 mm^2
b) 9221 mm^2
c) 14146.5 mm^2
d) 16609.5 mm^2

I know that the answer is A however I am not sure how to work it out. If you could provide an answer with your explanation I would be very grateful.
Title: Re: Assistance with a Geo - Trig question
Post by: keltingmeith on September 08, 2014, 08:41:13 pm: TSA of a sphere is $4\pi r^2$ , so the TSA of a hemisphere is half that.

The TSA of a cone is $\pi r^2 + \pi rh$ , minus the circle that's normally present, is $\pi r h$ .

So, the TSA of our shape is $2\pi r^2 + \pi rh$ . Plugging in r=28 and h=40, we get A.
Title: Re: Assistance with a Geo - Trig question
Post by: IntelxD on September 08, 2014, 10:09:20 pm: Quote from: EulerFan101 on September 08, 2014, 08:41:13 pm
TSA of a sphere is $4\pi r^2$ , so the TSA of a hemisphere is half that.

The TSA of a cone is $\pi r^2 + \pi rh$ , minus the circle that's normally present, is $\pi r h$ .

So, the TSA of our shape is $2\pi r^2 + \pi rh$ . Plugging in r=28 and h=40, we get A.

Thank you very much.

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