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Trigonometric Functions (1 Viewer)

mitchmiles

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Here's a trigonometric functions question that has everyone confused.

A sector of a circle with radius 5cm and an angle π/3 subtended at the centre is cut out of cardboard. it is then curved around to form a cone. Find its exact surface area and volume. (π=pi)

Answers are:
SA=175π/36 cm^2
and
V=[125(35)^1/2)]π/648 cm^3

Show working please :)
 

rumbleroar

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I don't have pen and paper with me right now, but treat the arc length of the circle as the circumference as the base? And the slant is the 5cm. As for the SA....are they using it as a closed or open cone? If it's open, subtract the area of the base.

Hopefully that helps a bit :/
 

Drongoski

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Here's a trigonometric functions question that has everyone confused.

A sector of a circle with radius 5cm and an angle π/3 subtended at the centre is cut out of cardboard. it is then curved around to form a cone. Find its exact surface area and volume. (π=pi)

Answers are:
SA=175π/36 cm^2
and
V=[125(35)^1/2)]π/648 cm^3

Show working please :)
Let R = radius of base circle of cone

arc length of cut-out = (pi/3)x5 cm

.: 2 x pi x R = (pi/3) x 5 cm ==> R = (5/6) cm

Area of cut-out = 0.5 x (5cm)^2 x (pi/3) = (25/6) x pi cm^2

Area of base of cone = pi x (5/6)^2 cm^2 = 25/36 pi cm^2

.: total surface area = sum of these 2 areas = (175/36) pi cm^2

[But the question was not clear as to whether or not the surface area of the base of the cone was to be included! If cone was made from the cut-out, then the base would have been 'empty', in which case the area = that of the cut-out only]
The vertical height of the cone, h cm, is given by: h^2 = 5^2 - (5/6)^2 = 5^2 x (35/36)

.: vol of cone = (1/3) x area of base x height = (1/3) x pi x (5/6)^2 x 5 x sqrt(35/36) cm^3

= (125/648)sqrt(35) pi cm^3
 
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