Im confused can any1 help me on this question ?? or try and put me on the right track?
(a) A circle has radius 2cm (in that circle there is an triangle AOB) and the chord AB has length 3cm. AO is 2cm and BO is 2cm. O is the centre of the circle.
(b) consider a paper disc whose boundary is the circle in part (a). The sector AOB is removed from the paper disc and the remaining paper is bent round to form a shallow cone by joining the edge OB to the edge OA.
(i) Which arc becomes the circle that is at the base of the cone? Find the angle subtended by this arc in the original circle (in radians), and hence find the length of the arc.
(ii) Hence find the radius of the circle that is at the base of the cone.
(iii) Find the perpendicular height of the cone.
(iv)Given that the formula for the volume of a cone is V = 1/3 pie r^2 h
(Where r is the radius of the base and h is the perpendicular
Height),
find the volume of the cone in this question
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Sure, I can help you with this question. Let's break it down step by step.
(i) To find which arc becomes the circle at the base of the cone, we need to identify the arc that is opposite to the chord AB. Since the length of AB is 3cm, this means that the arc opposite AB has a length of 3cm as well.
To find the angle subtended by this arc in the original circle, we can use the formula: angle = length of arc / radius. In this case, the length of the arc is 3cm and the radius is 2cm, so the angle subtended by this arc is 3/2 radians.
(ii) Now, to find the radius of the circle that is at the base of the cone, we need to use the fact that the circumference of a circle is equal to 2π times the radius. Since we know the length of the arc (3cm) is equal to 2π times the radius of the base circle, we can set up the equation:
3 = 2π * radius
radius = 3 / (2π)
(iii) To find the perpendicular height of the cone, we can use the Pythagorean theorem. In this case, the two sides of the right triangle are the radius (2cm) and the height h, and the hypotenuse is the slant height of the cone.
Using the Pythagorean theorem, we have:
radius^2 + height^2 = slant height^2
Since we removed a sector AOB with radius 2cm from the paper disc, the slant height of the cone is the length of the chord AB, which is 3cm.
So, we have:
2^2 + height^2 = 3^2
4 + height^2 = 9
height^2 = 9 - 4
height^2 = 5
height = √5
(iv) Finally, to find the volume of the cone, we can use the formula V = (1/3)π * radius^2 * height. We already calculated the radius to be 3 / (2π) and the height to be √5.
Substituting these values into the formula, we have:
V = (1/3)π * (3 / (2π))^2 * √5
Simplifying further, we get:
V = (1/3) * (9 / (4π)) * √5
V = 9√5 / (12π)
V = (3√5) / (4π)
So the volume of the cone in this question is (3√5) / (4π) cubic units.