Question

A sector of a circle of radius 6cm has an angle of 120 degree . It is rolled up so that the two bounding radii are joined together to form a cone . Find the total surface area of the cone and the volume of the cone.

Answer 1

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Answer 2

Answer:

The total surface area of the cone is $50.2857 cm^2$ and  the volume of the cone is $23.70489 cm^3$

Step-by-step explanation:

Given : A sector of a circle of radius 6 cm has an angle of 120 degrees.

To Find :  Find the total surface area of the cone and the volume of the cone.

Solution:

We are given that it is rolled up so that the two bounding radii are joined together to form a cone.

So, circumference of the base of cone = Length of the arc of the sector

                                                                 = $\frac{\theta}{360^{\circ} } \times 2 \pi r$

                                                                 = $\frac{120}{360} \times 2 \pi 6$

                                                                  = $4\pi$

We know that circumference of circle = 2πr

So, circumference of base of cone = 2πr

$\Rightarrow  2 \pi r= 4 \pi$

$\Rightarrow  2r= 4$

$\Rightarrow r=2$

So, the radius of base of cone  is r = 2 cm

Radius of the sector = 6 cm = Slant height of the cone l

Formula : $l^2 =h^2+r^2$

Substitute the values

$6^2 =h^2+2^2$

$36-4 =h^2$

$32 =h^2$

$\sqrt{32}=h$

$5.65685=h$

So, the height of the cone h = 5.65685 cm

Volume of cone =$\frac{1}{3}\pi r^2 h$

                          =$\frac{1}{3} \times \frac{22}{7} \times 2^2 \times 5.65685$

                          =$23.70489 cm^3$

Total Surface area of the cone = $\pi rl +\pi r^2$

                                                    = $\frac{22}{7} \times 2 \times 6 +\frac{22}{7} \times 2^2$

                                                   = $50.2857 cm^2$

Hence the total surface area of the cone is $50.2857 cm^2$ and  the volume of the cone is $23.70489 cm^3$