Question

a sector of circle of radius 15 cm has an angle of 120degree. it is rolled up so that two bounding radii are joined together to form a cone. find the volume of the cone

Answer 1

Given Radius of the circle = Slant height of the cone = 15cm.

Given Angle theta = 120.

We know that Length of the arc = theta/360 * 2pir

                                                      = 120/360 * 2pi * 15

                                                      =  2pi * 5

                                                      = 10pi.


Therefore the circumference = 10pi.

We know that Circumference of the base of cone = 2pir.

2pir = 10pi

2r = 10

r = 5cm.


We know that height of the cone h^2 = l^2 - r^2

                                                               = 15^2 - 5^2

                                                               = 225 - 25

                                                               = 200

                                                        $h = 10 \sqrt{2} cm$


Now,


Volume of the cone = 1/3pir^2h
 
                                  $= \frac{1}{3} * \frac{22}{7} * (5)^2 * 10 \sqrt{2}$

                                 $= \frac{22 * 25 * 10 * 1.414}{3 * 7}$

                                 $= \frac{7777}{21}$

                                 $= 370.33cm^3.$



Hope this helps!

Answer 2

Answer:

hope it helps

Step-by-step explanation:

370.33 is the correct answer