Question

A sector of radius 10.5 cm with the central angle 120 is folded to form a cone find volume

Answer 1

The volume of the cone is 125.76 cm³.

Step-by-step explanation:

It is given that,

The radius of the sector of a circle = 10.5 cm  

The central angle, θ = 120°

Step 1:

Now, it is rolled up to form a cone.

So, we get

The circumference of the base of the cone  

= Length of the arc of the sector

= (θ/360) * 2πr

= (120/360) * 2π * 10.5

= (1/3) * 2 * (22/7) * 10.5

= 22 cm

Step 2:

We know that the circumference of circle = 2πr

So, we can write

The circumference of the base of cone = 2πr

⇒ 22 = 2πr  

⇒ r = 3.5 cm ← radius of base of cone  

Also,  

The slant height of the cone (l) = radius of the sector = 10.5 cm

We know that slant height  

l = √(h² + r² )  

⇒ l² = (h² + r² )

⇒ 10.5² = h² + 3.5²  

⇒ h = √[110.25 – 12.25]

=> h = √[98]

=> h = 9.8 cm ← height of the cone  

Thus, substituting the value of r = 3.5 cm and h = 9.8 cm, we get

The volume of the cone is given by,

= $\frac{1}{3}$ * π * r² * h

= $\frac{1}{3} * \frac{22}{7}$ * 3.5² * 9.8

= $\frac{1}{3} * \frac{22}{7}$ * 12.25 * 9.8

= $\frac{2641.1}{21}$

= 125.76 cm³

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