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Maharashtra State BoardSSC (English Medium) 10th Standard

A sector of a circle of radius 15 cm has the angle 120º. It is rolled up so that two bounding radii are joined together to form a cone. Find the volume of the cone. (π = 22/7)

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Question

A sector of a circle of radius 15 cm has the angle 120º. It is rolled up so that two bounding radii are joined together to form a cone. Find the volume of the cone. `(π = 22/7)`

Sum
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Solution

Given: Radius of the sector (R) = 15 cm

Central angle (θ) = 120°

1. Finding Slant Height (l) and Base Radius (r):

When a sector of a circle is rolled up to form a cone:

Slant height of the cone (l) = Radius of the sector (R)

l = 15 cm

Base radius of the cone (r):

Circumference of cone base = Arc length of sector

2πR = `θ/(360^circ) xx 2πR`

Dividing both sides by 2π:

r = `(120^circ)/(360^circ) xx 15`

r = `1/3 xx 15`

r = 5 cm

2. Finding Perpendicular Height (h):

By Pythagoras’ theorem in a cone:

\(h=\sqrt{l^{2}-r^{2}}\)

= `sqrt(15^2 - 5^2)`

= `sqrt(225 - 25)`

= `sqrt(200)`

= `10sqrt(2)` cm

= 10 × 1.4142

≈ 14.14 cm

3. Finding Volume of the Cone (V):

`V = 1/3 πr^2h`

Substitute the given value `π = 22/7` along with r = 5 and h = `10sqrt(2)`:

`V = 1/3 xx 22/7 xx (5)^2 xx 10sqrt(2)`

`V = 1/3 xx 22/7 xx 25 xx 10sqrt(2)`

`V = (5500sqrt(2))/21`

`V ≈ (5500 xx 1.4142)/21`

`V ≈ 7778.1/21`

V ≈ 370.39 cm3

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Chapter 7: Mensuration - Exercise
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