Math, asked by Anuragsanjay, 10 months ago

A fristum of cone is made made from seven circular ring used in maths experiment the radius of the upper ring 4cm then the radius of the each ring is increased by 1cm the radius of the lower ring is 10cm the thickness of each ring is 3cm find the volume of such figure which is formed (Neglate the blank space near centre see figure below

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Answered by RvChaudharY50

||✪✪ QUESTION ✪✪||

A frustum of cone is made made from seven circular ring used in maths experiment the radius of the upper ring 4cm then the radius of the each ring is increased by 1cm the radius of the lower ring is 10cm the thickness of each ring is 3cm find the volume of such figure which is formed (Neglate the blank space near centre see figure below ?

|| ★★ FORMULA USED ★★ ||

→ Volume of frustum of cone = (1/3) * π * h [ R² + r² + Rr ]

where R is radius of lower part and r is radius of upper part .

|| ✰✰ ANSWER ✰✰ ||

we have ,

→ radius of upper ring part = 4cm = r

→ radius of lower ring part = 10cm = R

→ Height of frustum of cone = 7*3 = 21 cm ( as their as 7 rings with width as 3cm.) .

Putting all values we get,

→ Volume of frustum of cone = (1/3) * π * h [ R² + r² + Rr ]

→ Volume = (1/3) * (22/7) * 21 * [ 4² + 10² + 4*10 ]

→ Volume = 22 * [ 16 + 100 + 40 ]

→ Volume = 22 * 156

→ Volume = 3432 cm³ .

Therefore, the volume of frustum made is 3432 cm³.

Answered by Anonymous

QUESTION :-

FORMULA USED :-

=> Volume of frustum of cone = (1/3) * π * h [ R² + r² + Rr ]

ANSWER :-

Given that , radius of lower part is 10cm and upper part is 4cm.

=> we have , Height of frustum of cone = 7*3 = 21 cm ( as their as 7 rings with width as 3cm.) .

So,

=> Volume = (1/3) * (22/7) * 21 * [ 4² + 10² + 4*10 ]

=> Volume = 22 * [ 16 + 100 + 40 ]

=> Volume = 22 * 156

=> Volume = 3432 cm³ .

Hence, the volume of frustum so made is 3432 cm³.

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