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Given a right circular cone of base radius = r.
Let the height be H and slanting height be S.
See the diagram.
O is the center of the base of the cone and frustum. P is the center of the top surface of the frustum. Given OP = h and also the distance
OQ² = L² = h² + r²/9 .
Volume = ?
As PQ² = OQ² - OP²
= h² + r²/9 - h² = r²/9
=> PQ = r/3.
Using the similar triangles (rule for frustum & Cone):
(H-h)/H = PQ/OB = (r/3) / r
1 - h/H = 1/3
H = 3/2 * h
Volume of frustum
= volume of total cone - volume of small cone on top of frustum
= π/3 r² H - π/3 (r/3)² (H - h)
= π/3 * r² * 3/2 h - π/3 * r²/9 (h/2)
= π/54 r² h * [27 - 1]
= 13 π/27 * r² h
Let the height be H and slanting height be S.
See the diagram.
O is the center of the base of the cone and frustum. P is the center of the top surface of the frustum. Given OP = h and also the distance
OQ² = L² = h² + r²/9 .
Volume = ?
As PQ² = OQ² - OP²
= h² + r²/9 - h² = r²/9
=> PQ = r/3.
Using the similar triangles (rule for frustum & Cone):
(H-h)/H = PQ/OB = (r/3) / r
1 - h/H = 1/3
H = 3/2 * h
Volume of frustum
= volume of total cone - volume of small cone on top of frustum
= π/3 r² H - π/3 (r/3)² (H - h)
= π/3 * r² * 3/2 h - π/3 * r²/9 (h/2)
= π/54 r² h * [27 - 1]
= 13 π/27 * r² h
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kvnmurty:
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thank u so much
BY the way can i know which class do u read in
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