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Problems on frustum of a Cone
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Problems on frustum of a Cone
FRUSTUM: If a right circular cone is cut off by a plane parallel to its base, then the portion of the cone between the cutting plane and the base of the cone is called a frustum of the cone.
In Fig-1 right circular cone VAB is cut by a plane parallel to its circular base with centre O and diameter AB. The portion containing the vertex V is removed(Fig-2). The left out portion ABA'B' shown in Fig 3 is the frustum of the cone VAB. The circular faces AOB and A'O'B' arte called the circular ends of the frustum.

Height: The height or thickness of a frustum is the perpendicular distance between its two circular bases.In Fig-3 OO' is the height.

Slant Height: The slant height of a frustum of a right circular cone is the length of the line segment joining the extremities of two parallel radii, drawn in the same direction, of the circular bases. In Fig-3 Slant height of the frustum is AA' or BB'.
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Example: A tent is made in the form of a conic frustum surmounted by a cone. The diameters of the base and the top of the frustum are 20 m and 6 m respectively and the height is 24 m. If the height of the tent is 28 m,find the quantity of canvas required.
Solution: Let h be the height of the frustum and r1 and r2 be the radii of its circular bases.
you have, h = 24 m, r1 = 10 m and r2 = 3 m.
l = Slant height of the frustum
= {(r1 - r2)2 + h2}
= {(10 - 3)2 + 242}
= (49 + 576)
= 625 m = 25 m
For cone VA'B', you have
  l2 = slant height = (O'B'2 + VO'2) = (32 + 42) = 5 m.
Quantity of canvas required
= Lateral surface area of frustum + Lateral surface area of cone VA 'B'
= (r1 + r2)l + * r2* l2
= [(10 + 3) * 25 + * 3* 5]
= (325 + 15) m2
= 340m2
 
   
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