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Concept Detail
Maths / Surface Area and Volume / Volume of Frustum
(A Brief Glimpse of ABHYAS Content - Have aLook !!!!)

Volume of Frustum: If a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone. Let "h" be the height, "l" the slant height and $\dpi{120} \large r_{1}$ and $\dpi{120} \large r_{2}$ are the radii of the ends $\dpi{120} \large (r_{1}>r_{2})$ of the frustum of a cone. Then we can directly find the volume, the curved surface area and the total surface area of frustum by using the formulas given: Volume of frustum of cone $\dpi{120} \large =\frac{1}{3}\pi h(r^{2}_{2}+r_{1}r_{2})$

Example : If the radii of the circular ends of a conical bucket which is 45 cm high, are 28 cm and 7 cm. Find the capacity of the bucket ( Use $\dpi{120} \large \pi =22/7$ ).

Solution: Clearly, bucket forms a frustum of a cone such that the radii of its circular ends are $\dpi{120} \large r_{1}=28cm,r_{2}=7cm$  and height h = 45 cm.

Therefore, capacity of the bucket = Volumne of the frustum $\dpi{120} \large =\frac{1}{3}\pi h(r^{2}_{2}+r^{2}_{2}+r_{1}r_{2})$Capacity of the bucket $\dpi{120} \large =\frac{1}{3}\times \frac{22}{7}\times 45(28^{2}+7^{2}+28\times 7)=22\times \frac{15}{7}\times 7(4^2\times 7 +7+28)$= 22 x 15 x (16 x 7 +7 + 28) = 330 x 147 $\dpi{120} \large =48510cm^{3}$

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