Question

# A metallic bucket,open at the top, of height 24 cm is in the from of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and,respectively.Find (i) the volume of water which can completely fill the bucket; (ii) the care of the metal sheet used to make the bucket.

Solution

## Radius of lower circular end $$= r = 7 cm$$ Radius of upper circular end $$= R = 14 cm$$ height of bucket $$= h = 24 cm$$ Slant height of frustum:$$l = \sqrt { \left( R-r \right) ^ 2+h^ 2 }$$$$=\sqrt { \left( 14-7 \right) ^ 2+24^ 2 }$$$$=\sqrt {7^2+24^2}$$$$= \sqrt{625}$$ $$= 25 cm$$ (i) Volume of frustum of cone $$= \dfrac{1}{3} \pi h (R^2 + r^2 + Rr)cm^3$$ $$= \dfrac{1}{3} \times \dfrac{22}{7} \times 24 (14^2 + 7^2 + 14 \times 7)$$ $$=\dfrac{1}{3} \times \dfrac{22}{7} \times 24 \times (196 + 98+49)$$ $$=\dfrac{1}{3} \times \dfrac{22}{7} \times 24 \times (343)$$ $$= 8624 cm^3$$  (ii) Curved surface area $$= x (R+r)$$$$=22/7 \times 25 \times (14 + 7)$$ $$= 1650 cm^2$$ Area of the base of bucket = $$\pi r^2$$  (consider the lower base of a bucket) $$= 22/7 \times 7 \times 7$$ $$= 154 cm^2$$ Area of metal sheet used to make the bucket = curved surface area + Area of the base $$= 1650 + 154$$  $$= 1804$$ Area pf metal sheet used to make the bucket is $$1804 cm^2$$MathematicsRS AgarwalStandard X

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