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A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is 23\dfrac{2}{3} of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.

Mensuration

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Answer

The figure of the buoy made by surmounting a right cone on a hemisphere is shown below:

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is 2/3 of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct  to 2 places of decimal. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Radius of base of hemisphere = Radius of cone = 3.5 m.

Volume of hemisphere (V) = 23πr3\dfrac{2}{3}πr^3

Putting values,

V=23×227×(3.5)3=4421×42.875=44×42.87521=1886.521=89.8 m3V = \dfrac{2}{3} \times \dfrac{22}{7} \times \Big(3.5)^3 \\[1em] = \dfrac{44}{21} \times 42.875 \\[1em] = \dfrac{44 \times 42.875}{21} \\[1em] = \dfrac{1886.5}{21} \\[1em] = 89.8 \text{ m}^3

Volume of cone = 23\dfrac{2}{3} Volume of hemisphere.

∴ Volume of cone = 23×89.8\dfrac{2}{3} \times 89.8

=179.63=59.87 m3.= \dfrac{179.6}{3} \\[1em] = 59.87 \text{ m}^3.

Volume of cone = 13πr2h\dfrac{1}{3}πr^2h.

13πr2h=59.8713×227×(3.5)2×h=59.872221×12.25×h=59.87h=59.87×2122×12.25h=1257.27269.5h=4.67m.\therefore \dfrac{1}{3}πr^2h = 59.87 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times h = 59.87 \\[1em] \Rightarrow \dfrac{22}{21} \times 12.25 \times h = 59.87 \\[1em] \Rightarrow h = \dfrac{59.87 \times 21}{22 \times 12.25} \\[1em] \Rightarrow h = \dfrac{1257.27}{269.5} \\[1em] \Rightarrow h = 4.67 m.

Slant height of cone = l = h2+r2\sqrt{h^2 + r^2}

Putting values we get,

l=(4.67)2+(3.5)2l=21.81+12.25l=34.06l=5.84 ml = \sqrt{(4.67)^2 + (3.5)^2} \\[1em] l = \sqrt{21.81 + 12.25} \\[1em] l = \sqrt{34.06} \\[1em] l = 5.84 \text{ m}

Surface area of the buoy = Curved Surface area of cone + Curved Surface area of hemisphere = πrl + 2πr2.

∴ Surface area of buoy = πr(l+2r)πr(l + 2r)

=227×3.5×(5.84+2×3.5)=227×3.5×(5.84+7)=227×3.5×12.84=988.687=141.17 m2.= \dfrac{22}{7} \times 3.5 \times (5.84 + 2 \times 3.5) \\[1em] = \dfrac{22}{7} \times 3.5 \times (5.84 + 7) \\[1em] = \dfrac{22}{7} \times 3.5 \times 12.84 \\[1em] = \dfrac{988.68}{7} \\[1em] = 141.17 \text{ m}^2.

Hence, the height of cone = 4.67 m and surface area of buoy is 141.17 m2.

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