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This is one of our many quizzes on Solid shapes. Click on this to access more such competitive quizzes on Geometry.

Solid shapes Quiz 2

mathematics

 Determine the volume of a sphere of radius $5\sqrt 5 cm$.

 Apply volume of a sphere formula = $\dfrac{4}{3}\pi {r^3}$. 

  Volume of a sphere $ = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {(5\sqrt 5 )^3}$ $ \\    \Rightarrow \dfrac{4}{3} \times \dfrac{{22}}{7} \times 125 \times 5\sqrt 5    \\   \Rightarrow 5856.36c{m^3}   \\  $

 What is the curved surface area of a cone having radius $\dfrac{7}{3} \text{cm}$ and height is 3 times of the radius.

 Firstly calculate $l = \sqrt {{h^2} + {r^2}} $ and further apply curved surface area $\pi rl$. 

  $l = \sqrt {49 + \dfrac{{49}}{9}}  = \dfrac{{\sqrt {49 \times 9 + 49} }}{3} \\ $ Curved surface area $ = \dfrac{{22}}{7} \times \dfrac{7}{3} \times \dfrac{{\sqrt {490} }}{3}$ $ \\    \Rightarrow \dfrac{{22}}{9} \times 7\sqrt {10}    \\   \Rightarrow 54.10c{m^2}   \\  $

 The total surface area of a cuboid is $232c{m^2}$. Its length is 2 times the height and breadth is $10cm$. Find the height of the cuboid. 

 Apply total surface area of cuboid is $2(lb + bh + lh)$ and further solve quadratic equation by $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Let the height be $h$ and length be $2h$. $ \\ $ Now, total surface area of the cuboid: $ \\   \therefore Total{\text{ }}Surface{\text{ }}Area = 2(lb + bh + lh)   \\   \Rightarrow 232 = 2[(2h \times 10) + (10 \times h) + (2h \times h)]   \\   \Rightarrow \dfrac{{232}}{2} = 2{h^2} + 30h   \\   \Rightarrow {h^2} + 15h - 58 = 0   \\  $ The roots are: $ \\  \therefore h = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}   \\  \,\,\,\,\,\,\,\, = \dfrac{{ - 15 \pm \sqrt {225 - 4.1.( - 58)} }}{2}   \\  \,\,\,\,\,\,\,\, = \dfrac{{ - 15 + \sqrt {457} }}{2}, - \dfrac{{15 + \sqrt {457} }}{2}   \\  $ So, $h =  - \dfrac{{15 + \sqrt {457} }}{2}$ is not possible. $ \\ $ Hence, the height of the cuboid is $\dfrac{{ - 15 + \sqrt {457} }}{2}cm$.

 Surface area of a floor is given i.e. $196c{m^2}$ now a man has to cover the floor with tiles. Dimensions of tiles are given as $7cm$ of length and $2cm$ of breadth. So, how many tiles are required to cover the floor?

 ${\text{No. of tiles required}} = \dfrac{{\text{total area of floor}}}{{\text{area of one tiles}}}$

 ${\text{No. of tiles required}} = \dfrac{{\text{total area of floor}}}{{\text{area of one tiles}}} = \dfrac{{196}}{{7 \times 2}}   \\   \Rightarrow 14tiles   \\  $

 Find the total surface area of a cone whose height is $4cm$ and radius is $7cm$.

 First calculate silent highest cone and further use of total surface area of cone $\pi rl + \pi {r^2}$.

 $l = \sqrt {16 + 49}  = \sqrt {65} $ $ \\   {\text{Total surface area of cone}} = \dfrac{{22}}{7} \times 7 \times \sqrt {65}  + \dfrac{{22}}{7} \times 7 \times 7   \\   \Rightarrow 22(7 + \sqrt {65} )c{m^2}   \\  $