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

Solid geometry Quiz 2

mathematics

 What is the surface area of a cube if its space diagonal is given as $20\text{cm}$.

 By substituting space diagonal length in the formula $S=2{{d}^{2}}$, we get surface area of a cube.

  $\text{Surface area of a cube when diagonal is given as }S=2{{d}^{2}} \\ \text{Since d = space diagonal length }= 20cm \\ S=2{{\left( 20 \right)}^{2}}=800c{{m}^{2}}$

 What is $\sqrt{3}$  times the edge of a cube if the space diagonal is given by $12cm$.

 Use the relation between space diagonal and edge and substitute values in it.

  $\text{The space diagonal is }12cm \\ \text{The relation between space diagonal and } \text{edge of a cube is given by,} \\ d=a\sqrt{3} \\ \Rightarrow 12=a\sqrt{3} \\ \Rightarrow a=\dfrac{12}{\sqrt{3}} \\ \text{It is given to find }\sqrt{3}\text{ times, then} \\ \therefore \sqrt{3}a=\dfrac{12}{\sqrt{3}}\times \sqrt{3}=12cm$

 A tent is in the shape of an inverted cone. A worker is trying to reach halfway up the tent for maintenance work. What is the distance he has to climb if the radius and height of the tent is given as  $8 m$ and $14 m$ respectively? 

 Use Pythagoras theorem to define the relation between height, radius and slant length.

  $\text{Slant height of a cone can be given as,} \\ {{s}^{2}}={{r}^{2}}+{{h}^{2}} \\ \Rightarrow {{s}^{2}}={{8}^{2}}+{{(14)}^{2}} \\ \Rightarrow s=\sqrt{260}=2\sqrt{65} \\ \text{Since the person has to climb up to half distance,} \\ \therefore \dfrac{2\sqrt{65}}{2}=\sqrt{65}m$

 What is ${{\dfrac{3}{7}}^{th}}$ the volume of a sphere with diameter $5\sqrt{7}m$.

 Use formula of volume of sphere and substitute given values.

  $\text{Volume of sphere, }V=\dfrac{4}{3}\pi {{r}^{3}} \\ \text{Since the sphere of diameter is }5\sqrt{7}m \\  \therefore V=\dfrac{4}{3}\pi \times \left( \dfrac{5\sqrt{7}}{2}\times \dfrac{5\sqrt{7}}{2}\times \dfrac{5\sqrt{7}}{2} \right) \\ V=\dfrac{3}{7}\times \dfrac{1}{3}\times \pi \times \left( \dfrac{125\times 7\sqrt{7}}{2} \right) \\ V=\left( \dfrac{125\times \pi \times \sqrt{7}}{2} \right)$

 A child has two toys of shapes, pyramid and cylinder. What is the difference between their heights with the parameters given below? $ \\ Pyramid \\ Volume-456m \\ length-20\ m \\ width-36\ m \\ $     $ \\ Cylinder \\ Volume-120\ m \\ radius-4\ m $   

 Using the formula of volume of pyramid and cylinder we get differences between heights.

  $\text{Volume of pyramid, }V=\dfrac{\text{Length}\times \text{breadth}\times \text{height}}{3} \\ \therefore \dfrac{3V}{\text{length}\times \text{breadth}}=\text{height} \\ \Rightarrow \text{H=}\dfrac{3\times 456}{20\times 36}=1.9m \\ \text{Volume of cylinder, }V=\pi {{r}^{2}}h \\ \therefore h=\dfrac{V}{\pi {{r}^{2}}} \\ \Rightarrow h=\dfrac{V}{\pi {{(4)}^{2}}}=2.4m \\ \text{Differences between of height }2.4-1.9=0.5m$