question_answer

Find the volume of tetrahedron whose height is\[4\sqrt{3}\,cm\].

A)  \[36\,c{{m}^{3}}\]                 

B)  \[72\,c{{m}^{3}}\]

C)  \[70\,c{{m}^{3}}\]                 

D)  \[48\,c{{m}^{3}}\]

Correct Answer: B

Solution :

Height of equilateral triangle = Slant height of pyramid formed by joining three face of tetrahedron. \[\therefore \,\,\,\,\,\,{{\left( \frac{\sqrt{3}}{2}a \right)}^{2}}-{{\left( \frac{a}{2\sqrt{3}} \right)}^{2}}={{\left( 4\sqrt{3} \right)}^{2}}\] \[\left( \frac{a}{2\sqrt{3}}=~Incentre\text{ }of\,\Delta  \right)\Rightarrow \frac{3{{a}^{2}}}{4}-\frac{{{a}^{2}}}{12}=48\] \[\frac{9{{a}^{2}}-{{a}^{2}}}{12}=48\Rightarrow \frac{8{{a}^{2}}}{12}=48\] \[{{a}^{2}}=\frac{48\times 12}{8}=72\Rightarrow a=6\sqrt{2}\] \[\therefore \]  Volume of tetrahedron \[=\,\,\,\frac{1}{3}\times \frac{\sqrt{3}}{4}\times {{a}^{2}}\times \] height \[=\frac{1}{3}\times \frac{\sqrt{3}}{4}\times 6\sqrt{2}\times 6\sqrt{2}\times 4\sqrt{3}=\underline{\mathbf{72}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}}\]