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 \[=\frac{1}{3}\times \frac{\sqrt{3}}{4}\times 6\sqrt{2}\times 6\sqrt{2}\times 4\sqrt{3}=\underline{\mathbf{72}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{3}}}}\]You need to login to perform this action.
You will be redirected in
3 sec