question_answer

The base of a right pyramid is an equilateral triangle of side \[10\sqrt{3}\,cm.\] If the total surface area of the pyramid is \[270\,\sqrt{3}\,c{{m}^{2}},\] then its height is

A) 10 cm              

B) \[10\sqrt{3}\,cm\]

C) 12 cm  

D) \[12\sqrt{3}\,cm\]

Correct Answer: C

Solution :

Total surface area of pyramid

= Area of base + Slant area

\[\Rightarrow \]\[270\sqrt{3}=\frac{\sqrt{3}}{4}{{(10\sqrt{3})}^{2}}+\frac{1}{2}\times 3\times 10\sqrt{3}\times \text{Slant}\,\,\text{height}\]

\[\Rightarrow \] \[270\sqrt{3}=75\sqrt{3}+15\sqrt{3}\times l\]    [l = slant height]

\[\Rightarrow \] \[195\sqrt{3}=15\sqrt{3}\times l=13\,cm\]

In \[\Delta AOB,\]

\[O{{A}^{2}}=A{{B}^{2}}-O{{B}^{2}}\]\[\Rightarrow \]\[{{h}^{2}}={{l}^{2}}-O{{B}^{2}}\]

\[\Rightarrow \]   \[{{h}^{2}}={{13}^{2}}-{{\left( \frac{1}{3}\times \frac{\sqrt{3}}{2}\times 10\sqrt{3} \right)}^{2}}\]

\[\left[ \because BO=\frac{1}{3}\times Median\,\,\,\frac{1}{3}\times \frac{\sqrt{3}}{2}a \right]\]

\[\Rightarrow \]   \[169-25=144\]

\[\Rightarrow \]   \[h=12\,cm\]