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 \[\Delta AOB,\] |
\[O{{A}^{2}}=A{{B}^{2}}-O{{B}^{2}}\]\[\Rightarrow \]\[{{h}^{2}}={{l}^{2}}-O{{B}^{2}}\] |
\[\Rightarrow \] \[{{h}^{2}}={{13}^{2}}-{{\left( \frac{1}{3}\times \frac{\sqrt{3}}{2}\times 10\sqrt{3} \right)}^{2}}\] |
\[\left[ \because BO=\frac{1}{3}\times Median\,\,\,\frac{1}{3}\times \frac{\sqrt{3}}{2}a \right]\] |
\[\Rightarrow \] \[169-25=144\] |
\[\Rightarrow \] \[h=12\,cm\] |
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