| If the slant height of a right pyramid with square base is \[4\,\,m\] and the total slant surface of the pyramid is \[12\,\,{{m}^{2}},\] then the ratio of total lateral surface and area of the base is [SSC (10+2) 2012] |
A) 16 : 3
B) 24 : 5
C) 32 : 9
D) 12 : 3
Correct Answer: A
Solution :
| \[\because \]Lateral surface area \[=\frac{1}{2}\times \]Perimeter of base \[\times \]Slant height |
| \[\Rightarrow \] \[12=\frac{1}{2}\times \text{Perimeter}\,\,\text{of}\,\,\text{base}\times 4\] |
| \[\Rightarrow \]Perimeter of base \[=\frac{12\times 2}{4}=6\,\,cm\] |
| \[\therefore \] Side of square \[=\frac{6}{4}=\frac{3}{2}\,\,cm\] |
| Now, area of base\[={{(\text{side})}^{2}}=\frac{9}{4}\,\,c{{m}^{2}}\] |
| \[\therefore \] \[\text{Ratio}=\frac{12}{\frac{9}{4}}=\frac{4\times 4}{3}=16:3\] |
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