SATHEE: Quantitative Aptitude Ques 287

Quantitative Aptitude Ques 287

Question: 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},cm^{2}, $ then its height is

Options:

A) 10 cm

B) $ 10\sqrt{3},cm $

C) 12 cm

D) $ 12\sqrt{3},cm $

Show Answer

Answer:

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 Slantheight $

$ \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, $ $ OA^{2}=AB^{2}-OB^{2} $
$ \Rightarrow $ $ h^{2}=l^{2}-OB^{2} $

$ \Rightarrow $ $ h^{2}=13^{2}-{{( \frac{1}{3}\times \frac{\sqrt{3}}{2}\times 10\sqrt{3} )}^{2}} $ $ [ \because BO=\frac{1}{3}\times Median,\frac{1}{3}\times \frac{\sqrt{3}}{2}a ] $

$ \Rightarrow $ $ 169-25=144 $

$ \Rightarrow $ $ h=12,cm $

Quantitative Aptitude Ques 286

Quantitative Aptitude Ques 288

Quantitative Aptitude Ques 287

Question: 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},cm^{2}, $ then its height is

Options:

Answer:

Solution:

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