Question : The volume of a cone with a height equal to the radius and slant height of 5 cm is:
Option 1: $\frac{125 \pi}{12 \sqrt{3}} \mathrm{~cm}^3$
Option 2: $\frac{125 \pi}{6 \sqrt{3}} \mathrm{~cm}^3$
Option 3: $\frac{125 \pi}{12 \sqrt{2}} \mathrm{~cm}^3$
Option 4: $\frac{125 \pi}{6 \sqrt{2}} \mathrm{~cm}^3$

Question

Question : The volume of a cone with a height equal to the radius and slant height of 5 cm is:

Option 1: $\frac{125 \pi}{12 \sqrt{3}} \mathrm{~cm}^3$

Option 2: $\frac{125 \pi}{6 \sqrt{3}} \mathrm{~cm}^3$

Option 3: $\frac{125 \pi}{12 \sqrt{2}} \mathrm{~cm}^3$

Option 4: $\frac{125 \pi}{6 \sqrt{2}} \mathrm{~cm}^3$

Team Careers360 · Accepted Answer

Correct Answer: $\frac{125 \pi}{6 \sqrt{2}} \mathrm{~cm}^3$

Solution : Given: A cone with a height equal to the radius and a slant height of 5 cm.
Using Pythagoras theorem, we have,
⇒ $r^{2}+r^{2}=5^{2}$
⇒ $2r^{2}=25$
⇒ $r=\frac{5}{\sqrt{2}}$
Volume of the cone = $\frac{1}{3}\pi r^{2}h$ ---------(where $r=h$)
Putting the values, we have,
⇒ $\frac{1}{3}\pi ×\frac{5}{\sqrt{2}}×\frac{5}{\sqrt{2}}×\frac{5}{\sqrt{2}}$
⇒ $\frac{125}{6\sqrt{2}}\pi \mathrm{~cm}^3$
Hence, the correct answer is $\frac{125}{6\sqrt{2}}\pi \mathrm{~cm}^3$.

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Question : The volume of a cone with a height equal to the radius and slant height of 5 cm is:Option 1: 125π123 cm3Option 2: 125π63 cm3Option 3: 125π122 cm3Option 4: 125π62 cm3

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Question : The volume of a cone with a height equal to the radius and slant height of 5 cm is:
Option 1: $\frac{125 π}{12 \sqrt{3}} {cm}^{3}$
Option 2: $\frac{125 π}{6 \sqrt{3}} {cm}^{3}$
Option 3: $\frac{125 π}{12 \sqrt{2}} {cm}^{3}$
Option 4: $\frac{125 π}{6 \sqrt{2}} {cm}^{3}$