3. The circumference of the base of a 12 m high
conical tent is 66 m. Find the volume of the
air contained in it.
please answer it fast guys
please
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3. The circumference of the base of a 12 m high
conical tent is 66 m. Find the volume of the
air contained in it.
please answer it fast guys
please
please
please
please
please
please
please
please
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Verified answer
Given:
The circumference of the conical tent is 66m
The height of the conical tent is 12 m
To be found:
The volume of the air contained in it.
(so we have to find the volume of the conical tent.)
[tex]\underline{\sf{The\ formula\ for\ finding\ the\ volume\ of\ cone}}[/tex]
[tex]\boxed{\red{\rm{\frac{1}{3}\pi r^{2}h}\ \bf unit\ sq.}}[/tex]
So,
We have h = 12m
We need radius = r =?
As given that the circumference of the conical tent is 66m
So,
[tex]\implies 2\pi r = 66[/tex]
[tex]\implies 2\times \frac{22}{7} \times r = 66[/tex]
[tex]\implies \frac{44}{7} \times r = 66[/tex]
[tex]\implies r = 66 \times \frac{7}{44}[/tex]
[tex]\implies r = \frac{6 \times 7}{4}[/tex]
[66/11 = 6 and 44/11 = 4]
[tex]\implies r = \frac{3\times 7}{2}[/tex]
[tex]\therefore r = \boxed{\frac{21}{2}m}\ or\ \boxed{\bf 10.5m}[/tex]
Now,
Finding the Volume,
[tex]= \frac{1}{3} \pi r^{2}h[/tex] unit sq.
[tex]= \frac{1}{3} \times \frac{22}{7} \times (\frac{21}{2})^{2} \times 12[/tex]
[tex]= \frac{1}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times 12[/tex]
[tex]= \boxed{9702m^{2}}[/tex]
Hence,
The volume of the air contained in it is 9702 m sq.
Given ,
The circumference of the base of a 12 m high conical tent is 66 m
As we know that ,
The circumference of circle is given by
[tex] \boxed{ \tt{Circumference = 2\pi r}}[/tex]
Thus ,
66 = 2 × 22/7 × r
66 = 44/7 × r
6/4 = 1/7 × r
r = (3 × 7)/2
r = 21/2 m
Now , the volume of cone is given by
[tex] \boxed { \tt{Volume = \frac{1}{3} \pi {(r)}^{2} h} } [/tex]
Thus ,
V = 1/3 × 22/7 × 21/2 × 21/2 × 12
V = 9702 sq. m