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2 years ago

a cone with a radius of 3” has a total area of 24pie square inches. Find the volume of the cone. 12pie, 24 pie or 15 pie

Mathematics

1 answer:

Ratling [72]2 years ago

7 0

Answer:

$12\pi (inches)^{2}$

Step-by-step explanation:

Given:

radius of cone = 3 inch

Total area of cone = $24\pi (inch)^{2}$

Find the volume of the cone?

We know the area of cone = $\pi r(r+s)$ -------(1)

Where r = radius

Ans s = side of the cone

Put the area value in equation 1

$24\pi=\pi r(r+s)$ -------(1)

$\frac{24\pi }{\pi r} =r+s$

$\frac{24}{ r}-r =s$

Put r value in above equation.

$s =\frac{24}{ 3}-3$

$s=8-3$

$s=5$

The side s = 5 inches

We know the side of the cone formula

$s^{2}=r^{2}+ h^{2}$

$h^{2}=s^{2}- r^{2}$

Put r and s value in above equation.

$h^{2}=5^{2}- 3^{2}$

$r^{2} = 25-9$

$r^{2} = 16$

$r = 4$

The volume of cone is

$V = \frac{1}{3} \pi r^{2} h$

Put r and h value in above equation.

$V = \frac{1}{3} \pi (3)^{2}\times 4$

$V = 12\pi (inches)^{2}$

The Volume of the cone is $12\pi (inches)^{2}$

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In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

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$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

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${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

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${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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Step-by-step explanation:

"4 of his friends ate s tacos" is unclear. I assume it should be

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That implies that Ricky ate 2s tacos, instead of 2 tacos. If 2s = 2 then s = 1, but then 6s ≠ 18.

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