Question

If a cone shaped hole is 3 feet deep and the circumferenceOf the base of the hole is 44 feet what is the volume of the hole. Use 22/7 for pi

Answer 1

Answer:

154 feet³

Step-by-step explanation:

circumference = 2πr

radius = 44×7/2×22 = 7 feet

volume = 1/3πr²h

= 22×7×7×3/3×7

= 154 feet³

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Answer 2

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Question :

If a cone shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet . What is the volume of the hole ? ( use pi(π) = 22/7 )

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Answer :

Given :-

Cone shaped hole is 3 feet deep .

Circumference of the base of the hole is 44 feet .

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Required to find :-

• Volume of the hole ?

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Formulae used :-

$\longrightarrow{\boxed{\mathrm{Volume \: of \; Cone }{\tt{ \;=\; \dfrac{1}{3} \pi {r}^{2} h}}}}$

$\boxed{\mathrm{Circumference \; of \; the \; circle = 2 \pi r}}$

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Solution :-

In the question it is given that ,

Cone shaped hole is 3 feet deep

This above statement means that,

Height of the cone = 3 feet

Similarly,

Circumference of the base of the hole is 44 feet .

Now,

we know that

$\boxed{\mathrm{Circumference \; of \; the \; circle = 2 \pi r}}$

However,

Let the radius of the base be " x " feet .

According to problem,

$\tt{ 2 \times \dfrac{22}{7} \times x \; = \; 44 }$

$\tt{ \dfrac{22}{7} \times x = \dfrac{44}{2} }$

$\tt{ x = \dfrac{44 \times 7}{2 \times 22}}$

$\tt{ x = \dfrac{ 308}{44}}$

$\tt{\rightarrow{ x = 7 \; feet}}$

So,

The radius of the cone is 7 feet .

Now we got the required values in order to find the volume of the cone .

Here, we should the formula .

The formula is ,

$\longrightarrow{\boxed{\mathrm{Volume \: of \; Cone }{\tt{ \;=\; \dfrac{1}{3} \pi {r}^{2} h}}}}$

Now substitute the required values in it .

Hence, we get

$\rightarrow{\mathrm{Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times {7}^{2} \times 3 }}$

$\rightarrow{\mathrm{Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 3 }}$

$\rightarrow{\mathrm{Volume = \dfrac{22 \times 49 \times 3}{3 \times 7 }}}$

$\rightarrow{\mathrm{Volume = \dfrac{3234}{21}}}$

$\implies{\red{\underline{\underline{\mathrm{Volume = 154\; {ft}^{3} }}}}}$

$\huge{\underline{\boxed{\tt{\orange{\therefore{Volume = 154 \; {ft}^{3}}}}}}}$

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✅ Hence Solved .

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