We know,
[tex] { \longrightarrow \bf \qquad { { Volume_{(cone) }= \dfrac{1}{3} \pi {r}^{2}h }}} [/tex]
Where,
- r is the base radius of the cone.
- h is the height of thr cone.
Here,
- Radius of the cone is 6 .
- Height of the cone is 8 .
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Substituting the value in the formula :
We will take the value of π as 3.14 .
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[tex]{ \longrightarrow \sf \qquad { { Volume_{(cone) }= \dfrac{1}{3} \times 3.14 \times {6}^{2} \times 8 }}} [/tex]
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[tex]{ \longrightarrow \sf \qquad { { Volume_{(cone) }= \dfrac{1}{ \cancel{3}} \times 3.14 \times \cancel{36} \times 8 }}} [/tex]
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[tex]{ \longrightarrow \sf \qquad { { Volume_{(cone) }= {1} \times 3.14 \times 12 \times 8 }}} [/tex]
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[tex]{ \longrightarrow \sf \qquad { { Volume_{(cone) }= {1} \times 3.14 \times 96 }}} [/tex]
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[tex]{ \longrightarrow \bf \qquad { { Volume_{(cone) }= 301.44}}} [/tex]
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Therefore,
- The volume of the cone is 301.44 units³.
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