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Volume of Snub Dodecahedron given Total Surface Area Formula

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Volume of Snub Dodecahedron is the total quantity of three dimensional space enclosed by the surface of the Snub Dodecahedron. Check FAQs

V = \frac{((12 \cdot ((3 \cdot [phi]) + 1)) \cdot ({({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot [phi]) + 7) \cdot ({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot [phi]) + 6)}{6 \cdot {(3 - {({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{TSA}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

V - Volume of Snub Dodecahedron?TSA - Total Surface Area of Snub Dodecahedron?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?[phi] - Golden ratio?

Volume of Snub Dodecahedron given Total Surface Area Example

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Here is how the Volume of Snub Dodecahedron given Total Surface Area equation looks like with Values.

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Here is how the Volume of Snub Dodecahedron given Total Surface Area equation looks like.

37324.3814 = \frac{((12 \cdot ((3 \cdot 1.618) + 1)) \cdot ({({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot 1.618) + 7) \cdot ({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot 1.618) + 6)}{6 \cdot {(3 - {({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{5500}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

37324.3814m³ = \frac{((12 \cdot ((3 \cdot 1.618) + 1)) \cdot ({({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot 1.618) + 7) \cdot ({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot 1.618) + 6)}{6 \cdot {(3 - {({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{5500m²}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

37324.3814 = \frac{((12 \cdot ((3 \cdot 1.618) + 1)) \cdot ({({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot 1.618) + 7) \cdot ({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot 1.618) + 6)}{6 \cdot {(3 - {({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{5500}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

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Volume of Snub Dodecahedron given Total Surface Area Solution

Follow our step by step solution on how to calculate Volume of Snub Dodecahedron given Total Surface Area?

FIRST Step Consider the formula

V = \frac{((12 \cdot ((3 \cdot [phi]) + 1)) \cdot ({({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot [phi]) + 7) \cdot ({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot [phi]) + 6)}{6 \cdot {(3 - {({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{TSA}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

Next Step Substitute values of Variables

∴

V = \frac{((12 \cdot ((3 \cdot [phi]) + 1)) \cdot ({({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot [phi]) + 7) \cdot ({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot [phi]) + 6)}{6 \cdot {(3 - {({(\frac{[phi]}{2} + \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{[phi]}{2} - \frac{\sqrt{[phi] - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{5500m²}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

Next Step Substitute values of Constants

∴

V = \frac{((12 \cdot ((3 \cdot 1.618) + 1)) \cdot ({({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot 1.618) + 7) \cdot ({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot 1.618) + 6)}{6 \cdot {(3 - {({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{5500m²}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

Next Step Prepare to Evaluate

∴

V = \frac{((12 \cdot ((3 \cdot 1.618) + 1)) \cdot ({({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2}) - (((36 \cdot 1.618) + 7) \cdot ({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}}))) - ((53 \cdot 1.618) + 6)}{6 \cdot {(3 - {({(\frac{1.618}{2} + \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}} + {(\frac{1.618}{2} - \frac{\sqrt{1.618 - \frac{5}{27}}}{2})}^{\frac{1}{3}})}^{2})}^{\frac{3}{2}}} \cdot {(\sqrt{\frac{5500}{(20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})}})}^{3}

Next Step Evaluate

∴

V = 37324.3813534105m³

LAST Step Rounding Answer

∴

V = 37324.3814m³

Volume of Snub Dodecahedron given Total Surface Area Formula Elements

Variables

Constants

Functions

Volume of Snub Dodecahedron

Volume of Snub Dodecahedron is the total quantity of three dimensional space enclosed by the surface of the Snub Dodecahedron.

Symbol: V

Measurement: VolumeUnit: m³

Note: Value should be greater than 0.

Formulas to find Volume of Snub Dodecahedron

Total Surface Area of Snub Dodecahedron

Total Surface Area of Snub Dodecahedron is the total quantity of plane enclosed by the entire surface of the Snub Dodecahedron.

Symbol: TSA

Measurement: AreaUnit: m²

Note: Value should be greater than 0.

Formulas to find Total Surface Area of Snub Dodecahedron

Golden ratio

The Golden ratio occurs when the ratio of two numbers is the same as the ratio of their sum to the larger of the two numbers.

Symbol: [phi]