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Total Surface Area of Snub Dodecahedron given Volume Formula

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

TSA = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{V \cdot 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}}}{((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)})}^{\frac{2}{3}}

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Total Surface Area of Snub Dodecahedron given Volume Example

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

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

5566.1727 = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{38000 \cdot 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}}}{((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)})}^{\frac{2}{3}}

5566.1727m² = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{38000m³ \cdot 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}}}{((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)})}^{\frac{2}{3}}

5566.1727 = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{38000 \cdot 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}}}{((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)})}^{\frac{2}{3}}

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Total Surface Area of Snub Dodecahedron given Volume Solution

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

FIRST Step Consider the formula

TSA = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{V \cdot 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}}}{((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)})}^{\frac{2}{3}}

Next Step Substitute values of Variables

∴

TSA = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{38000m³ \cdot 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}}}{((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)})}^{\frac{2}{3}}

Next Step Substitute values of Constants

∴

TSA = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{38000m³ \cdot 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}}}{((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)})}^{\frac{2}{3}}

Next Step Prepare to Evaluate

∴

TSA = ((20 \cdot \sqrt{3}) + (3 \cdot \sqrt{25 + (10 \cdot \sqrt{5})})) \cdot {(\frac{38000 \cdot 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}}}{((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)})}^{\frac{2}{3}}

Next Step Evaluate

∴

TSA = 5566.17267755386m²

LAST Step Rounding Answer

∴

TSA = 5566.1727m²

Total Surface Area of Snub Dodecahedron given Volume Formula Elements

Variables

Constants

Functions

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

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

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]