Pentagonal pyramid

Find the volume and surface of a regular pentagonal pyramid with a base edge a = 12.8 cm and a height v = 32.1 cm.

Correct result:

V = 3016.1483 cm³
S = 1347.0581 cm²

Solution:

a = 12.8 cm v = 32.1 cm n = 5 S_{1} = \frac{1}{4} \cdot \sqrt{5 \cdot (5 + 2 \cdot \sqrt{5})} \cdot a^{2} = \frac{1}{4} \cdot \sqrt{5 \cdot (5 + 2 \cdot \sqrt{5})} \cdot 12. 8^{2} ≐ 281.883 {cm}^{2} V = \frac{1}{3} \cdot S_{1} \cdot v = \frac{1}{3} \cdot 281.883 \cdot 32.1 = 3016.1483 {cm}^{3}

S_{1} = n \cdot \frac{a \cdot h_{1}}{2} h_{1} = 2 \cdot S_{1} / n / a = 2 \cdot 281.883 / 5 / 12.8 ≐ 8.8088 cm h_{2} = \sqrt{v^{2} + h_{1}^{2}} = \sqrt{32. 1^{2} + 8.808 8^{2}} ≐ 33.2867 cm S_{2} = \frac{a \cdot h_{2}}{2} = \frac{12.8 \cdot 33.2867}{2} ≐ 213.035 {cm}^{2} S = S_{1} + n \cdot S_{2} = 281.883 + 5 \cdot 213.035 = 1347.0581 {cm}^{2}

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