Triangular pyramid

It is given perpendicular regular triangular pyramid: base side a = 5 cm, height v = 8 cm, volume V = 28.8 cm³. What is it content (surface area)?

Correct result:

S = 71.79 cm²

Solution:

a = 5 cm h = 8 cm V = 28.8 {cm}^{3} h_{1} = \sqrt{a^{2} - (a / 2)^{2}} = \sqrt{5^{2} - (5 / 2)^{2}} ≐ 4.3301 cm S_{1} = \frac{a \cdot h_{1}}{2} = \frac{5 \cdot 4.3301}{2} ≐ 10.8253 {cm}^{2} h_{2}^{2} = h^{2} + (h_{1} / 3)^{2} h_{2} = \sqrt{h^{2} + (h_{1} / 3)^{2}} = \sqrt{8^{2} + (4.3301 / 3)^{2}} ≐ 8.1292 cm S_{2} = \frac{a \cdot h_{2}}{2} = \frac{5 \cdot 8.1292}{2} ≐ 20.3229 {cm}^{2} S = S_{1} + 3 \cdot S_{2} = 10.8253 + 3 \cdot 20.3229 ≐ 71.7941 = 71.79 {cm}^{2} Verifying Solution: V_{2} = \frac{1}{3} \cdot S_{1} \cdot h = \frac{1}{3} \cdot 10.8253 \cdot 8 ≐ 28.8675 {cm}^{3} V_{2} = V

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