Tetrahedral pyramid

A regular tetrahedral pyramid is given. Base edge length a = 6.5 cm, side edge s = 7.5 cm. Calculate the volume and the area of its face (side area).

Correct result:

V = 83.4668 cm³
S = 87.8703 cm²

Solution:

a = 6.5 cm s = 7.5 cm S_{1} = a^{2} = 6. 5^{2} = \frac{169}{4} = 42.25 {cm}^{2} u_{1} = \sqrt{2} \cdot a = \sqrt{2} \cdot 6.5 ≐ 9.1924 cm h = \sqrt{s^{2} - (u_{1} / 2)^{2}} = \sqrt{7. 5^{2} - (9.1924 / 2)^{2}} ≐ 5.9266 cm V = \frac{1}{3} \cdot S_{1} \cdot h = \frac{1}{3} \cdot 42.25 \cdot 5.9266 = 83.4668 {cm}^{3}

h_{2} = \sqrt{h^{2} + (a / 2)^{2}} = \sqrt{5.926 6^{2} + (6.5 / 2)^{2}} ≐ 6.7593 cm S_{2} = \frac{1}{2} \cdot a \cdot h_{2} = \frac{1}{2} \cdot 6.5 \cdot 6.7593 ≐ 21.9676 {cm}^{2} S = 4 \cdot S_{2} = 4 \cdot 21.9676 = 87.8703 {cm}^{2}

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