Triangular pyramid

Determine the volume and surface area of a regular triangular pyramid having a base edge a=20 cm and a lateral edge b = 35 cm

Correct result:

V = 1848.4228 cm³
S = 1179.4357 cm²

Solution:

a = 20 cm b = 35 cm a_{2} = a / 2 = 20 / 2 = 10 cm h_{1} = \sqrt{b^{2} - a_{2}^{2}} = \sqrt{3 5^{2} - 1 0^{2}} = 15 \sqrt{5} cm ≐ 33.541 cm h = \sqrt{h_{1}^{2} - a_{2}^{2}} = \sqrt{33.54 1^{2} - 1 0^{2}} = 5 \sqrt{41} cm ≐ 32.0156 cm S_{1} = a \cdot h_{1} / 2 = 20 \cdot 33.541 / 2 = 150 \sqrt{5} {cm}^{2} ≐ 335.4102 {cm}^{2} S_{2} = \frac{\sqrt{3}}{4} \cdot a^{2} = \frac{\sqrt{3}}{4} \cdot 2 0^{2} = 100 \sqrt{3} {cm}^{2} ≐ 173.2051 {cm}^{2} V = \frac{1}{3} \cdot S_{2} \cdot h = \frac{1}{3} \cdot 173.2051 \cdot 32.0156 = 1848.4228 {cm}^{3}

S = 3 \cdot S_{1} + S_{2} = 3 \cdot 335.4102 + 173.2051 = 1179.4357 {cm}^{2}

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