Pyramid four sides

In a regular tetrahedral pyramid is a body height 38 cm and a wall height 42 cm. Calculate the surface area of the pyramid; the result round to square centimeters.

Result

S =  4285 cm2

Solution:

h=38 s=42 h2=s2−(a/2)2 a=2⋅ s2−h2=2⋅ 422−382=16 5≐35.7771 S1=a2=35.77712=1280 S2=a⋅ s/2=35.7771⋅ 42/2=336 5≐751.3188 S=S1+4⋅ S2=1280+4⋅ 751.3188≐4285.2754=4285 cm2h = 38 \ \\ s = 42 \ \\ h^2 = s^2 - (a/2)^2 \ \\ a = 2 \cdot \ \sqrt{ s^2-h^2 } = 2 \cdot \ \sqrt{ 42^2-38^2 } = 16 \ \sqrt{ 5 } \doteq 35.7771 \ \\ S_{ 1 } = a^2 = 35.7771^2 = 1280 \ \\ S_{ 2 } = a \cdot \ s/2 = 35.7771 \cdot \ 42/2 = 336 \ \sqrt{ 5 } \doteq 751.3188 \ \\ S = S_{ 1 } + 4 \cdot \ S_{ 2 } = 1280 + 4 \cdot \ 751.3188 \doteq 4285.2754 = 4285 \ cm^2

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