A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is 5 5, its base's sides have lengths of 5 5, and its base has a corner with an angle of (5 pi)/8 5π8. What is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 12, 2017

T S A = 78.999

Explanation:

AB = BC = CD = DA = a = 5
Height OE = h = 5
OF = a/2 = 5/2 = 2.5
EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(5^2+2.5^2) = color(red)(5.5902)EF=√EO2+OF2=√h2+(a2)2=√52+2.52=5.5902

Area of DCE = (1/2)*a*EF = (1/2)*5*5.5902 = color(red)(13.9755)DCE=(12)⋅a⋅EF=(12)⋅5⋅5.5902=13.9755
Lateral surface area = 4*Delta DCE = 4*13.9755 = color(blue)(55.902)

/_C =pi - (5pi)/8 = (3pi)/8
Area of base ABCD = a* a * sin /_C = 5^2 sin ((3pi)/8) = 23.097

T S A = Lateral surface area + Base area
T S A =55.902 + 23.097 = color(purple)(78.999)

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