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 3 3, its base has sides of length 9 9, and its base has a corner with an angle of pi/4 π4. What is the pyramid's surface area?

1 Answer
sankarankalyanam
Oct 18, 2017

Total Surface Area = 200.2119=200.2119

Explanation:

AB = BC = CD = DA = a = 9
Height OE = h = 3
OF = a/2 = 9/2
EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(3^2+(9/2)^2)=5.4083EF=√EO2+OF2=√h2+(a2)2=√32+(92)2=5.4083

Area of DCE = (1/2)*a*EF = (1/2)*9*5.4083 = 24.3375DCE=(12)⋅a⋅EF=(12)⋅9⋅5.4083=24.3375
Lateral surface area = 4*Delta DCE = 4*24.3375 = 94.3799

/_C = pi/4, /_C/2 = pi/8
diagonal AC = d_1 & diagonal BD = d_2
OB = d_2/2 = BC*sin (C/2)=9*sin(pi/8) = 6.364
OC = d_1/2 = BC cos (C/2) = 9* cos (pi/8) = 8.3149

Area of base ABCD = (1/2)*d_1*d_2 = (1/2)(2*6.364)(2*8.3149) = 105.832

Total Surface Area = Lateral surface area + Base area. T S A 94.3799 + 105.832 = 200.2119#
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