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 #, its base's sides have lengths of #2 #, and its base has a corner with an angle of #(7 pi)/8 #. What is the pyramid's surface area?

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Answer 1 · 2017-11-29T14:11:48

Total Surface Area of the pyramid = #color(purple)(14.1799)#

AB = BC = CD = DA = a = 2
Height OE = h = 3
OF = a/2 = 2/2 = 1
# EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(3^2+1^2) = color(red)3.1623#

Area of #DCE = (1/2)*a*EF = (1/2)*2*3.1623= color(red)(3.1623)#
Lateral surface area #= 4*Delta DCE = 4*3.1623= color(blue)(12.6491)#

#/_C = pi/8, /_C/2 = pi/16#
diagonal #AC = d_1# & diagonal #BD = d_2#
#OB = d_2/2 = BC*sin (C/2)=2*sin(pi/16)= **0.3902**#

#OC = d_1/2 = BC cos (C/2) = 2* cos (pi/16) = 1.9616

Area of base ABCD #= (1/2)*d_1*d_2 = (1/2)(2*0.3902)(2*1.9616) = color (blue)(1.5308)#

Total Surface Area #= Lateral surface area + Base area#
T S A # =12.6491 + 1.5308 = color(purple)(14.1799)#

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