A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 6 6 and 4 4 and the pyramid's height is 7 7. If one of the base's corners has an angle of (5pi)/65π6, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 20, 2017

T S A = 116.4108

Explanation:

CH = 4 * sin ((5pi)/6) = 8CH=4⋅sin(5π6)=8
Area of parallelogram base = a * b1 = 6*8 = color(red)(48)=a⋅b1=6⋅8=48

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (6/2)^2)= 7.6158EF=h1=√h2+(a2)2=√72+(62)2=7.6158
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 7.6158= color(red)(15.2316)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(4/2)^2 )= 6.3246
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*6*6.3246 = color(red)( 18.9738)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 15.2316)+ (2* 18.9738) = color(red)(68.4108)

Total surface area =Area of parallelogram base + Lateral surface area = 48 + 68.4108 = 116.4108

Total Surface Area # T S A = **116.4108**#enter image source here

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