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

1 Answer
sankarankalyanam
Dec 12, 2017

T S A = 87.541

Explanation:

CH = 5 * sin (pi/6) = 2.5CH=5⋅sin(π6)=2.5
Area of parallelogram base = a * b1 = 6*2.5 = color(red)(15 )=a⋅b1=6⋅2.5=15

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (6/2)^2)= 6.7082EF=h1=√h2+(a2)2=√62+(62)2=6.7082
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*5* 6.7082= color(red)(16.7705)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(5/2)^2 )= 6.5
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*6*6.5 = color(red)( 19.5)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 16.7705)+ (2* 19.5) = color(red)(72.541)

Total surface area =Area of parallelogram base + Lateral surface area = 15 + 72.541 = 87.541

Total Surface Area # T S A = **87.541**#enter image source hereTotal Surface Area T S A = **87.541**enter image source here

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