A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 6 6 and 7 7 and the pyramid's height is 2 2. If one of the base's corners has an angle of (5pi)/65π6, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 7, 2017

T S A = 70.4258

Explanation:

CH = 6 * sin ((5pi)/6) = 3CH=6⋅sin(5π6)=3
Area of parallelogram base = 7* b1 = 7*3 = color(red)(21 )=7⋅b1=7⋅3=21

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (7/2)^2)= 4.011EF=h1=√h2+(a2)2=√22+(72)2=4.011
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*6* 4.0311= color(red)(12.0933)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(6/2)^2 )= 3.6056
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*3.6056 = color(red)( 12.6196)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 12.0933)+ (2* 12.6196) = color(red)(49.4258)

Total surface area =Area of parallelogram base + Lateral surface area = 21 + 49.4258 = 70.4258

Total Surface Area # T S A = **70.4258**#enter image source hereTotal Surface Area T S A = **70.4258**enter image source here

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