A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 4 4 and 5 5 and the pyramid's height is 6 6. If one of the base's corners has an angle of pi/4 π4, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 6, 2017

T S A = 71.765

Explanation:

CH = 4 * sin (pi/4) = 4 sin (45) = 4/sqrt2 = 2.8284CH=4⋅sin(π4)=4sin(45)=4√2=2.8284
Area of parallelogram base = 5* b1 = 5*2.8284 = color(red)(14.142 )=5⋅b1=5⋅2.8284=14.142

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (5/2)^2)= 6.5EF=h1=√h2+(a2)2=√62+(52)2=6.5
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.5 = color(red)(13)#

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)*5*6.3246 = color(red)( 15.8115)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 13)+ (2* 15.8115) = color(red)(57.623)

Total surface area =Area of parallelogram base + Lateral surface area = 14.142 + 57.623 = 71.765

Total Surface Area # T S A = **71.765**#enter image source here

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