A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 3 3 and 9 9 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 7, 2017

T S A = 89.3341

Explanation:

CH = 3 * sin (pi/4) = 2.1213CH=3⋅sin(π4)=2.1213
Area of parallelogram base = a * b1 = 9*2.1213 = color(red)(19.0917 )=a⋅b1=9⋅2.1213=19.0917

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (9/2)^2)= 7.5EF=h1=√h2+(a2)2=√62+(92)2=7.5
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 7.5= color(red)(11.25)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(3/2)^2 )= 6.1847
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*6.1847 = color(red)( 23.8712)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 11.25)+ (2* 23.8712) = color(red)(70.2424)

Total surface area =Area of parallelogram base + Lateral surface area = 19.0917 + 70.2424 = 89.3341

Total Surface Area # T S A = **89.3341**#enter image source hereTotal Surface Area T S A = **89.3341**enter image source here

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