A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 2 2 and 1 1 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 25, 2017

T S A = 7.4788

Explanation:

CH = 1 * sin ((pi)/6) = 0.5CH=1⋅sin(π6)=0.5
Area of parallelogram base = a * b1 = 2*0.5 = color(red)(1)=a⋅b1=2⋅0.5=1

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (2/2)^2)= 2.2361EF=h1=√h2+(a2)2=√22+(22)2=2.2361
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 2.2361= color(red)(1.1181)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(1/2)^2 )= 2.1213
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*2*2.1213 = color(red)( 2.1213)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 1.1181)+ (2* 2.1213) = color(red)(6.4788)

Total surface area =Area of parallelogram base + Lateral surface area = 1 + 6.4788 = 7.4788

Total Surface Area # T S A = **7.4788**#enter image source here

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