A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is #8 #, its base has sides of length #6 #, and its base has a corner with an angle of #(3 pi)/4 #. What is the pyramid's surface area?

Answer 1

Surface area of pyramid with rhombus base = 127.9839

Opposite base angles of the rhombus are #(3pi)/4 & pi/4# respy.
#d1 & d2# be the two diagonals intersecting at right angle.
#sin (theta/2) = sin ((pi/4)/2) = sin (pi/8)=((d1/2)/6)#
#sin(pi/8)=d1/12 or d1=12*sin(pi/8)=#4.5922

Similarly,#cos(pi/8)=((d2/6)/2)=(d2/12)#
#d2=12*cos(pi/8)=#11.0866

Area of Rhombus base #=(d1*d2)/2=(4.5922*11.0866)/2=#25.4559

Height of one side surface #h1=sqrt((b/2)^2+h^2)#
As #b/2=6/2=3, h1=sqrt(3^2+8^2)=8.544#

Area of 4 sides of pyramid #=4*(1/2)b*h1=2*6*8.544=#102.588

Surface area of pyramid = base area + 4 side areas
#=25.4559+102.588=127.9839

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Answer 2

The surface area of the pyramid is 96 square units.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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