Question

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

Explanation:

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