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 `6`, its base's sides have lengths of `5`, and its base has a corner with an angle of `(5 pi)/8`. What is the pyramid's surface area?

Answer 1

Total surface area of the pyramid `= 70.4308`

Explanation:

Side of Rhombus = 5
Vertical height = 6

Slant length of pyramid = sqrt(6^2 + (5/2)^2) = 6.5
Lateral surface area of pyramid = 4(1/2)5 * 6.5 = 65#

Now to find the base area
Let the diagonals be `d_1 & d_2`
`d_1 /2 = 5*sin (((5pi)/8)/2) = 5*sin ((5pi)/16) = 0.9755`
`d_1 = 2*.9775 = 1.955`
`d_2 / 2 = 5* cos((5pi)/16) = 2.7779`
`d_2 = 2.7779*2 = 5.5558`
Area of Rhombus = (d_1d_2)/2 =(1.9555.5558)/2=5.4308#

Total surface area `= 65 + 5.4308 = 70.4308`