Given: Right Pyramid, ABCD is a rhombus

Question

AB = 6, mBAD = 50, mSKO = 40
Segment SO is perpendicular to ABCD, Segment OK is perpendicular to segment DC

Find: Volume and Surface Area of the figure

Answer 1

The area of the base is 36sin50 = 27.58
KS is the slant height of the pyramid.
OK is the height of the base, 6sin50 = 4.60
So, OK/KS=sin40 ==> KS=4.6/sin40 = 7.15
OS is the height of the pyramid, and OS^2 = KS^2-OK^2 = 29.96, so OS=5.47

The volume of the pyramid is 1/3 Bh = 1/3 * 27.58 * 5.47 = 50.33
The area is the base plus four sides.
Each side is an isosceles triangle with base=6 and height=KS=7.15
So, the surface area is 27.58+4*(6*7.15)/2 = 113.38

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