Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `6` and `3` and the pyramid's height is `7`. If one of the base's corners has an angle of `(5pi)/6`, what is the pyramid's surface area?

Answer 1

`V_("pyr")=1/3|6*3|sin(5/6pi)*7 = 21 "units"^3`

Explanation:

Given : parallelogram (quadrilateral) pyramid with sides, height and angle between sides as follows -
`AB=6; AC=3, EF=7` and `/_BAC=5/6pi`

Required: Volume?

Solution Strategy: Use the general pyramid volume formula.
a) `V_("pyr") = 1/3("Base Area" xx "Height")=1/3A_("pllgm")xxH`
b) `A_("pllgm") = |s_1*s_2|*sintheta`

So inserting b) into a) we get the volume
`V_("pyr")=1/3|bar(AB)*bar(AC)|sintheta*bar(EF)` substituting
`V_("pyr")=1/3|6*3|sin(5/6pi)*7 = 21 "cubic units"`