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?

1 Answer
Yonas Yohannes
Apr 5, 2016

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

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