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

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

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Answer 1 · 2016-04-05T18:15:13

$V_{pyr} = \frac{1}{3} | 6 \cdot 3 | sin (\frac{5}{6} π) \cdot 7 = 21 {units}^{3}$ $V_("pyr")=1/3|6*3|sin(5/6pi)*7 = 21 "units"^3$
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Explanation:

Given : parallelogram (quadrilateral) pyramid with sides, height and angle between sides as follows -
$A B = 6; A C = 3, E F = 7$ $AB=6; AC=3, EF=7$ and $∠ B A C = \frac{5}{6} π$ $/_BAC=5/6pi$

Required: Volume?

Solution Strategy: Use the general pyramid volume formula.
a) $V_{pyr} = \frac{1}{3} (Base Area \times Height) = \frac{1}{3} A_{pllgm} \times H$ $V_("pyr") = 1/3("Base Area" xx "Height")=1/3A_("pllgm")xxH$
b) $A_{pllgm} = | s_{1} \cdot s_{2} | \cdot sin θ$ $A_("pllgm") = |s_1*s_2|*sintheta$

So inserting b) into a) we get the volume
$V_{pyr} = \frac{1}{3} ∣ ∣ ¯ ¯¯¯¯ ¯ A B \cdot ¯ ¯¯¯¯ ¯ A C ∣ ∣ sin θ \cdot ¯ ¯¯¯¯ ¯ E F$ $V_("pyr")=1/3|bar(AB)*bar(AC)|sintheta*bar(EF)$ substituting
$V_{pyr} = \frac{1}{3} | 6 \cdot 3 | sin (\frac{5}{6} π) \cdot 7 = 21 cubic units$ $V_("pyr")=1/3|6*3|sin(5/6pi)*7 = 21 "cubic units"$

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

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A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 6 66 and 3 33 and the pyramid's height is 7 77 . If one of the base's corners has an angle of (5pi)/65π6(5pi)/6, what is the pyramid's surface area?

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