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

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Answer 1 · 2016-09-13T02:02:37

$82.319$

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Total surface area = base area (a1) + side area 1 (a2) + side area 2 (a3).

1) base area (a1) = parallelogram area = BxxH
$=7xx3xxsin60 = 18.186$ , ( $H=3sin60$ )

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2) side area 1 (a2) $=(1/2)xxB1xxS1$ , ( $B1=7$ )

$S1=sqrt((6^2+1.5^2))=6.185$
$a2=(1/2)xx7xx6.185xx2=43.295$ , (2 sides)

3) side area 2 (a3) $=(1/2)xxB2xxS2$ , ( $B2=3$ )

$S2=sqrt((6^2+3.5^2))=6.946$
$a3=(1/2)xx3xx6.946xx2=20.838$ , (2 sides)

Fianlly, total surface area = a1+a2+a3

$=18.186+43.295+20.838$
$=82.319$

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