A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $5$ and $5$ and the pyramid's height is $8$ . If one of the base's corners has an angle of $pi/4$ , what is the pyramid's surface area?

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Answer 1 · 2018-06-10T08:55:29

$color(indigo)("Total Surface Area of pyramid " A_T = 101.5 " sq units"$

$"Total Surface Area of Pyramid " (A_T) = "Base Area of parallelogram " (A_p) + "Lateral Surface Area of Pyramid "( A_L)$

$"Since base = 5, it's a rhombus"$

$A_p = b^2 sin theta = 5^2 * sin (pi/4) = 17.68$

$L S A " " A_L = 4 * (1/2) b * sqrt((b/2)^2 + h^2)$

$A_L = 2 * 5 * sqrt((5/2)^2 + 8^2) = 83.82$

$color(indigo)("Total Surface Area " A_T = A_p + A_L = 17.68 + 83.82 = 101.5 " sq units"$

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 5 5 and 5 5 and the pyramid's height is 8 8 . If one of the base's corners has an angle of pi/4pi/4, what is the pyramid's surface area?