A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $4$ and $1$ 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?

Question

1522 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-04T03:00:13

$T S A$ $A_T = color(green)(37.1424)$

enter image source here

Total Surface Area of Pyramid $A_T$ = Area of parallelogram base ABCD + 2 * ( Area of Triangles (AED + CED))

Let Area of ABCD as $A_p$ , Area of AED as $A_(t1)$ , Area of CED as $A_(t2)$

$A_p = a * b sin theta = 4 * 1 * sin ((5pi)/6) = 2$

$A_(t1) = (1/2) * 1 * l_1 = (1/2) * 1 * sqrt((a/2)^2 + h^2)$

$A_(l1) = (1/2) * 1 * sqrt(2/2)^2 + 7^2) = 3.5355$

Similarly, $A_(l2) = (1/2) * 4 * sqrt((1/2)*2 + 7^2) = 14.0357$

$T S A$ $A_T= A_p + 2 * (A-(l1) + A_(l2)) = 2 + 2 * (3.5355 + 14.0357)$

$A_T = color(green)(37.1424)$

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 4 4 and 1 1 and the pyramid's height is 7 7 . If one of the base's corners has an angle of (5pi)/6(5pi)/6, what is the pyramid's surface area?