A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #9 # and #6 # and the pyramid's height is #9 #. If one of the base's corners has an angle of #(5pi)/12#, what is the pyramid's surface area?

Question

A pyramid has a parallelogram shaped base and a peak directly above its center.  Its base's sides have lengths of #9 # and #6 # and the pyramid's height is #9 #.  If one of the base's corners has an angle of #(5pi)/12#, what is the pyramid's surface area?

Theodore Abad · Accepted Answer

#color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93#

#l = 9, b = 6, theta = (5pi)/12, h = 9#

#"To find the Total Surface Area T S A"#

#"Area of parallelogram base " A_B = l b sin theta#

#A_B = 9 * 6 * sin ((5pi)/12) = 52.16#

#S_1 = sqrt(h^2 + (b/2)^2) = sqrt(9^2 + 3^2) = 9.49#

#S_2 = sqrt(h^2 + (l/2)^2) = sqrt(9^2 + (9/2)^2) = 10.06#

#"Lateral Surface Area " A_L = 2 * ((1/2) l * S_1 + (1/2) b * S_2#

#A_L = (cancel2 * cancel(1/2)) * (l * S_1 + b * S_2) = (9 * 9.49 + 6 * 10.06)#

#A_L = 145.77#

#color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93#

Evelyn Arboleda · Answer

undefined To find the surface area of the pyramid, first calculate the area of the base. Then, find the area of each triangular face and sum them up.

The area of the parallelogram base can be calculated using the formula for the area of a parallelogram: $ \text{base} \times \text{height} $.

The area of each triangular face can be found using the formula for the area of a triangle: $ \frac{1}{2} \times \text{base} \times \text{height} $. The base of each triangle is the side length of the parallelogram base, and the height can be found using the Pythagorean theorem with the given height of the pyramid.

After finding the areas of the base and the triangular faces, sum them up to get the total surface area of the pyramid.