A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #9 # 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?

1 Answer

#160.334\ \text{unit}^2#

Explanation:

Area of parallelogram base with sides #8# & #9# & an interior angle #{5\pi}/6#

#=8\cdot 9\sin({5\pi}/6)=36#

The parallelogram shaped base of pyramid has its semi-diagonals

#1/2\sqrt{8^2+9^2-2\cdot 8\cdot 9\cos({5\pi}/6)}=8.211# &

#1/2\sqrt{8^2+9^2-2\cdot 8\cdot 9\cos({\pi}/6)}=2.252#

Now, two unequal lateral edges of each triangular lateral face of pyramid are given as

#\sqrt{7^2+(8.211)^2}=10.79# &

#\sqrt{7^2+(2.252)^2}=7.353#

There are two pairs of opposite identical triangular lateral faces of pyramid. One pair of two opposite triangular faces has the sides #8, 10.79# & #7.353# and another pair of two opposite triangular faces has the sides #9, 10.79# & #7.353#

1) Area of each of two identical triangular lateral faces with sides #8, 10.79# & #7.353#

semi-perimeter of triangle, #s={8+10.79+7.353}/2=13.0715#

Now, using heron's formula the area of lateral triangular face of pyramid

#=\sqrt{13.0715(13.0715-8)(13.0715-10.79)(13.0715-7.353)}#

#=29.409#

2) Area of each of two identical triangular lateral faces with sides
#9, 10.79# & #7.353#

semi-perimeter of triangle, #s={9+10.79+7.353}/2=13.5715#

Now, using heron's formula the area of lateral triangular face of pyramid

#=\sqrt{13.5715(13.5715-9)(13.5715-10.79)(13.5715-7.353)}#

#=32.758#

Hence, the total surface area of pyramid (including area of base)

#=2(\text{area of lateral triangular face of type-1})+2(\text{area of lateral triangular face of type-2})+\text{area of parallelogram base}#

#=2(29.409)+2(32.758)+36#

#=160.334\ \text{unit}^2#