A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is 6 6, its base's sides have lengths of 5 5, and its base has a corner with an angle of (5 pi)/8 5π8. What is the pyramid's surface area?

1 Answer
Oct 18, 2017

Total surface area of the pyramid = 70.4308=70.4308

Explanation:

Side of Rhombus = 5
Vertical height = 6

Slant length of pyramid = sqrt(6^2 + (5/2)^2) = 6.5
Lateral surface area of pyramid = 4(1/2)5 * 6.5 = 65#

Now to find the base area
Let the diagonals be d_1 & d_2d1&d2
d_1 /2 = 5*sin (((5pi)/8)/2) = 5*sin ((5pi)/16) = 0.9755d12=5sin(5π82)=5sin(5π16)=0.9755
d_1 = 2*.9775 = 1.955d1=2.9775=1.955
d_2 / 2 = 5* cos((5pi)/16) = 2.7779d22=5cos(5π16)=2.7779
d_2 = 2.7779*2 = 5.5558d2=2.77792=5.5558
Area of Rhombus = (d_1d_2)/2 =(1.9555.5558)/2=5.4308#

Total surface area = 65 + 5.4308 = 70.4308=65+5.4308=70.4308