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 #7 #, its base's sides have lengths of #1 #, and its base has a corner with an angle of #(5 pi)/8 #. What is the pyramid's surface area?

1 Answer

Pyramid surface area #=14.9543" "#square units

Explanation:

We need to calculate the edges of the lateral faces of the pyramid. We start by calculating the diagonals of the base.

By Cosine Law with base side #a=1#

Let diagonal #d_1=sqrt(a^2+a^2-2a*a*cos ((5pi)/8)#

diagonal #d_1=sqrt(1^2+1^2-2(1)*(1)*cos ((5pi)/8)#

diagonal #d_1=1.6629392246051#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let diagonal #d_2=sqrt(a^2+a^2-2a*a*cos ((3pi)/8)#

diagonal #d_2=sqrt(1^2+1^2-2(1)*(1)*cos ((3pi)/8)#

diagonal #d_2=1.1111404660392#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compute for the lateral edges #b# and #c#
#b=sqrt((d_1/2)^2+h^2)=sqrt((1.6629392246051/2)^2+7^2)#
#b=7.0492085879326#

compute c:

#c=sqrt((d_2/2)^2+h^2)=sqrt((1.1111404660392/2)^2+7^2)#
#c=7.0220124098307#

Compute for the Areas of the base, and Lateral faces.

For the Rhombus base:

#S_r=2*(1/2*a*a*sin (5pi/8))=2(1/2*(1)(1)*sin ((5pi)/8))=0.9238795325113#

For the Lateral Face Area #S_f # by Heron's Area Formula:

start with the half-perimeter #s=1/2(a+b+c)#

#s=1/2(1+7.0492085879326+7.0220124098307)=7.5356104988815#

#S_f=sqrt(s(s-a)(s-b)(s-c))#

#S_f=sqrt(7.5356104988815(7.5356104988815-1)(7.5356104988815-7.0492085879326)(7.5356104988815-7.0220124098307))#

#S_f=3.5076127333136#

~~~~~~~~~~~~~~~~~~~~~~~~
Surface Area of the Pyramid #S#

#S=4*S_f+S_r=4(3.5076127333136)+0.9238795325113#

#S=14.954330465765#

God bless....I hope the explanation is useful.