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=1a=1
Let diagonal d_1=sqrt(a^2+a^2-2a*a*cos ((5pi)/8)d1=√a2+a2−2a⋅a⋅cos(5π8)
diagonal d_1=sqrt(1^2+1^2-2(1)*(1)*cos ((5pi)/8)d1=√12+12−2(1)⋅(1)⋅cos(5π8)
diagonal d_1=1.6629392246051d1=1.6629392246051
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Let diagonal d_2=sqrt(a^2+a^2-2a*a*cos ((3pi)/8)d2=√a2+a2−2a⋅a⋅cos(3π8)
diagonal d_2=sqrt(1^2+1^2-2(1)*(1)*cos ((3pi)/8)d2=√12+12−2(1)⋅(1)⋅cos(3π8)
diagonal d_2=1.1111404660392d2=1.1111404660392
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compute for the lateral edges bb and cc
b=sqrt((d_1/2)^2+h^2)=sqrt((1.6629392246051/2)^2+7^2)b=√(d12)2+h2=√(1.66293922460512)2+72
b=7.0492085879326b=7.0492085879326
compute c:
c=sqrt((d_2/2)^2+h^2)=sqrt((1.1111404660392/2)^2+7^2)c=√(d22)2+h2=√(1.11114046603922)2+72
c=7.0220124098307c=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))=0.9238795325113Sr=2⋅(12⋅a⋅a⋅sin(5π8))=2(12⋅(1)(1)⋅sin(5π8))=0.9238795325113
For the Lateral Face Area S_f Sf by Heron's Area Formula:
start with the half-perimeter s=1/2(a+b+c)s=12(a+b+c)
s=1/2(1+7.0492085879326+7.0220124098307)=7.5356104988815s=12(1+7.0492085879326+7.0220124098307)=7.5356104988815
S_f=sqrt(s(s-a)(s-b)(s-c))Sf=√s(s−a)(s−b)(s−c)
S_f=sqrt(7.5356104988815(7.5356104988815-1)(7.5356104988815-7.0492085879326)(7.5356104988815-7.0220124098307))Sf=√7.5356104988815(7.5356104988815−1)(7.5356104988815−7.0492085879326)(7.5356104988815−7.0220124098307)
S_f=3.5076127333136Sf=3.5076127333136
~~~~~~~~~~~~~~~~~~~~~~~~
Surface Area of the Pyramid SS
S=4*S_f+S_r=4(3.5076127333136)+0.9238795325113S=4⋅Sf+Sr=4(3.5076127333136)+0.9238795325113
S=14.954330465765S=14.954330465765
God bless....I hope the explanation is useful.
