The center of the parallelogram is at the intersection of the diagonals.
Compute half-length of the longer diagonal d_LdL by cosine law for sides
d_L=sqrt(8^2+7^2-2*(7)*(8)*cos (135^@))dL=√82+72−2⋅(7)⋅(8)⋅cos(135∘)
d_L/2=1/2*sqrt(113+56sqrt(2))dL2=12⋅√113+56√2
Let x be one of the lengths of edges of the triangular face which is directly on top of the longer diagonal as seen on top of the pyramid
x=sqrt((d_L/2)^2+h^2)x=√(dL2)2+h2
x=sqrt((1/2*sqrt(113+56sqrt(2)))^2+5^2)x=√(12⋅√113+56√2)2+52
x=8.54687018x=8.54687018
Compute half-length of the shorter diagonal d_SdS by cosine law for sides
d_S=sqrt(8^2+7^2-2*(7)*(8)*cos (45^@))dS=√82+72−2⋅(7)⋅(8)⋅cos(45∘)
d_S/2=1/2*sqrt(113-56sqrt(2))dS2=12⋅√113−56√2
Let y be one of the lengths of edges of the triangular face which is directly on top of the shorter diagonal as seen on top of the pyramid
y=sqrt((d_S/2)^2+h^2)y=√(dS2)2+h2
y=sqrt((1/2*sqrt(113-56sqrt(2)))^2+5^2)y=√(12⋅√113−56√2)2+52
y=5.783684823y=5.783684823
Compute area A_8A8 of triangular face with sides 8, x, and y, by Heron's Formula
where s_8=1/2*(8+x+y)=11.1652775s8=12⋅(8+x+y)=11.1652775
A_8=sqrt(s_8*(s_8-8)*(s_8-x)*(s_8-y))A8=√s8⋅(s8−8)⋅(s8−x)⋅(s8−y)
A_8=sqrt(11.1652775*(11.1652775-8)*(11.1652775-8.54687018)(11.1652775-5.783684823))A8=√11.1652775⋅(11.1652775−8)⋅(11.1652775−8.54687018)(11.1652775−5.783684823)
A_8=sqrt(498)A8=√498
Compute area A_7A7 of triangular face with sides 7, x, and y, by Heron's Formula
where s_7=1/2*(7+x+y)=10.6652775s7=12⋅(7+x+y)=10.6652775
A_7=sqrt(s_7*(s_7-7)*(s_7-x)*(s_7-y))A7=√s7⋅(s7−7)⋅(s7−x)⋅(s7−y)
A_7=sqrt(10.6652775*(10.6652775-7)*(10.6652775-8.54687018)(10.6652775-5.783684823))A7=√10.6652775⋅(10.6652775−7)⋅(10.6652775−8.54687018)(10.6652775−5.783684823)
A_7=(7*sqrt(33))/2A7=7⋅√332
Compute Total Area T_A:TA:
T_A=area base+2*A_8+2*A_7TA=areabase+2⋅A8+2⋅A7
T_A=(7*sin 45^@)*(8)+2*sqrt(498)+2*(7*sqrt(33))/2TA=(7⋅sin45∘)⋅(8)+2⋅√498+2⋅7⋅√332
T_A=28sqrt2+2sqrt(498)+7sqrt(33)TA=28√2+2√498+7√33
T_A=124.4417455" "TA=124.4417455 square units
God bless....I hope the explanation is useful.