A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 7 and 8 and the pyramid's height is 5 . If one of the base's corners has an angle of pi/4, what is the pyramid's surface area?

1 Answer

total surface area pyramid =28sqrt2+2sqrt(498)+7sqrt(33)
total surface area pyramid =124.4417455" "square units

Explanation:

The center of the parallelogram is at the intersection of the diagonals.

Compute half-length of the longer diagonal d_L by cosine law for sides

d_L=sqrt(8^2+7^2-2*(7)*(8)*cos (135^@))
d_L/2=1/2*sqrt(113+56sqrt(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=sqrt((1/2*sqrt(113+56sqrt(2)))^2+5^2)
x=8.54687018

Compute half-length of the shorter diagonal d_S by cosine law for sides

d_S=sqrt(8^2+7^2-2*(7)*(8)*cos (45^@))
d_S/2=1/2*sqrt(113-56sqrt(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=sqrt((1/2*sqrt(113-56sqrt(2)))^2+5^2)
y=5.783684823

Compute area A_8 of triangular face with sides 8, x, and y, by Heron's Formula

where s_8=1/2*(8+x+y)=11.1652775

A_8=sqrt(s_8*(s_8-8)*(s_8-x)*(s_8-y))
A_8=sqrt(11.1652775*(11.1652775-8)*(11.1652775-8.54687018)(11.1652775-5.783684823))
A_8=sqrt(498)

Compute area A_7 of triangular face with sides 7, x, and y, by Heron's Formula

where s_7=1/2*(7+x+y)=10.6652775

A_7=sqrt(s_7*(s_7-7)*(s_7-x)*(s_7-y))
A_7=sqrt(10.6652775*(10.6652775-7)*(10.6652775-8.54687018)(10.6652775-5.783684823))
A_7=(7*sqrt(33))/2

Compute Total Area T_A:

T_A=area base+2*A_8+2*A_7

T_A=(7*sin 45^@)*(8)+2*sqrt(498)+2*(7*sqrt(33))/2
T_A=28sqrt2+2sqrt(498)+7sqrt(33)
T_A=124.4417455" "square units

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