Question

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?

Answer 1

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.