Question

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?

Answer 1

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.