Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `2` and `7` and the pyramid's height is `7`. If one of the base's corners has an angle of `(5pi)/6`, what is the pyramid's surface area?

Answer 1

`7+(7sqrt(197))/2+(7sqrt(17))/2~=70.555`

Explanation:

Views of the solid

Data:
`AB=CD=7`
`AD=BC=2`
`EM=7`
`B hat A D=150^@`

`S_("parallelogram")=b*h_1=7*2*cos60^@=7*1=7`

It can be proven that `AM=CM` and `BM=DM` (by using angle-side-angle). This is why:
`triangle_(AEM) -= triangle_(CEM) => AE=CE`
and `triangle_(BEM) -= triangle_(DEM)=> BE=DE`
what means that:
`triangle_(ABE) -= triangle_(CDE)`
and `triangle_(ADE) -= triangle_(BCE)`

So what we need to solve the problem is to find:
height of `triangle_(ABE)`, `h_2` in the 4th figure above
height of `triangle_(ADE)`, `h_3` in the 5th figure above

• Finding the diagonals of the parallelogram (using Law of Cosines)
  `BD^2=AB^2+AD^2-2*AB*AD*cos150^@`
  `BD^2=4+49-2*2*7*(-sqrt(3)/2)`
  `BD^2=53+14sqrt3 => BD=sqrt(53+14sqrt3)`

`AC^2=AB^2+BC^2-2*AB*BC*cos30^@`
`AC^2=4+49-2*2*7*(sqrt3/2)`
`AC^2=53-14sqrt3 => AC=sqrt(53-14sqrt3)`

Finding the slant edge AE using `triangle_(AEM)` in which `AM=(AC)/2`
`AE^2=EM^2+((AC)/2)^2`
`AE^2=7^2+(53-14sqrt3)/4=(196+53-14sqrt3)/4=(249-14sqrt3)/4`

Finding the slant edge BD using `triangle_(BEM)` in which `BM=(BD)/2`
`BE^2=EM^2+((BD)/2)^2`
`BE^2=7^2+(53+14sqrt3)/4=(196+53+14sqrt3)/4=(249+14sqrt3)/4`

• Finding the height `h_2` using `triangle_(ABE)`, 4th figure
  `AE^2=h_2^2+m^2 => h_2^2=AE^2-m^2`
  `BE^2=h_2^2+(7-m)^2`
  `BE^2=AE^2-cancel(m^2)+49-14m+cancel(m^2)`
  `14m=AE^2-BE^2+49 => m=(7-sqrt3)/2`
  `-> m^2=(49-14sqrt3+3)/4=(52-14sqrt3)/4`

`h_2^2=AE^2-m^2=(249-cancel(14sqrt3))/4-(52-cancel(14sqrt3))/4`
`h_2^2=197/4 => h_2=sqrt197/2`
We can also find `h_2` in this way:
`h_2^2=(h_1/2)^2+EM^2=(1/2)^2+7^2=1/4+49=197/4` => `h_2=sqrt197/2`

• Finding the height `h_3` using `triangle_(ADE)`, 5th figure
  `AE^2=h_3^2+n^2 => h_3^2=AE^2-n^2`
  `DE^2=h_3^2+(2+n)^2`
  `DE^2=AE^2-cancel(n^2)+4+4n+cancel(n^2)`
  `4n=DE^2-AE^2-4=(249+14sqrt3)/4-(249-14sqrt3)/4-4`
  `n=7*sqrt3/4-1`
  `-> n^2=147/16-7*sqrt3/2+1 => n^2=163/16-7*sqrt3/2`

`h_3^2=AE^2-n^2=(249-14sqrt3)/4-163/16+7*sqrt3/2`
`h_3^2=833/16 => h_3=(7sqrt17)/4`
We can also find `h_3` in this way:
`h_3^2=( (7cos60^@)/2)^2+EM^2=(7/4)^2+7^2=49/16+49=833/16 => h_3=(7sqrt17)/4`

Finally:

`S_T=S_"paralellogram" +2*S_(triangle_(ABE))+2*S_(triangle_(ADE))`
`S_T=7+cancel(2)*(7*sqrt197/2)/cancel(2)+cancel(2)*(cancel(2)*7*sqrt17/cancel(4))/2`
`S_T=7+(7sqrt197)/2+(7sqrt17)/2`