Question

What is the surface area of a 11 cm high pyramid whose base is an equilateral triangle with a 62 cm perimeter? Show work.

Answer 1

´`961/sqrt(3)cm^2~=554.834 cm^2`

Explanation:

To a better understanding refer to the figures below

We are dealing with a solid of 4 faces, i.e., a tetrahedron.

Conventions (see Fig.1)

I called

• `h` the height of the tetrahedron,
• `h"'"` the slanted height or height of the slanted faces,
• `s` each of the sides of the equilateral triangle of the base of the tetrahedron,
• `e` each of the edges of the slanted triangles when not `s`.

There are also

• `y`, the height of equilateral triangle of the base of the tetrahedron,
• and `x`, the apothegm of that triangle.

The perimeter of `triangle_(ABC)` is equal to 62, then:
`s=62/3`

In Fig. 2, we can see that

`tan 30^@=(s/2)/y` => `y=(s/2)*1/(sqrt(3)/3)=31/cancel(3)*cancel(3)/sqrt(3)=31/sqrt(3)~=17.898`
So
`S_(triangle_(ABC))=(s*y)/2=(62/3*31/sqrt(3))/2=961/(3sqrt(3))~=184.945`
and that
`s^2=x^2+x^2-2x*x*cos 120^@`
`s^2=2x^2-2x^2(-1/2)`
`3x^2=s^2` => `x=s/sqrt(3)=62/(3sqrt(3)`

In Fig. 3, we can see that

`e^2=x^2+h^2=(62/(3sqrt(3)))^2+11^2=3844/27+121=(3844+3267)/27=7111/27` => `e=sqrt(7111)/(3sqrt(3))`

In Fig. 4, we can see that

`e^2=h"'"^2+(s/2)^2`
`h"'"^2=e^2-(s/2)^2=(sqrt(7111)/(3sqrt(3)))^2-(31/3)^2=(7111-3*1089)/27=3844/27`
`h"'"=62/(3sqrt(3))~=11.932`

Area of one slanted triangle
`S_("slanted "triangle)=(s*h"'")/2=(62/3*62/(3sqrt(3)))/2=1922/(9sqrt(3))~=123.296`

Then the total area is
`S_T=S_(triangle_(ABC))+3*S_("slanted "triangle)=961/(3sqrt(3))+1922/(3sqrt(3))=961/sqrt(3)cm^2~=554.834 cm^2`