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

1 Answer
Feb 11, 2016

´961/sqrt(3)cm^2~=554.834 cm^2

Explanation:

To a better understanding refer to the figures below

I created this figure using MS ExcelI created this figure using MS Excel
I created this figure using MS ExcelI created this figure using MS Excel
I created this figure using MS ExcelI created this figure using MS Excel
I created this figure using MS ExcelI created this figure using MS Excel

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