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
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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
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
Then the total area is
