Given:
Tetrahedron with vertices at
(0,0,0)
(5,0,0)
(0,5,0)
(0,0,5)
The surface area consists of 4 triangles, three of which each has an area
A1=bh/2=5*5/2=12.5
The slanted surface is an equilateral triangle of side
s=sqrt(5^2+5^2)=5sqrt(2).
The corresponding area is
A2=(sqrt(3)/4)s^2=sqrt(3)/4*(5sqrt(2))^2=sqrt(3)/4*50
=12.5sqrt(3)
So the overall surface area of the tetrahedron is
A=3A1+A2=3*12.5+12.5sqrt(3)
=12.5(3+sqrt(3)) ft ²
=59.15 ft ² to two decimal places.
Number of cans =A/25=2.366 cans, so three cans are needed.