How to find the surface area of a tetrahedron - ISEE Upper Level Mathematics Achievement
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A regular tetrahedron has edges of length 4. What is its surface area?
A regular tetrahedron has edges of length 4. What is its surface area?
A regular tetrahedron has four faces, each of which is an equilateral triangle. Therefore, its surface area, given sidelength 
, is
.
Substitute 
:
A regular tetrahedron has four faces, each of which is an equilateral triangle. Therefore, its surface area, given sidelength
Substitute
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In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates 
.
What is the surface area of this tetrahedron?
In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates
What is the surface area of this tetrahedron?
The tetrahedron looks like this:
is the origin and 
are the other three points, which are each twelve units away from the origin on one of the three (mutually perpendicular) axes.
Three of the surfaces are right triangles with two legs of length 12, so the area of each is
.
The fourth surface, 
, has three edges each of which is the hypotenuse of an isosceles right triangle with legs 12, so each has length 
by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'
The surface area is therefore
.
The tetrahedron looks like this:
Three of the surfaces are right triangles with two legs of length 12, so the area of each is
The fourth surface,
The surface area is therefore
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A regular tetrahedron comprises four faces, each of which is an equilateral triangle. Each edge has length 16. What is its surface area?
A regular tetrahedron comprises four faces, each of which is an equilateral triangle. Each edge has length 16. What is its surface area?
The area of each face of a regular tetrahedron, that face being an equilateral triangle, is
Substitute edge length 16 for 
:
The tetrahedron has four faces, so the total surface area is
The area of each face of a regular tetrahedron, that face being an equilateral triangle, is
Substitute edge length 16 for
The tetrahedron has four faces, so the total surface area is
Compare your answer with the correct one above
In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates 
.
In terms of 
, give the surface area of this tetrahedron.
In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates
In terms of
The tetrahedron looks like this:
is the origin and 
are the other three points, which are 
units away from the origin, each along one of the three (perpendicular) axes.
Three of the surfaces are right triangles with two legs of length 12, so the area of each is
.
The fourth surface, 
, has three edges each of which is the hypotenuse of an isosceles right triangle with legs 
, so each has length 
by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'
The surface area is therefore
.
The tetrahedron looks like this:
Three of the surfaces are right triangles with two legs of length 12, so the area of each is
The fourth surface,
The surface area is therefore
Compare your answer with the correct one above