Geometry - How to find the surface area of a tetrahedron

Find the surface area of a regular tetrahedron with a side length of $\frac{1}{2}$ .

$\frac{\sqrt3}{4}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{8}$

$\frac{\sqrt3}{6}$

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3\left(\frac{1}{2}\right)^2=\frac{\sqrt3}{4}$

What is the surface area of the following tetrahedron? Assume the figure is a regular tetrahedron.

Tetrahedron

$4\sqrt{3}m^2$

$\frac{\sqrt{3}}{2}m^2$

$8\sqrt{3}m^2$

$2\sqrt{3}m^2$

$3\sqrt{3}m^2$

Explanation

A tetrahedron is a three-dimensonal figure where each side is an equilateral triangle. Therefore, each angle in the triangle is $60^{\circ}$ .

In the figure, we know the value of the side and the value of the base . Since dividing the triangle by half creates a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, we know the value of must be $\sqrt{3}: m$ .

Therefore, the area of one side of the tetrahedron is:

$A = (b)(h) = (1: m)(\sqrt{3}: m) = \sqrt{3}: m^2$

Since there are four sides of a tetrahedron, the surface area is:

$SA = 4 (b)(h) = 4\sqrt{3}: m^2$

The surface area of a regular tetrahedron is $16\sqrt3$ . If each side length is , find the value of .

$4\sqrt2$

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$16\sqrt3=\sqrt3(2x)^2$

Now, solve for .

$x=\sqrt4=2$

The surface area of a regular tetrahedron is $36\sqrt3$ . If the length of each side is , find the value of .

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$36\sqrt3=\sqrt3(3a)^2$

Now, solve for .

In terms of , find the surface area of a regular tetrahedron that has a side length of .

$25x^2\sqrt3$

$25x\sqrt3$

$5x^2\sqrt3$

$125x^2\sqrt3$

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3(5x)^2=25x^2\sqrt3$

Give the surface area of a regular tetrahedron with edges of length 60.

$3,600 \sqrt{3}$

$7,200 \sqrt{3}$

$1,800 \sqrt{3}$

$60 \sqrt{3}$

$120\sqrt{3}$

Explanation

A tetrahedron comprises four triangular surfaces; if the tetrahedron is regular, then each surface is an equilateral triangle. The area of an equilateral triangle with sides of length can be computed using the formula

$A = \frac{s^{2}\sqrt{3}}{4}$ ;

The total surface area of the tetrahedron is four times this, or

$S = 4A = 4 \cdot \frac{s^{2}\sqrt{3}}{4} = s^{2}\sqrt{3}$

Set and substitute:

$S = s^{2}\sqrt{3} = 60 ^{2}\sqrt{3} = 3,600 \sqrt{3}$ .

The surface area of a regular tetrahedron is . If each side length is , find the value of . Round to the nearest tenths place.

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$64=\sqrt3(4x)^2$

$16x^2=\frac{64}{\sqrt3}$

$x^2=\frac{64}{16\sqrt3}=2.309$

What is the surface area of a regular tetrahedron when its volume is 27?

Explanation

The problem is essentially asking us to go from a three-dimensional measurement to a two-dimensional one. In order to approach the problem, it's helpful to see how volume and surface area are related.

This can be done by comparing the formulas for surface area and volume:

$V= \frac{a^3}{6\sqrt{2}}$