Tetrahedrons

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Geometry › Tetrahedrons

Questions 1 - 10

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.

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

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 volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

The volume of a tetrahedron is found with the formula:

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

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

The volume of a tetrahedron is found with the formula:

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

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

A regular tetrahedron has a surface area of $684cm^{2}$ . Each face of the tetrahedron has a height of . What is the length of the base of one of the faces?

Explanation

A regular tetrahedron has 4 triangular faces. The area of one of these faces is given by:

$\small A=\frac{1}{2}bh$

Because the surface area is the area of all 4 faces combined, in order to find the area for one of the faces only, we must divide the surface area by 4. We know that the surface area is $\small 684cm^2$ , therefore:

$\small A=\frac{S.A.}{4}$

$A=\frac{684cm^{2}}{4}=171cm^2$

Since we now have the area of one face, and we know the height of one face is , we can now plug these values into the original formula:

$\small A=\frac{1}{2}bh$

$\small 171=\frac{1}{2}b(18)$

$\small 171=9b$

$\small b=19$

Therefore, the length of the base of one face is $\small 19cm$ .

A regular tetrahedron has a surface area of $684cm^{2}$ . Each face of the tetrahedron has a height of . What is the length of the base of one of the faces?

Explanation

A regular tetrahedron has 4 triangular faces. The area of one of these faces is given by:

$\small A=\frac{1}{2}bh$

$\small A=\frac{S.A.}{4}$

$A=\frac{684cm^{2}}{4}=171cm^2$

Since we now have the area of one face, and we know the height of one face is , we can now plug these values into the original formula:

$\small A=\frac{1}{2}bh$

$\small 171=\frac{1}{2}b(18)$

$\small 171=9b$

$\small b=19$

Therefore, the length of the base of one face is $\small 19cm$ .

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:

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

Given a regular tetrahedron with an edge of $\small \small 20 cm$ , what is the height (or diagonal)? The height is the line drawn from one vertex perpendicular to the opposite face.

Tetrahedron_20

$\small 20\sqrt{\frac{2}{3}}cm$

$\small 10\sqrt{\frac{2}{3}}cm$

$\small 20\sqrt{2}cm$

$\small 10\sqrt{2}cm$

None of the above.

Explanation

The height of a regular tetrahedron can be derived from the formula $\small h= \sqrt{\frac{2}{3}}a$ where $\small a$ is the length of one edge.

Plugging in we can solve for .

$\small h= \sqrt{\frac{2}{3}}\cdot 20$

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 $60^{\circ}$ .

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:

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

Given a regular tetrahedron with an edge of $\small \small 20 cm$ , what is the height (or diagonal)? The height is the line drawn from one vertex perpendicular to the opposite face.

Tetrahedron_20

$\small 20\sqrt{\frac{2}{3}}cm$

$\small 10\sqrt{\frac{2}{3}}cm$

$\small 20\sqrt{2}cm$

$\small 10\sqrt{2}cm$

None of the above.