Tetrahedrons - Geometry

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A tetrahedron has a volume that is twice the surface area times the edge. What is the length of the edge? (In the answer choices, represents edge.)

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The problem states that the volume is:

$V= 2($\sqrt{3}$\cdot $a^2$) \cdot a$

The point of the problem is to solve for the length of the edge. Becasuse there are no numbers, the final answer will be an expression.

In order to solve for it, we will have to rearrange the formula for volume in terms of .

$V = 2$\sqrt{3}$\cdot $a^3$$

$a= \sqrt[3]{$\frac{V}{2\sqrt3}$}$

Given a regular tetrahedron with an edge of $\small 10 cm$ , what is the height (or diagonal)? The height is the line drawn from one vertex perpendicular to the opposite face.

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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.

Therefore, plugging in the side length of ,

$h=\sqrt$\frac{2}{3}$\cdot 10$ .

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.

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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$

Find the height of this regular tetrahedron:

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The height of a regular tetrahedron can be found using the formula $h=$\sqrt{\frac{2}{3}$}\cdot s$ where s is the length of the sides.

In this case, the sides have length $2$\sqrt{3}$$ , so we are multiplying $\sqrt{\frac{2}{3}$}\cdot 2$\sqrt{3}$ .

We can simplify this by multiplying the numbers inside the radical:

$2$\sqrt{\frac{2}{3}$\cdot 3}$ , which simplifies to $2$\sqrt{2}$$ .

What is the length of one edge of a regular tetrahedron whose volume equals $\small \small \small $\frac{4}{3\sqrt{2}$} $cm^3$$ ?

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The formula for the volume of a tetrahedron is:

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

When $\small V= $\frac{4}{3\sqrt{2}$}$ we have $\small $$\frac{4}{3\sqrt{2}$}=\frac{a^3$}{6\sqrt{2}$}$ .

Multiplying the left side by $\small $\frac{2}{2}$$ gives us,

$\small $$\frac{8}{6\sqrt{2}$}=\frac{a^3$}{6\sqrt{2}$}$ , or $\small $8=a^3$$ .

Finally taking the third root of both sides yields $\small 2=a$

A regular tetrahedron has a total surface area of $\small 16$\sqrt{3}$ $ft^2$$ . What is the combined length of all of its edges?

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A regular tetrahedron has four faces of equal area made of equilateral triangles.

Therefore, we know that one face will be equal to:

$\small \left($\frac{1}{4}\right)\left(16$\sqrt{3}\right) $ft^2$$ , or $\small 4$\sqrt{3}$ $ft^2$$

Since the surface of one face is an equilateral triangle, and we know that,

$\small Area=\left($\frac{1}{2}\right)b\times h$ , the problem can be expressed as:

$\small 4$\sqrt{3}$=\left($\frac{1}{2}\right)b\times h$

In an equilateral triangle, the height $\small h$ , is equal to $\small $\frac{1}{2}$(b)$\sqrt{3}$$ so we can substitute for $\small h$ like so:

$\small 4$\sqrt{3}$=\left($\frac{1}{2}\right)b \times \left($\frac{1}{2}\right)b$\sqrt{3}$$

Solving for $\small b$ gives us the length of one edge.

$\small $4$\sqrt{3}$=\frac{b^2$\sqrt{3}$}{4}$

$\small $(4)(4$\sqrt{3}$)=(4)\left($\frac{b^2$\sqrt{3}$}{4}\right)$

$\small $16$\sqrt{3}$=b^2$$\sqrt{3}$$

$\small $16=b^2$$

$\small b=\pm 4$

However, we know that the edge of the tetrahedron is a positive number so $\small b=4$ .

Since the base $\small b$ is the same as one edge of the tetrahedron, and a tetrahedron has six edges we multiply $\small 6 \times 4$ to arrive at $\small 24 $ft^2$$

What is the length of one edge of a regular tetrahedron when the total surface area equals $\small \small $49$\sqrt{3}$cm^2$$ ?

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A regular tetrahedron has four faces of equal area made of equilateral triangles.

Therefore, we know that one face will be equal to,

$\left($\frac{1}{4}\right)49$\sqrt{3}$$ cm , or $\small \small $\frac{49}{4}$$\sqrt{3}$$ cm.

Since the surface of one face is an equilateral triangle, and we know that,

$\small Area=\left($\frac{1}{2}\right)b\times h$ , the problem can be expressed as:

$\small \small $\frac{49}{4}$$\sqrt{3}$=\left($\frac{1}{2}\right)b\times h$

In an equilateral triangle, the height $\small h$ is equal to $\small $\frac{1}{2}$(b)$\sqrt{3}$$ so we can substitute for $\small h$ like so:

$\small \small $\frac{49}{4}$$\sqrt{3}$=\left($\frac{1}{2}\right)b \times \left($\frac{1}{2}\right)b$\sqrt{3}$$

Solving for $\small b$ gives us the length of one edge.

$\small \small $$\frac{49}{4}$$\sqrt{3}$=\frac{b^2$\sqrt{3}$}{4}$

$\small \small $49$\sqrt{3}$=b^2$$\sqrt{3}$$

$\small \small $49=b^2$$

$\small b=\pm7$

However, we know that the edge of the tetrahedron is a positive number so

$\small \small b=7$ .

What is the length of one edge of a regular tetrahedron whose volume equals $\small \small \small 36$ $\small $m^3$$ ?

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The formula for the volume of a tetrahedron is $\small $V=\frac{a^3$}{6\sqrt{2}$}$ . When $\small \small V= 36$ we have $\small $36=\frac{a^3$}{6\sqrt{2}$}$ .

We simply solve for $\small a$ ...

$\small $6$\sqrt{2}$\times36=a^3$$

$\small $216$\sqrt{2}$=a^3$$ .

Take the cube root of both sides to find the answer for a.

$\small \small 6\left ( \left ( $2\right^{1/2}$ ) \right $)^{1/3}$=a$

$\small \small a=6 \sqrt[6]{2}$

What would the length of one edge of a regular tetrahedron be if the area of one side was $\small \small 25$ $\small $m^2$$ ?

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The area of one side is given as $\small 25$ $\small $m^2$$ . The side of a regular tetrahedron is an equilateral triangle so area is determined by:

$\small Area= $\frac{1}{2}$b\times h$ .

In an equilateral triangle, $\small h=\frac{1}{2}$b$\sqrt{3}$$ so we can substitute for $\small h$ into the area formula:

$\small Area = $\frac{1}{2}$b\times $\frac{1}{2}$b$\sqrt{3}$$ .

Plugging in the value of the area which was given yields.

$\small 25= $\frac{1}{2}$b\times $$\frac{1}{2}$b$\sqrt{3}$=\frac{b^2$\sqrt{3}$}{4}$

Solve for $\small b$ will give us the length of an edge.

$\small $100=b^2$$\sqrt{3}$$

$\small $b^2$=\frac{100}{\sqrt{3}$}$

$\small b=\frac{10}{\sqrt[4]{3}$}$

A regular tetrahedron has a surface area of . Each face of the tetrahedron has a height of . What is the length of the base of one of the faces?

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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}$$

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 length of an edge of a regular tetrahedron if its surface area is 156?

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The only given information is the surface area of the regular tetrahedron.

This is a quick problem that can be easily solved for by using the formula for the surface area of a tetrahedron:

$SA= $\sqrt{3}$\cdot $a^2$$

If we substitute in the given infomation, we are left with the edge being the only unknown.

$156 = $\sqrt{3}$ \cdot $a^2$$

$$\sqrt{\frac{156}{\sqrt{3}$}}=a$

$a=9.49 \approx{\color{Blue} 9.5}$

What is the length of a regular tetrahedron if one face has an area of 43.3 squared units and a slant height of $\frac{10\sqrt3}{2}$ ?

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The problem provides the information for the slant height and the area of one of the equilateral triangle faces.

The slant height merely refers to the height of this equilateral triangle.

Therefore, if we're given the area of a triangle and it's height, we should be able to solve for it's base. The base in this case will equate to the measurement of the edge. It's helpful to remember that in this case, because all faces are equilateral triangles, the measure of one length will equate to the length of all other edges.

We can use the equation that will allow us to solve for the area of a triangle:

$A=\frac{1}{2}$ \cdot b\cdot h$

where is base length and is height.

Substituting in the information that's been provided, we get:

$43.3 = $\frac{1}{2}$ \cdot b \cdot $\frac{10\sqrt{3}$}{2}$

$b \cdot $\frac{10\sqrt{3}$}{2}= 2 \cdot 43.3$

$b = 2 \cdot 43.3 \cdot $\frac{2}{10\sqrt{3}$}$

$b =9.99971\approx {\color{Blue} 10}$

The volume of a regular tetrahedron is 94.8. What is the measurement of one of its edges?

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This becomes a quick problem by just utilizing the formula for the volume of a tetrahedron.

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

Upon substituting the value for the volume into the formula, we are left with , which represents the edge length.

$94.8= $$\frac{a^3$}{6\sqrt{2}$}$

$a= \sqrt[3]{94.8 \cdot 6$\sqrt{2}$}$

$a= {\color{Blue} 9.3}$

In order for the height of a regular tetrahedron to be one, what should the lengths of the sides be?

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The formula for the height of a regular tetrahedron is $h=$\sqrt{\frac{2}{3}$}\cdot s$ , where s is the length of the sides.

In this case we want h to be 1, so we need something that multiplies to 1 with $$\sqrt{\frac{2}{3}$}$ .

We know that $$\frac{2}{3}$\cdot $\frac{3}{2}$=1$ , so then we know that

$\sqrt{\frac{2}{3}$}\cdot $\sqrt{\frac{3}{2}$}=$\sqrt{1}$ , which equals 1.

Therefore s should be

$$\sqrt{\frac{3}{2}$}$ .

The volume of a regular tetrahedron is $\sqrt{200}$ . Find the length of one side.

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The formula for the volume of a regular tetrahedron is $V = \frac{s^3 }{6\sqrt{2}}$ .

In this case we know that the volume, V, is $\sqrt{200}$ , so we can plug that in to solve for s, the length of each edge:

$\sqrt{200}=\frac{s^3}{6\sqrt{2}}$ \[multiply both sides by $6\sqrt{2}$ \]

$6\sqrt{400}=s^3$ \[evaluate $\sqrt{400}=20$ and multiply\]

\[take the cube root of each side\]

$\sqrt[3]{120} = s$ .

We can simplify this by factoring 120 as the product of 8 times 15. Since the cube root of 8 is 2, we get:

$2\sqrt[3]{15}$ .

A regular tetrahedron has surface area 1,000. Which of the following comes closest to the length of one edge?

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A regular tetrahedron has six congruent edges and, as its faces, four congruent equilateral triangles. If we let be the length of one edge, each face has as its area

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

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

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

$S = $s^{2}$ $\sqrt{3}$$

Set and solve for :

Divide by $\sqrt{3}$ :

Take the square root of both sides:

$s \approx $\sqrt{577.3503}$$

$s \approx 24.0281$

Of the given choices, 20 comes closest.

The above figure shows a triangular pyramid, or tetrahedron, on the three-dimensional coordinate axes. The tetrahedron has volume 1,000. Which of the following is closest to the value of ?

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If we take the triangle on the -plane to be the base of the pyramid, this base has legs both of length ; its area is half the product of the lengths which is

$B = $\frac{1}{2}$ ab = $\frac{1}{2}$ r ^{2}$

Its height is the length of the side along the -axis, which is also of length .

The volume of a pyramid is equal to one third the product of its height and the area of its base, so

$V = $\frac{1}{3}$ \cdot r \cdot $\frac{1}{2}$ r ^{2}$

$= $\frac{1}{3}$ \cdot $\frac{1}{2}$\cdot r \cdot r ^{2}$

$= $\frac{1}{6}$ r ^{3}$

Setting the volume equal to 1,000, we can solve for :

$$\frac{1}{6}$ r ^{3} = 1,000$

Multiply both sides by 6:

$6\cdot $\frac{1}{6}$ r ^{3} =6\cdot 1,000$

$r ^{3} =6 ,000$

Take the cube root of both sides:

$r =\sqrt[3]{6 ,000}$

$r \approx 18.2$

The closest choice is 20.

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

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A tetrahedron is a three-dimensonal figure where each side is an equilateral triangle. Therefore, each angle in the triangle is .

In the figure, we know the value of the side and the value of the base . Since dividing the triangle by half creates a 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$$

A regular tetrahedron has side lengths . What is the surface area of the described solid?

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The area of one face of the triangle can be found either through trigonometry or the Pythagorean Theorem.

Since all the sides of the triangle are , the height is then $$\frac{\sqrt{3}$}{2}$ , so the area of each face is:

$$\frac{bh}{2}$=\frac{1}{2}$ \cdot 1 \cdot $\frac{\sqrt{3}$}{2}=\frac{\sqrt{3}$}{4}$

There are four faces, so the area of the tetrahedron is:

$4 \cdot $\frac{bh}{2}$=\sqrt3$

If the edge length of a tetrahedron is $\sqrt3$ , what is the surface area of the tetrahedron?

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Write the formula for finding the surface area of a tetrahedron.

$A=\sqrt3 $s^2$$

Substitute the edge and solve.

$A=\sqrt3 $(\sqrt3)^2$ = \sqrt3 (3) = 3\sqrt3$