Tetrahedrons - ACT Math

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Question

Calculate the diagonal of a regular tetrahedron (all of the faces are equilateral triangles) with side length .

Answer

The diagonal of a shape is simply the length from a vertex to the center of the face or vertex opposite to it. With a regular tetrahedron, we have a face opposite to the vertex, and this basically amounts to calculating the height of our shape.

We know that the height of a tetrahedron is ${\large }s\sqrt{\frac{2}{3}}$ where s is the side length, so we can put into this formula:

${\large} 9\sqrt{2/3} = 3(3/\sqrt3)(\sqrt2) = 3\sqrt{3\cdot2} = 3\sqrt{6}$

which gives us the correct answer.

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Question

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?

Answer

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 $\small 19cm$ .

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Question

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

Answer

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.

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

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

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

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

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Question

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

Answer

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:

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

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

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

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

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Question

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

Answer

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.

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

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

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

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

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Question

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.)

Answer

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 .

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

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

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

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Question

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

Answer

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

$A=\sqrt3 s^2$

Substitute the edge and solve.

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

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Question

Each of the faces of a regular tetrahedron has a base of $\small 11cm$ and a height of . What is the surface area of this tetrahedron?

Answer

The surface area is the area of all of the faces of the tetrahedron. To begin, we must find the area of one of the faces. Because a tetrahedron is made up of triangles, we simply plug the given values for base and height into the formula for the area of a triangle:

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

$A=\frac{1}{2}(11cm)(9cm)$

Therefore, the area of one of the faces of the tetrahedron is . However, because a tetrahedron has 4 faces, in order to find the surface area, we must multiply this number by 4:

$S.A.=4 \cdot 49.5cm^2$

$\small S.A.=198cm^2$

Therefore, the surface area of the tetrahedron is .

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Question

What is the surface area of a regular tetrahedron with a slant height of $\frac{\sqrt3}{2}$ ?

Answer

If this is a regular tetrahedron, then all four triangles are equilateral triangles.

If the slant height is $\frac{\sqrt3}{2}$ , then that equates to the height of any of the triangles being $\frac{\sqrt3}{2}$ .

In order to solve for the surface area, we can use the formula

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

where in this case is the measure of the edge.

The problem has not given the edge; however, it has provided information that will allow us to solve for the edge and therefore the surface area.

Picture an equilateral triangle with a height $\frac{\sqrt3}{2}$ .

Drawing in the height will divide the equilateral triangle into two 30/60/90 right triangles. Because this is an equilateral triangle, we can deduce that finding the measure of the hypotenuse will suffice to solve for the edge length ().

In order to solve for the hypotenuse of one of the right triangles, either trig functions or the rules of the special 30/60/90 triangle can be used.

Using trig functions, one option is using $cos(30^{\circ})=\frac{\frac{\sqrt3}{2}}{a}$ .

Rearranging the equation to solve for ,

$a \cdot cos(30^{\circ}) = \frac{\sqrt3}{2}$

$a = \frac{\frac{\sqrt3}{2}}{cos(30^{\circ})}$

Now that has been solved for, it can be substituted into the surface area equation.

$SA= \sqrt{3}\cdot (1)^2$

$SA= \sqrt{3} \cdot 1$

$SA = {\color{Blue} \sqrt{3}}$

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Question

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

Answer

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:

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

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

We can see that both calculation revolve around the edge length.

That means, if we can solve for (edge length) using volume, we can solve for the surface area.

$27=\frac{a^3}{6\sqrt{2}}$

$27 \cdot 6\sqrt{2}= a^3$

$\sqrt[3]{27 \cdot 6\sqrt{2}}=a$

${\color{Green} 6.12}=a$

Now that we know , we can substitute this value in for the surface area formula:

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

$SA = \sqrt{3} \cdot (6.12)^2$

$SA={\color{Blue} 64.9}$

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Question

Find the volume of a regular tetrahedron if one of its edges is $\sqrt[3]{6}:cm$ long.

Answer

Write the volume equation for a tetrahedron.

$V=\frac{e^3}{6\sqrt2}$

In this formula, stands for the tetrahedron's volume and stands for the length of one of its edges.

Substitute the given edge length and solve.

$V=\frac{(\sqrt[3]6:cm)^3}{6\sqrt2} = \frac{6:cm^3}{6\sqrt2}= \frac{1}{\sqrt2}:cm$

Rationalize the denominator.

$\frac{1}{\sqrt2}:cm\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{2}:cm$

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Question

Find the volume of a tetrahedron if the side length is $\frac{1}{6}$ .

Answer

Write the equation to find the volume of a tetrahedron.

$V=\frac{a^3}{6\sqrt2}$

Substitute the side length and solve for the volume.

$V=\frac{(\frac{1}{6})^3}{6\sqrt2}= \frac{1}{216}\left(\frac{1}{6\sqrt2}\right)= \frac{1}{1296\sqrt2}$

Rationalize the denominator.

$V=\frac{1}{1296\sqrt2} \cdot \frac{\sqrt2}{\sqrt2}= \frac{\sqrt2}{1296\cdot 2} = \frac{\sqrt2}{2592}$

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Question

What is the volume of a regular tetrahedron with an edge length of 6?

Answer

The volume of a tetrahedron can be solved for by using the equation:

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

where is the measurement of the edge of the tetrahedron.

This problem can be quickly solved by substituting 6 in for .

$V=\frac{6^3}{6\sqrt{2}}=\frac{216}{6\sqrt{2}}=\frac{36}{\sqrt{2}}$

${\color{Blue} V=25.5}$

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