Geometry - How to find the length of an edge

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

Explanation

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 :

$s^{2} \sqrt{3} = 1,000$

Divide by $\sqrt{3}$ :

$\frac{s^{2} \sqrt{3} }{ \sqrt{3} }= \frac{1,000}{ \sqrt{3} }$

$s^{2} = \frac{1,000}{1.7321 }$

$s^{2} \approx 577.3503$

Take the square root of both sides:

$s \approx \sqrt{577.3503}$

$s \approx 24.0281$

Of the given choices, 20 comes closest.

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

$\small \small 2 cm$

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

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

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

None of the above.

Explanation

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 $\small 2=a$

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

$2\sqrt[3]{15}$

$2\sqrt{30}$

$6\sqrt{400}$

$2\sqrt[3]{20}$

Explanation

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

$\small 24 ft$

$\small 12 ft$

$\small 4 ft$

$\small 20 ft$

None of the above.

Explanation

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 $\small \small 49\sqrt{3}cm^2$ ?

$\small 7cm$

$\small 7\sqrt{3}cm$

$\small \frac{7}{\sqrt{3}}cm$

$\small \frac{7}{3}cm$

None of the above.

Explanation

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

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

$\small \frac{5}{\sqrt{3}}$ $\small m$

$\small 10\sqrt{3}$ $\small m$

$\small 10\sqrt[3]{3}$ $\small m$

None of the above.

Explanation

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.

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

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

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

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

Explanation

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.

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

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

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

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

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

$\small 6\sqrt[6]{2}m$

$\small 216\sqrt{2}m$

$\small 6\sqrt[5]{2}m$

$\small 36\sqrt{2}m$

None of the above.

Explanation

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.

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

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

Tetra

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 ?

Explanation

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