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

\(\displaystyle A = \frac{s^{2} \sqrt{3}}{4}\);

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

\(\displaystyle S = 4A\)

\(\displaystyle S = 4 \cdot \frac{s^{2} \sqrt{3}}{4}\)

\(\displaystyle S = s^{2} \sqrt{3}\)

Set \(\displaystyle S = 1,000\) and solve for \(\displaystyle s\):

\(\displaystyle s^{2} \sqrt{3} = 1,000\)

Divide by \(\displaystyle \sqrt{3}\):

\(\displaystyle \frac{s^{2} \sqrt{3} }{ \sqrt{3} }= \frac{1,000}{ \sqrt{3} }\)

\(\displaystyle s^{2} = \frac{1,000}{1.7321 }\)

\(\displaystyle s^{2} \approx 577.3503\)

Take the square root of both sides:

\(\displaystyle s \approx \sqrt{577.3503}\)

\(\displaystyle s \approx 24.0281\)

Of the given choices, 20 comes closest.

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

\(\displaystyle B = \frac{1}{2} ab = \frac{1}{2} r ^{2}\)

Its height is the length of the side along the \(\displaystyle z\)-axis, which is also of length \(\displaystyle r\). 

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

\(\displaystyle V = \frac{1}{3} \cdot r \cdot \frac{1}{2} r ^{2}\)

\(\displaystyle = \frac{1}{3} \cdot \frac{1}{2}\cdot r \cdot r ^{2}\)

\(\displaystyle = \frac{1}{6} r ^{3}\)

Setting the volume \(\displaystyle V\) equal to 1,000, we can solve for \(\displaystyle r\):

\(\displaystyle \frac{1}{6} r ^{3} = 1,000\)

Multiply both sides by 6:

\(\displaystyle 6\cdot \frac{1}{6} r ^{3} =6\cdot 1,000\)

\(\displaystyle r ^{3} =6 ,000\)

Take the cube root of both sides:

\(\displaystyle r =\sqrt[3]{6 ,000}\)

\(\displaystyle r \approx 18.2\)

The closest choice is 20.

The height of a regular tetrahedron can be derived from the formula 

\(\displaystyle \small h= \sqrt{\frac{2}{3}}a\) where \(\displaystyle \small a\) is the length of one edge.

Therefore, plugging in the side length of \(\displaystyle 10cm\),

\(\displaystyle h=\sqrt\frac{2}{3}\cdot 10\).

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

Plugging in \(\displaystyle a=20\) we can solve for \(\displaystyle h\).

\(\displaystyle \small h= \sqrt{\frac{2}{3}}\cdot 20\)

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

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

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

\(\displaystyle 2\sqrt{\frac{2}{3}\cdot 3}\), which simplifies to \(\displaystyle 2\sqrt{2}\).

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

In the figure, we know the value of the side \(\displaystyle (2\: m)\) and the value of the base \(\displaystyle (1\: m)\). Since dividing the triangle by half creates a \(\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}\) triangle, we know the value of \(\displaystyle h\) must be \(\displaystyle \sqrt{3}\: m\).

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

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

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

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

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 \(\displaystyle 1\), the height is then\(\displaystyle \frac{\sqrt{3}}{2}\), so the area of each face is:

\(\displaystyle \frac{bh}{2}=\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}\)

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

 \(\displaystyle 4 \cdot \frac{bh}{2}=\sqrt3\)

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

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

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

\(\displaystyle \text{Surface Area}=\sqrt3(4^2)=16\sqrt3\)

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

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

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

\(\displaystyle \text{Surface Area}=\sqrt3(12^2)=144\sqrt3\)

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

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

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

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