Advanced Geometry - Geometry

Card 0 of 1888

Question

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

Answer

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:

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

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Question

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

Answer

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:

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

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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

Find the surface area of a regular tetrahedron with a side length of .

Answer

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.

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

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Question

Find the surface area of a regular tetrahedron with a side length of $\frac{1}{2}$ .

Answer

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.

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

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Question

Find the surface area of a regular tetrahedron with a side length of .

Answer

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.

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

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Question

In terms of , find the surface area of a regular tetrahedron that has a side length of .

Answer

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.

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

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Question

In terms of , find the surface area of a regular tetrahedron with a side length of $\frac{1}{4}z$ .

Answer

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.

$\text{Surface Area}=\sqrt3\left(\frac{z}{4}\right)^2=\frac{z^2\sqrt3}{16}$

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Question

In terms of , find the surface area of a regular tetrahedron with side lengths of .

Answer

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.

$\text{Surface Area}=\sqrt3(3q)^2=9q^2\sqrt3$

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Question

The surface area of a regular tetrahedron is $16\sqrt3$ . If each side length is , find the value of .

Answer

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.

$16\sqrt3=\sqrt3(2x)^2$

Now, solve for .

$x=\sqrt4=2$

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Question

The surface area of a regular tetrahedron is $36\sqrt3$ . If the length of each side is , find the value of .

Answer

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.

$36\sqrt3=\sqrt3(3a)^2$

Now, solve for .

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Question

The surface area of a regular tetrahedron is . If each side length is , find the value of . Round to the nearest tenths place.

Answer

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.

$54=\sqrt3(6x)^2$

Solve for .

$36x^2\sqrt3=54$

$x^2\sqrt3=1.5$

$x^2=\frac{1.5}{\sqrt3}=0.866$

$x=\sqrt{0.866}=0.9$

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Question

The surface area of a regular tetrahedron is . If each side length is , find the value of . Round to the nearest tenths place.

Answer

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.

$64=\sqrt3(4x)^2$

$16x^2=\frac{64}{\sqrt3}$

$x^2=\frac{64}{16\sqrt3}=2.309$

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Question

The surface area of a regular tetrahedron is $100\sqrt3$ . If each side length is , find the value of .

Answer

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 and solve for .

$100\sqrt3=\sqrt3(x-1)^2$

Since we are dealing with a 3-dimensional shape, only is valid.

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Question

In terms of , find the surface area of a regular tetrahedron with side lengths .

Answer

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.

$\text{Surface Area}=\sqrt3(9c)^2=81c^2\sqrt3$

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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

Give the surface area of a regular tetrahedron with edges of length 60.

Answer

A tetrahedron comprises four triangular surfaces; if the tetrahedron is regular, then each surface is an equilateral triangle. The area of an equilateral triangle with sides of length can be computed using the formula

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

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

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

Set and substitute:

$S = s^{2}\sqrt{3} = 60 ^{2}\sqrt{3} = 3,600 \sqrt{3}$ .

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Question

Give the -intercept(s) of the parabola with equation $f(x) = x^{2}+4x+7$ . Round to the nearest tenth, if applicable.

Answer

The -coordinate(s) of the -intercept(s) are the real solution(s) to the equation $f(x) = x^{2}+4x+7 = 0$ . We can use the quadratic formula to find any solutions, setting - the coefficients of the expression.

An examination of the discriminant $b^{2} - 4ac$ , however, proves this unnecessary.

$b^{2} - 4ac = 4^{2} - 4\cdot 1\cdot 7 = -12$

The discriminant being negative, there are no real solutions, so the parabola has no -intercepts.

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