Solid Geometry - Advanced Geometry

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Question

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.

Answer

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 ,

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

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Question

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.

Answer

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 .

$\small h= \sqrt{\frac{2}{3}}\cdot 20$

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Question

Find the height of this regular tetrahedron:

Answer

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 $2\sqrt{2}$ .

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Question

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

Answer

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 $\small 2=a$

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Question

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?

Answer

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 $\small 24 ft^2$

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Question

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

Answer

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

$\small \small b=7$ .

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Question

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

Answer

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.

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

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

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Question

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

Answer

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.

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

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

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

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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

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

Answer

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

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

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Question

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

Answer

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:

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

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Question

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

Answer

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.

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Question

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 ?

Answer

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.

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Question

The lateral area is twice as big as the base area of a cone. If the height of the cone is 9, what is the entire surface area (base area plus lateral area)?

Answer

Lateral Area = LA = π(r)(l) where r = radius of the base and l = slant height

LA = 2B

π(r)(l) = 2π(r2)

rl = 2r2

l = 2r

From the diagram, we can see that r2 + h2 = l2. Since h = 9 and l = 2r, some substitution yields

r2 + 92 = (2r)2

r2 + 81 = 4r2

81 = 3r2

27 = r2

B = π(r2) = 27π

LA = 2B = 2(27π) = 54π

SA = B + LA = 81π

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Question

A right cone has a radius of 4R and a height of 3R. What is the ratio of the total surface area of the cone to the surface area of just the base?

Answer

We need to find total surface area of the cone and the area of the base.

The area of the base of a cone is equal to the area of a circle. The formula for the area of a circle is given below:

$A=\pi r^2$ , where r is the length of the radius.

In the case of this cone, the radius is equal to 4R, so we must replace r with 4R.

$A=\pi(4R)^2=\pi(4)^2(R)^2=16\pi R^2$

To find the total area of the cone, we need the area of the base and the lateral surface area of the cone. The lateral surface area (LA) of a cone is given by the following formula:

$LA=\frac{1}{2}(2\pi r)(l)$ , where r is the radius and l is the slant height.

We know that r = 4R. What we need now is the slant height, which is the distance from the edge of the base of the cone to the tip.

In order to find the slant height, we need to construct a right triangle with the legs equal to the height and the radius of the cone. The slant height will be the hypotenuse of this triangle. We can use the Pythagorean Theorem to find an expression for l. According to the Pythagorean Theorem, the sum of the squares of the legs (which are 4R and 3R in this case) is equal to the square of the hypotenuse (which is the slant height). According to the Pythagorean Theorem, we can write the following equation:

Let's go back to the formula for the lateral surface area (LA).

$LA=\frac{1}{2}(2\pi r)(l)$

$LA=\frac{1}{2}(2\pi(4R))(5R)$

$LA=20\pi R^2$

To find the total surface area (TA), we must add the lateral area and the area of the base.

$TA=20\pi R^2+16\pi R^2=36\pi R^2$

The problem requires us to find the ratio of the total surface area to the area of the base. This means we must find the following ratio:

$\frac{36\pi R^2}{16\pi R^2}$

We can cancel $\pi R^2$ , which leaves us with 36/16.

Simplifying 36/16 gives 9/4.

The answer is 9/4.

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Question

What is the surface area of a cone with a radius of 6 in and a height of 8 in?

Answer

Find the slant height of the cone using the Pythagorean theorem: _r_2 + _h_2 = _s_2 resulting in 62 + 82 = _s_2 leading to _s_2 = 100 or s = 10 in

SA = πrs + πr_2 = π(6)(10) + π(6)2 = 60_π + 36_π_ = 96_π_ in2

60_π_ in2 is the area of the cone without the base.

36_π_ in2 is the area of the base only.

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