Pyramids

Help Questions

Math › Pyramids

Questions 1 - 10

Pyramid 1 has a square base with sidelength ; its height is .

Pyramid 2 has a square base with sidelength ; its height is .

Which is the greater quantity?

(a) The volume of Pyramid 1

(b) The volume of Pyramid 2

(b) is greater.

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

Use the formula $V = \frac {1}{3} Bh$ on each pyramid.

(a) $B = x^{2}; h = 2x$

$V = \frac {1}{3} Bh = \frac {1}{3} \cdot x^{2} \cdot 2x = \frac {2}{3} x^{3}$

(b) $B = (2 x)^{2} = 4x^{2}; h = x$

$V = \frac {1}{3} Bh = \frac {1}{3} \cdot 4x^{2} \cdot x = \frac {4}{3} x^{3}$

Regardless of , (b) is the greater quantity.

Pyramid 1 has a square base with sidelength ; its height is .

Pyramid 2 has a square base with sidelength ; its height is .

Which is the greater quantity?

(a) The volume of Pyramid 1

(b) The volume of Pyramid 2

(b) is greater.

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

Use the formula $V = \frac {1}{3} Bh$ on each pyramid.

(a) $B = x^{2}; h = 2x$

$V = \frac {1}{3} Bh = \frac {1}{3} \cdot x^{2} \cdot 2x = \frac {2}{3} x^{3}$

(b) $B = (2 x)^{2} = 4x^{2}; h = x$

$V = \frac {1}{3} Bh = \frac {1}{3} \cdot 4x^{2} \cdot x = \frac {4}{3} x^{3}$

Regardless of , (b) is the greater quantity.

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.

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 $\small 24 ft^2$

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:

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

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 $\small 2=a$

A cube has sidelength one and one-half feet; a rectangular prism of equal volume has length 27 inches and height 9 inches. Give the width of the prism in inches.

$24 \textrm{ in}$

$36\textrm{ in}$

$27 \textrm{ in}$

$18 \textrm{ in}$

$30\textrm{ in}$

Explanation

One and one half feet is equal to eighteen inches, so the volume of the cube, in cubic inches, is the cube of this, or

$V = 18^{3} = 5,832$ cubic inches.

The volume of a rectangular prism is

Since its volume is the same as that of the cube, and its length and height are 27 and 9 inches, respectively, we can rewrite this as

$5,832= 27 \cdot 9 \cdot W$

$5,832= 243 \cdot W$

$W = 5,832 \div 243 = 24$

The width is 24 inches.

A cube has sidelength one and one-half feet; a rectangular prism of equal surface area has length 27 inches and height 9 inches. Give the width of the prism in inches.

$20 \frac{1}{4} \textrm{ in}$