SAT Math - 3-Dimensional Geometry

Find the surface area of a sphere with a radius of 3.

$36\pi$

$27\pi$

$72\pi$

$18\pi$

$48\pi$

Explanation

Write the formula for the surface area of a sphere.

$A=4\pi r^2$

Substitute the radius into the equation.

$A=4\pi(3)^2 = 4\pi (9) = 36\pi$

The answer is: $36\pi$

A convex polyhedron has twenty faces and thirty-six vertices. How many edges does it have?

Explanation

The number of vertices , edges , and faces of any convex polyhedron are related by By Euler's Formula:

Setting and solving for :

The polyhedron has 54 edges.

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

A convex polyhedron has twenty faces and thirty-six vertices. How many edges does it have?

Explanation

The number of vertices , edges , and faces of any convex polyhedron are related by By Euler's Formula:

Setting and solving for :

The polyhedron has 54 edges.

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

Find the surface area of a sphere with a radius of 3.

$36\pi$

$27\pi$

$72\pi$

$18\pi$

$48\pi$

Explanation

Write the formula for the surface area of a sphere.

$A=4\pi r^2$

Substitute the radius into the equation.

$A=4\pi(3)^2 = 4\pi (9) = 36\pi$

The answer is: $36\pi$

What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

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

Explanation

The volume of a tetrahedron is found with the formula:

$\small V= \frac{a^3}{6\sqrt{2}}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

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

Explanation

The volume of a tetrahedron is found with the formula:

$\small V= \frac{a^3}{6\sqrt{2}}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

Which of the following numbers comes closest to the length of line segment in three-dimensional coordinate space whose endpoints are the origin and the point ?

Explanation

Use the three-dimensional version of the distance formula:

$d= \sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

$d= \sqrt{(3-0)^{2} +(4-0)^{2}+(5-0)^{2}}$

$d= \sqrt{3^{2} +4^{2}+5^{2}}$

$d= \sqrt{9+16+25}$

$d= \sqrt{50}$

$d \approx 7.1$

The closest of the five choices is 7.

The surface area of cone $\small E$ is $\small 70\pi$ . If the radius of the base of the cone is , what is the height of the cone?

$\small \small 2\sqrt{14}$

$\small 45$

$\small 9$

$\small 3$

$\small 2\sqrt{5}$

Explanation

To figure out $\small h$ , we must use the equation for the surface area of a cone, $\small A=\pi r^2+\pi rs$ , where $\small r$ is the radius of the base of the cone and $\small s$ is the length of the diagonal from the tip of the cone to any point on the base's circumference. We therefore first need to solve for $\small s$ by plugging what we know into the equation:

$\small 70\pi=(5^2)\pi +5s\pi=25\pi +5s\pi$

This equation can be reduced to:

$\small 45\pi=5s\pi$

$\small 9=s$

For a normal right angle cone, $\small s$ represents the line from the tip of the cone running along the outside of the cone to a point on the base's circumference. This line represents the hypotenuse of the right triangle formed by the radius and height of the cone. We can therefore solve for $\small h$ using the Pythagorean theorem: