Other 3-Dimensional Geometry

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SAT Math › Other 3-Dimensional Geometry

Questions 1 - 4

1

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.

2

What is the volume of a triangular prism having a base of 2, a height of 8, and second height of 14?

Explanation

To find the volume of a triangular prism we use the equation

$V= \frac{1}{2}(b*h)*h_2$

In our case our

Therefore,

$V=\frac{1}{2}(2*8)*14$

$V=\frac{1}{2}(16)*14$

3

A rectangular prism has a height of 4 in., length of 8in., and a width of 7in. Find the volume of the prism.

$224\ \textup{in.}$

$17\ \textup{in.}$

$230\ \textup{in.}$

$146\ \textup{in.}$

Explanation

To find the volume of a rectangular prism we use the equation of:

Now substituting in our values we get:

4

If a cube has a surface area of then what is the the length of its longest diagonal?

$3\sqrt{2}$

$\frac{1}{3}\sqrt{3}$

$\sqrt{6}$

Explanation

The longest diagonal of a cube transverses the interior of the figure:

Cube diagonal

This distance is defined by the super pythagorean theorem :

$distance = \sqrt{a^2+b^2+c^2}$

where , , and are the length, width, and height. Because the figure is a cube, all three are the same measure, and each is the side of the cube.

We can use the given surface area to find the length of the side:

$area = 6\cdot side^2$

$\sqrt{6} = side$

We can use this value for the side to plug into the super pythagorean theorem

$distance = \sqrt{a^2+b^2+c^2}$

$distance = \sqrt{\sqrt{6}^2+\sqrt{6}^2+\sqrt{6}^2}$

$distance = \sqrt{18}$

Which can be simplified to

$3\sqrt{2}$

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