# SAT Solid Geometry

To successfully solve solid geometry problems on the SAT, you need to know:

- the contents of the Reference Information at the beginning of each math section
- how to find the volume and surface area of a rectangular solid
- how to find the volume and surface area of a cylinder

#### Contents

## Examples

How many cubes each with edge length of 2 cm would fit in a rectangular box with dimensions 20 cm by 10 cm by 8 cm?

(A) \(\ \ 8\)

(B) \(\ \ 200\)

(C) \(\ \ 400\)

(D) \(\ \ 800\)

(E) \(\ \ 1600\)

Correct Answer: B

Solution:Each cube has volume \(V = s^3 = 2^3 =8\) cm\(^3.\)

The rectangular solid has volume \(V=lwh = 20\times 10\times8 = 1600\) cm\(^3.\)

Therefore, \(\frac{1600}{8}=200\) cubes of the given dimensions will fit in the rectangular box.

Incorrect Choices:

(A)

This is the volume of one cube, not the number of cubes that will fit in the box.

(C)

Tip: The volume of a cube with edge length \(s\): \(V = s^3.\)

If you use the wrong formula \(V=s^2\) to find the volume of the cube, you will get this wrong answer.

(D)

If you divide the volume of the box by the edge length of the cube, you will get this wrong answer.

(E)

Tip: Read the entire question carefully.

If you think each cube has edge length of 1, you will get this wrong answer.

A white die has gray dots. If the area of each of the die's faces is \(a\) and the area of each of the die's dots is \(b,\) what is the area of the white surface on the die in terms of \(a\) and \(b?\)

(A) \(\ \ 4a-21b\)

(B) \(\ \ 6a - 6b\)

(C) \(\ \ 6a-21b\)

(D) \(\ \ 4a^2 - 21\pi b\)

(E) \(\ \ 6a^2 - 21\pi b^2\)

What is the surface area of the triangular prism above?

(A) \(\ \ 120\)

(B) \(\ \ 128\)

(C) \(\ \ 150\)

(D) \(\ \ 180\)

(E) \(\ \ 240\)

Correct Answer: D

Solution:

Tip: Pythagorean Theorem: \(a^2 + b^2 = c^2.\)

By the Pythagorean theorem, the longer leg of the right triangle is\[\sqrt{13^2-5^2} = \sqrt{169-25} = \sqrt{144} = 12.\]

We have:

\[A_{\text{triangle}} = \frac{1}{2}bh = \frac{1}{2} \cdot 12 \cdot 5 = 30\]

\[A_{\text{base}} = lw = 4 \cdot 12 = 48\]

\[A_{\text{side}} = lw = 5 \cdot 4 = 20\]

\[A_{\text{diagonal rectangle}} = lw = 4 \cdot 13 = 52\]

The surface area of the whole shape is

\[SA = 2 \cdot A_{\text{triangle}} + A_{\text{base}} + A_{\text{side}} + A_{\text{diagonal rectangle}} = 2 \cdot 30 + 48 + 20 + 52= 180.\]

Incorrect Choices:

(A)

Tip: Read the entire question carefully.

If you find the volume of the given figure, you will get this wrong answer.

(B)

Tip: Read diagrams carefully.

If you forget to account for the diagonal rectangle (dimensions 4 by 13), you will get this wrong answer.

(C)

Tip: Read diagrams carefully.

If you find the area of only one triangle and the three rectangles, then you will get this wrong answer.

(E)

Tip: Read the entire question carefully.

If you find the volume of a rectangular solid with dimensions \(5 \times 12 \times 4,\) you will get this wrong answer.

## Review

If you thought these examples difficult and you need to review the material, these links will help:

## SAT Tips for Solid Geometry

- Area of a triangle with height \(h\) and base \(b\): \(A_{\triangle} = \frac{1}{2}bh.\)
- Know the \(30^\circ-60^\circ-90^\circ\) and the \(45^\circ-45^\circ-90^\circ\) Theorems.
- Area of a circle with radius \(r: A_{\bigodot} = \pi r^2.\)
- The perimeter of a square with side length \(s\): \(P_{\square} = 4s.\)
- The volume of a cube with edge length \(s\): \(V = s^3.\)
- The volume of a rectangular solid with length \(l,\) width \(w,\) and height \(h: V = l \cdot w \cdot h.\)
- The surface area of a cube with edge length \(s\): \(SA = 6s^2.\)
- Volume of a cylinder with base radius \(r\) and height \(h: V = \pi r^2 h.\)
- SAT General Tips

**Cite as:**SAT Solid Geometry.

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