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Geometry – Study Guide

Table of Contents

Shapes

Triangles
Angles

Equations to Solve Angles and Sides of Triangles

Law of Sines and Cosines
Pythagorean Theorem
Special Right Triangles

Area of Basic Shapes

Areas of Triangles and Rectangles
Circles
Area of Parallelograms, Trapezoids, Rhombuses and Kties

Shapes

Triangles

Right triangles have one angle that equals $90^\circ$ . The symbol represents $90\deg$ . Here are some examples:

Isosceles triangles have two sides of the same length and two congruent angles. Here are some examples:

Equilateral triangles have 3 sides of equal length and three congruent angles.

This is true because all angles of the triangle add up to $180\deg$ . With this given information, $180\deg / 3$ angles equal to $60\deg$

Angles:

Right angles are angles that measure exactly $90^\circ$ . When forming perpendicular lines, right angles are formed. Here are some examples:

Acute angles are angles that measure less than $90^\circ$ . Here are some examples:

Obtuse angles are angles that measure greater than and less than $180^\circ$ . Here are some examples:

Complementary angles are angles that add up to $90^\circ$ . Here are some examples

Supplementary angles are angles that add up to $180^\circ$ . Here are some examples:

Equations to Solve Angles and Sides of Triangles

Law of Sines and Cosines

The Law of Siens and Law of Cosines can be used on any triang,e regardless of whether it is a right triangle to find a missing angle or side length.

Law of Sines
$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$

Law of Cosines
$a^2=b^2+c^2-2bc \cos{A}$
$b^2=a^2+c^2-2ac \cos{B}$
$c^2=a^2+b^2-2ab \cos{C}$

Pythagorean Theorem:

The Pythagorean Theorem can be applied to right triangles. Given any 2 sides of a right triangle, we can use the Pythagorean Theorem to determine the third side.

It states: $a^2+b^2=c^2$
Where a and b are the legs of the triangle, and c is the hypotenuse (the longest side).

Example:
Solve for x.

Step 1 Identify that $a^2+b^2=c^2$ (Pythagorean Theorem) for right triangles
Step 2 Substitute so that $a^2+b^2=c^2$ becomes $3^2+4^2=x^2$
Step 3 Simplify to $9+16=x^2$
Step 4 Add to get $25=x^2$
Step 5 Take the square root of both sides to solve for x. x=5

How would this problem change if you were asked to solve for x in the triangle below?
Example:

Step 1 Identify that $a^2+b^2=c^2$ (Pythagorean Theorem) for right triangles
Step 2 Substitute so that $a^2+b^2=c^2$ becomes $6^2+x^2=10^2$

Note that the difference between this equation and the one in the previous example, because 10 is the hypotenuse.

Step 3 Simplify to $36+x^2=100$
Step 4 Subtract to get $x^2=64$
Step 5 Square root both sides to solve for x. x=8

Special Right Triangles

These common types of right triangles have certain side length ratios, which can be useful in solving geometry problems.
Example:
Find x.

We know that this is a $40^{\circ} - 45^{\circ} - 90^{\circ}$ triangle because the legs are of equal length. The hypotenuse of a $45^{\circ} -45^{\circ} - 90^{\circ}$ triangle is equal to $\sqrt{2}$ times the length of a leg. We can use this fact to calculate $x=3\sqrt{2}$ .

Example:
Find x.

Because this triangle has a right angle and an angle of $60^{\circ}$ , the other angle must be $30^{\circ}$ . (The angles must sum to $180^{\circ}$ . Since the long leg of a $30^{\circ} - 60^{\circ} - 90^{\circ}$ right triangle is equal to $\sqrt{3}$ times the short leg, we can calculate x to equal $2\sqrt{3}$ .

Area of Basic Shapes

Area of Triangles and Rectangles

Two of the most fundamental shapes in geometry are triangles and rectangles. We can often find the area of more complex shapes by breaking them down into smaller rectangles and triangles. So it is very important that we can find the area of these basic shapes.

The area of any rectangle is the length multiplied by the width. The area of any triangle is one half of the base multiplied by the height.

Examples:

Circles

Another basic shape in geometry is the circle. To find the area of a circle we need $\pi (\approx 3.14)$ , which is the ratio of the circumference of a circle to its diameter. The area of a circle is $\pi$ multiplied by the radius squared.

$A=\pi r^2$

Example:
Find the area of the circle.

Area= $\pi r^2$
$=\pi \cdot(5)^2$
$=\pi \cdot(25)$
=78.54 units $^2$

Area of Parallelograms, Trapazoids, Rhombuses, and Kites

There are lots of four-sided shapes besides rectangles and squares.
Hierarchy of Four-Sided Shapes

Quadrilaterals –all four-sided polygons

Trapezoids — one pair of parallel sides
Parallelograms — two pairs of parallel sides

Rectangles — four right angles, four sides.
Rhombus — four equal length sides, with perpendicular diagonals

Kites — two adjacent pairs of equal length sides
Square — four-sided polygon having equal-length sides meeting at right angles, with perpendicular diagonals

Note: A square is both a rectangle and a rhombus.

To find the area of a parallelogram we need to multiply the base by the height.

Parallelogram: $A=bh$

The area of a trapezoid is one half of the height multiplied by the sum of the upper and lower bases.

Trapezoid: $A=\frac{1}{2}\cdot h (b_1+b_2)$

Example: Find the area of the parallelogram.

To solve this problem we will need to find the height of the parallelogram. The Pythagorean Theorem can be used with the right triangle formed by the vertical line to find h.

$c^2=a^2+b^2$
$10^2=6^2+h^2$
$h^2=100-36$
$h=\sqrt{64}$
$h=8$

Next, use the formula $A=bh$ to calculate the area.

$A=(15 \cdot 8)$
$A=120$ units $^2$

The area of a Rhombus is equal to one half of the product of the two diagonal lengths. The area of a kite can be found in exactly the same way.

$A=\frac{1}{2} d_1 d_2$

Examples:


$A=\frac{1}{2}(6)(4)$	$A=\frac{1}{2}(4)(9)$
$A=12$ units $^2$	$A=18$ units $^2$