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1 Introduction to Basic Geometry
1.1 Euclidean Geometry and Axiomatic Systems
1.1.1 Points, Lines, and Line Segments
Geometry is one of the oldest branches of mathematics. The word geometry in the Greek language translates the words for ”Earth” and ”Measure”. The Egyptians were one of the first civilizations to use geometry. The Egyptians used right triangles to measure and survey land. In our modern times, geometry is used to in fields such as engineering, architecture, medicine, drafting, astronomy, and geology. To begin this chapter on Geometry, we will describe two basic concepts which are a point and a line. A point is used to denote a specific location in space. In this section, everything that we do will be viewed in two dimensions. For example, we could draw a point in two dimensional space and label it as point A.
A line is determined by two distinct points and extends to infinity in both directions. Now suppose that we define two points in space and label them as A and B. We could pass a line through these points in space and the resulting line would look the next illustration. We
will label this line as
A line segment is part of a line that lies between two points. These two points are referred to as endpoints. In the next figure below there is an illustration of a line segment. We will label this line segment as 𝐴𝐵
1.1.2 Distance
Now that we have given a basic description a line and line segment, let’s use some properties of distance to find the missing length of a segment. In the next example we will find the distance between two points. To find missing distances of a line segment, we use a postulate called the segment addition postulate.
Segment Addition Postulate If point B lies between points A and C on 𝐴𝐵 , then AB + BC = AC
In the next example will find the distance between two points.
Example 1
Given AB = 2x + 3, BC = 3x +7, and AC = 25, find the value of x, AB, and BC
Solution Since the point B lies between point A and C on 𝐴𝐶, it must be true that AB + BC = AC. Substituting the values for AB, BC and AC into the above equation, we get the following equation that can be solved for x. 2𝑥 + 3 + 3𝑥 + 7 = 25 5 𝑥 + 10 = 25 5 𝑥 + 10 − 10 = 25 − 10 5 𝑥 = 15 5 𝑥 5 =^
15 5 𝑥 = 3 Now, use the value of x to find the values of and AB and BC. Therefore, 𝐴𝐵 = 2(3) + 3 = 6 + 3 = 9 and 𝐵𝐶 = 3(3) + 7 = 9 + 7 = 16
Example 2
Given AB = 10, BC = 2x + 4, CD = 12, and AD = 36, find the length of BC.
Since points B and C lie between points A and D on 𝐴𝐷, AB + BC + CD = AD
𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 = 𝐴𝐷 10 + 2𝑥 + 4 + 12 = 36 2 𝑥 + 26 = 36 2 𝑥 + 26 − 26 = 36 − 26 2 𝑥 = 10 𝑥 = 5
Now, use the value of x to find the length of BC: 𝐵𝐶 = 2(5) + 4 = 10 + 4 = 14
1.1.3 Rays and Angles
A ray starts at a point called an endpoint and extends to infinity in the other direction. A ray that has an endpoint at A and extends indefinitely through another point B is denoted
by
𝐴𝐵 Here is an example of a ray with an endpoint A that lies in a plane.
An angle is the union of two rays with a common endpoint called a vertex.
1.1.5 Special Types of Angles
A right angle is an angle whose measure is 90 degrees.
A straight angle is an angle whose measure is 180 degrees.
Special Angle Pairs There are two types of angle pairs which are complementary angles and supplementary angles. A pair of complementary angles are two angles whose sum is 90 degrees. Meanwhile, A pair of supplementary angles are two angles whose sum is 180 degrees. Adjacent Angles are two angles who share a common endpoint and common side, but share no interior points.
The Angle Addition Postulate If point D lies in the interior of ∠𝐴𝐵𝐶, then 𝑚∠𝐴𝐵𝐶 = 𝑚∠𝐴𝐵𝐷 + 𝑚∠𝐷𝐵𝐶
1.1.6 Finding Missing Angle Values
To find the value of the missing angle, we will use the angle addition postulate along with the definition of complementary angles and supplementary angles.
Example 3 Find the complement of angle measuring 36^0
Solution If the angles are complementary, their sum is 90 degrees. Now, let are missing angle be ∠𝐴 Therefore, ∠𝐴 + 36^0 = 90^0 → ∠𝐴 = 90^0 − 360 = 54^0
Example 4 Find the supplement of angle measuring 86^0
Solution If the angles are supplementary, their sum is 180 degrees. Now, let are missing angle be ∠𝐴 Therefore, ∠𝐴 + 86^0 = 180^0 → ∠𝐴 = 180^0 − 860 = 94^0
Example 5 Given ∠𝐴𝐶𝐵 = 54^0 and ∠𝐴𝐶𝐷 = 112^0 , find the measure of ∠𝐴𝐶𝐷
Solution ∠𝐵𝐶𝐷 can be found by subtracting ∠𝐴𝐶𝐵 from ∠𝐴𝐶𝐷 ∠𝐵𝐶𝐷 = 112^0 − 540 = 58^0
Example 6 Given that ∠𝐴𝐵𝐶 is a right angle, find the value of x, ∠𝐴𝐵𝐷, ∠𝐷𝐵𝐶
Solution:
- Any finite straight line can be extended in a straight line.
- A circle can be determined from any center and any radius.
- All right angles are equal.
- If two straight lines in a plane are crossed by a transversal, and sum the interior angle of the same side of the transversal is less than two right angles, then the two lines extended will intersect.
According to Euclid, the rest of geometry could be deduced from these five postulates. Euclid’s fifth postulate, often referred to as the Parallel Postulate, is the basis for what are called Euclidean Geometries or geometries where parallel lines exist. There is an alternate version to Euclid fifth postulate which is usually stated as ”Given a line and a point not on the line, there is one and only one line that passed through the given point that is parallel to the given line.” This is a short version of the Parallel Postulate called Fairplay’s Axiom which is named after the British math teacher who proposed to replace the axiom in all of the schools textbooks. Some individuals have tried to prove the parallel postulate, but after more than two thousand years it still remains unproven. For many centuries, these postulates have assumed to be true. However, some mathematics believed that the Euclid Fifth Postulate was suspect or incomplete. As a result, mathematicians have written alternate postulates to the Parallel Postulate. These postulates have led the way to new geometries called Non-Euclidean Geometries.
1.1.8 Exercises
- Which angle represents the compliment of 54^0?
(A) 126^0 (B) 46^0 (C) 36^0 (D) 26^0
- What angle represents the supplement of 65^0?
(A) 125^0 (B) 135^0 (C) 25^0 (D) 35^0
- Find the supplement of 45^0?
- Find the supplement of 45^0?
- Use the diagram below to find the value of x.
- Given that ∠𝑀 𝑂𝑁 is a right angle, find the measure of ∠𝑀 𝑂𝑃 and ∠𝑃 𝑂𝑁
- Use the diagram below to find the value of x.
There isn’t a special formula to find the perimeter of a formula. So, simply find the sum of the sides of the triangle.
1.2.2 Understanding Area
The next property of two dimensional objects we will investigate is area. The area of an object is the amount of surface that the object occupies. The area of object depends on its shape. Different shapes use different formulas to compute the area. We will start by finding the area of a rectangle. The area of a rectangle can be found by multiplying the length of the rectangle by the width of the rectangle. Let’s examine rectangles further to see why the formula of a rectangle is length times width. Suppose we had a rectangle that was 5 blocks by 4 blocks. This would mean that we would have 4 rows of blocks that each have 5 blocks in them. We could count the number of blocks or find the area by multiply the 4 rows of blocks by the 5 block that are in each rows. Therefore, the formula to find the area of a triangle would be 𝐴 = 𝑙𝑤
Using the formula 𝐴 = 𝑙𝑤, we get that the area of the rectangle is: 𝐴𝑟𝑒𝑎 = (𝑙𝑒𝑛𝑔𝑡ℎ)(𝑤𝑖𝑑𝑡ℎ) = (5𝑢𝑛𝑖𝑡𝑠)(4𝑢𝑛𝑖𝑡𝑠) = 20 square units
1.2.3 Computing Areas
Similarity, formula for the area of other objects can be used to find the area.
Key Formulas
Here are some of the key area formulas that will used in this section.
Object Shape Formula
Square 𝐴 = 𝑠^2
Rectangle 𝐴 = 𝑙𝑤
Triangle 𝐴 = 12 𝑏ℎ
Example 3 Find the area of the square.
Solution To get the answer, substitute the value of the length of the side of square into the area of a square formula. 𝐴 = (5𝑓 𝑡)^2 = 25𝑓 𝑡^2
Example 4 Find area of triangle with a base of 4 meters and a height of 6 meters.
Solution To get the answer, substitute the values of the length and width of the triangle into the area of a triangle formula. 𝐴 = 12 (5𝑚)(4𝑚) = 12 (20𝑚^2 ) = 10𝑚^2
Since the two rectangles are the same, 1 square yard = 9 square feet.
Example 6
Suppose you wanted to put down hardwood floors in your living room which is roughly 12 feet by 10 feet. If the cost of the hardwood flooring is $5.00 per square foot, find the approximate cost not including labor to put down hardwood flooring in your living room?
Solution
First, the area of the room, then multiply the area of the rectangle by $5.00 per square foot.
𝐴 = (12𝑓 𝑡)(10𝑓 𝑡) = 120𝑓 𝑡^2
𝐶𝑜𝑠𝑡 = ($5 𝑓 𝑡.^002 )(120𝑓 𝑡^2 )
Example 7
Given the floor plans for the 1st floor of a house below, find the area or square footage of the 1st floor of this house.
Solution
Simply divide the rectangle up into two rectangles and find the area of each rectangle
Area of Rectangle 1: 𝐴 = (25𝑓 𝑡)(25𝑓 𝑡) = 625𝑓 𝑡^2 Area of Rectangle 2: 𝐴 = (15𝑓 𝑡)(20𝑓 𝑡) = 300𝑓 𝑡^2
Total Area=Area of Rectangle 1 + Area of Rectangle 2 = 625𝑓 𝑡^2 + 300^2 = 925𝑓 𝑡^2
1.2.5 Circles
A circle can be defined the set of all points that are equidistance from a point called the center of the circle. Circles are defined by the value of their radius and their center. The radius of a circle is the defined as the straight line distance from the center of the circle to the edge of the circle. The diameter of a circle is a line segment that passes through the center of the circle and has endpoints that lie on the circle
The circumference of a circle is the distance around the circle. The circumference of a circle can thought of as the perimeter of the circle. The circumference of a circle can be found by the following formula: 𝐶 = 2𝜋𝑟 or 𝐶 = 𝜋𝑑 The area of the circle is given by the formula: 𝐴 = 𝜋𝑟^2 where pi is approximately 3.1416. Pi is defined as the ratio between the circumference of the circle and the diameter of the circle.
Example 8
Find the area and circumference of the following circle.
Example 9 Use the Egyptian Approximation for 𝜋, 25681 , to find the area and circumference in problem
Solution Use the area of a circle formula to find the area of the circle. 𝐴 = 𝜋𝑟^2 = 𝜋(4𝑚)^2 = 16𝜋𝑚^2 = 16(^25681 ) = 409681 = 50. 57 𝑚^2 Next, use the circumference formula to find the circumference of the circle. 𝐶 = 𝜋𝑑 = 𝜋(8𝑚) = 8𝜋 = 8(^25681 ) = 284881 = 35.16 m
1.2.7 Exercises
- Find the perimeter and area of the following rectangle.
- Find the circumference and area of the following circle.
- Find the area of the following triangle.
- Find the area of a rectangular shaped lot with a length of 40 yards and a width of 60 yards.
- Find the area and perimeter of a square shaped lot with a length of 100 meters.
- You want to put down hardwood floors in your living room that is roughly 18 feet by 16 feet. If the cost of the hardwood flooring is $5.00 per square foot, find the approximate cost not including labor to put down hardwood flooring in your living room?
- You want to put down carpet in your living room that is roughly 6 yards by 2 and half yards. If the cost of the carpet is $4.00 per square yard, find the approximate cost not including labor to put down hardwood flooring in your living room?
1.3 Volume
1.3.1 Understanding Volume
The main topic of this section is volume. You will specifically look at how to find the volume of various three dimension geometric objects such as rectangular solids and cylinders. The volume of an object is the amount space occupied by the object. We will use formulas to find the volume of an object in the same way we used formulas to find the area of an object. Let’s begin by classifying three dimensional objects into two categories. The two categories are prism and pyramids. A prisms is a three dimension object that has two bases that are identical and a pyramid is a three dimensional object with just one base.
1.3.2 Prisms
In the next diagram there are pictures of a rectangular solid, a prism with a triangular base, and a cylinder. These are just a few of the many possibilities that we have for objects that we classify as prisms. Notice that in each example, the objects have a base at the top and the bottom are identical.
To find the area of any prism, we simply multiply the base of the prism by the height of the prism. 𝑉 𝑜𝑙𝑢𝑚𝑒 = (𝐵𝑎𝑠𝑒)(𝐻𝑒𝑖𝑔ℎ𝑡) In next few examples, we will investigate how to find the volume of a prism such as a rectangular solid or cylinder.
Example 1: Rectangular Solids Find the following of the following rectangular solid.
Solution
Since the object is a prism, we will use the formula for the volume of a prism which is 𝑉 = 𝐵ℎ. The base of the rectangular solid is a rectangle. Therefore, the formula to find the volume would be 𝑉 = 𝐵ℎ = (𝑙𝑤)ℎ = 𝑙𝑤ℎ 𝑉 = (4𝑓 𝑡)(3𝑓 𝑡)(2𝑓 𝑡) = (12𝑓 𝑡^2 )(2𝑓 𝑡) = 24𝑓 𝑡^3
Example 2: Cylinders Find the volume of a cylinder with a radius of 3 inches and height of 4 inches.
A cylinder is a prism with a circular base. Using the formula for the prism along with the formula of the area a circle, we get the following formula to find the area of a cylinder. 𝑉 = 𝐵ℎ = (𝜋𝑟^2 )ℎ = 𝜋𝑟^2 ℎ
𝑉 = 𝜋(3𝑖𝑛)^2 (4𝑖𝑛) = 𝜋(9𝑖𝑛^2 )(4𝑖𝑛) = 36𝜋𝑖𝑛^3
Using the value of 3.14 for Pi we would get 𝑉 = 36𝜋𝑖𝑛^3 = 113. 04 𝑖𝑛^3
1.3.3 Pyramids
Pyramids like prism can have number of different shape provided the object has only one base. Here is a diagram that illustrates a few of the types of pyramids that exist.
Volume of Pyramids To find the volume of a pyramid, we multiply one third times the base multiplied by the height. The formula to compute the volume of a pyramid is 𝑉 = 13 𝐵ℎ In next few examples, we will investigate how to find the volume of some pyramid.
Example 3 Find the area of the following pyramid with a square base
