Circle Geometry: Anlge Measures Formed By Intersecting Lines

Intersecting Lines in a Circle Theorem

Figure 1

As I think about circles in a plane and intersecting lines, I can think of more triangle proofs and work with the angle and segment addition postulates. This will lead to both algebraical and geometric types of problems. By algebraic, I mean you will have to use a theorem or property to set up the equation. On the other hand, a geometric type of problem will consist of a proof of a stated given or already known theorem. Before getting into solving any problems, I should discuss the theorem of topic for the geometry article.

Theorem – Two Lines Intersecting Inside of a Circle

The measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs. Figure 1 illustrates the following equation to represent this theorem:

m \angle{1} = \frac{1}{2}(x + y).

Example 1 – Find the measure of the missing angle measure.

The first example ends up being a an easy three step equation to solve.

intersecting lines in a circle ex 1

Example 1

To set up the equation I HAVE to use geometry, in this case, Two Lines Intersecting Inside of a Circle Theorem (TLIICT), then I solve the equation using an algebra 1 skill.

m  \angle{1} = \frac{1}{2}(x + y)TLIICT

(1)55 = \frac{1}{2}(67 + y). Substitution Property of Equality

(2)110 = 67 + y Multiplication Property of Equality

(3)43 = y  Subtraction Property of Equality

(4)y = 43 Symmetric Property of Equality

Example 2 – Find the measure of the missing angle measure.

intersecting lines in a circle ex 2

Example 2

The solution to this problem may not present itself right away if one too focused on the current theorem. There is no variable for what you have to solve for first. To find the value of the variable, I must keep in mind that the measure of the angle I am solving for is supplementary to the measure of the angle I can find. Since it is not labeled in the original picture, in figure 2, I label what I can find with what is given, x.

Figure 2

Let x = to the value of the vertical angles formed not labeled y

m  \angle{1} = \frac{1}{2}(x + y)

x = \frac{1}{2}(60 + 70)

x = \frac{1}{2}(130)

x = 65

y= 180 - x

y= 180 - 65

y= 115

In this post, I have discussed the Lines Intersecting Inside of a Circle Theorem and worked out two problems. I worked out the problems differently in each example. Which method do you prefer to read and why? I look forward to your answers.

Art Meets Geometry – Two Point Perspective Drawing

Mr. Pi's Two Point Perspective "My Perspective of a Box"

"My Perspective of a Box"

Here is a drawing I did to show to my High School Geometry classes. At the time of this post, the class is starting a unit on 3-Dimensional shapes. Prior to this chapter, most of our work has been in a single plane. Learning how to draw simple 3-D shapes in both one point and two point perspective makes the transition to working with intersecting planes smoother for some students. Other students would do just as well without actually making the drawings. The image to the right is my little sketch after I drew my box. Be sure to click the image to see it in full size. Below is a picture I drew back in 2008 sometime.

How to Draw a Box in Two-Point Perspective:

  1. On a sheet of paper, draw horizontal line the entire length of a sheet of paper, place one point to the left of center and one point the right of center. Preferably closer to the side of the paper.
  2. Draw a vertical line of about 1 inch long above or below the box. The bigger the vertical line, the bigger the box.
  3. Draw light line segments connecting each end of the vertical line to both points. You should draw 4 line segments in this step.
  4. Draw two vertical segments, one on each side of the first vertical segment.
  5. Connect the new vertical segments two the vanishing points. You should draw four line segments for this step.
  6. Clean up the unnecessary lines and used dashed segments for the hidden edges.

Mr. Pi's Double Two Point Perspective

Mr. Pi's Double Two Point Perspective

How to Find the Area of Regular Polygons

Finding the area of regular polygons is the topic of this geometry math video.

First, the video discusses the parts of a regular polygon: apothem and radius. The center of the polygon is the same as the center of the circumscribed circle. The radius of the polygon is the distance from the center of the polygon to its vertex. The number of radii is determined by the number of sides. The apothem it the perpendicular distance from the center of the polygon to a side.

The first example models how to find the different angles of a regular polygon formed by the radius and the apothem. First, you divide the number in interior angles into 360 degrees. Once you have that you can figure out the other two angles quite easily.

Before applying the the area formula of a regular polygon, the video reviews the formula:

A = \frac{1}{2}ap

It is good to note that a = length of the apothem and p = perimeter of the polygon. The perimeter may not be calculated. In that case you multiply the number of sides by the length of each side.

This example is similar to the video, but it is different:

What is the area of a polygon with sixteen 36 in. sides and an apothem of 18\sqrt{3} in.

A = \frac{1}{2}ap

A = \frac{1}{2}(18 \sqrt{3}(16*36))

A = 5184 \sqrt{3} in^2 \approx 8979.0 in^2

Questions for the comment section:

Comparing example 2T on the video and the problem above, how are the problems alike? How are the problems the same?

Example 3T is finding the area of a regular polygon, but you need to find the length of the apothem using the 30-60-90 triangle relationship.

Area of a Triangle – Hands on Tactile Activity

I have been making videos to use in class, so the blog is going to be videos of me teaching. If you have related questions, post your questions in the comment section. I will do my best to answer you in a timely manner. This geometry video math lesson develops the area of a triangle from the area of a parallelogram. Grid or graph paper,  a pencil and a straight edge will be necessary if you want to complete the task yourself. If you want to really make the connection between the heights of the parallelogram and the triangle, you should create your own triangle with a different base and height. To use this as a hands on tactile geometry lesson, download grid paper here.

The connection is made through the heights of the the objects. The height of the parallelogram is one half the height of the triangle. To use this as a hands on tactile geometry lesson, download grid paper here.

Area of a Parallelogram Hands On Activity

This lesson focuses on the development of the area formula for a parallelogram. The development and activity is tactile or hands on in manner. To fully participate, download grid paper here.

Geometry Test Review – Similiary

This post is for my geometry students. Below is the link to the Chapter 7 practice test with answer key. Remember, chapter 7 was all about similarity.

If you have a question, post a comment and I will get back with an answer to your geometry question as soon as possilbe.

Geometry Chapter 7 Test Review and Answer Key

The Diagonals of a Rhombus are Perpendicular

This post is dedicated to proving the diagonals of a rhombus are perpendicular. To complete this or any proof, it is good to make a plan. In this case, I am going to establish that the two triangles on the top of the rhombus are congruent. This is seen in steps 2 through 4 of the proof below. Next, I show that angles AEB and CEB right angles, which is modeled in steps 6 through 8. That is enough to state the diagonals of a rhombus are perpendicular.

I encourage you to click on the image to see it at full size. It looks much better at full size. I scanned the piece of notebook paper that I work this proof out on and messed around in photoshop.

Proof - Diagonals of a Rhombus are Perpendicular

Proof - Diagonals of a Rhombus are Perpendicular

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