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| 9 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
The definition of three noticeable points will be explored below.
The incenter of a triangle is the point of intersection of the triangle's angle bisectors. The incenter is typically represented by the letter I. This point is considered to be the center of the triangle. For every triangle, the incenter is always inside the triangle.
Recall that by the Incenter Theorem, the incenter of a triangle is equidistant from each side of the triangle.
The circumcenter of a triangle is the point of intersection of the triangle's perpendicular bisectors. Circumcenter of a triangle is denoted by the letter S. It can be inside, outside, or on a triangle's side, depending on the triangle type.
The circumcenter is equidistant from the vertices of the triangle by the Circumcenter Theorem.
The centroid of a triangle is the point of intersection of the triangle's medians. The centroid is typically represented by the letter G. This point is always inside the triangle.
Begin by defining the segments that form the points.
To match the points with their definitions, the segments that form the points should be defined. Looking at the graph for point A, it can be seen that the segments do not bisect the sides of the triangle. However, they bisect the interior angles of the triangle.
With this information, it can be concluded that the segments are the angle bisectors of the triangle. Therefore, point A is the incenter of the triangle. Next, the second diagram will be considered.
The segments bisect the sides of the triangle and connect the midpoints of the sides with their opposite vertices. Therefore, they are the medians of the triangle, and point B is the centroid. Lastly, the segments that form point C will be defined.
Each segment is perpendicular to and bisects a side of the triangle. This means that they are perpendicular bisectors of the sides, and point C is the circumcenter of the triangle.
The concepts previously investigated can also be used in real life!
In the Mile High City, Denver, the cities transportation department is planning to pave three roads that connect three neighborhoods.
Magdalena and Vincenzo are the owners of two competing hotel chains. They see this as an opportunity to expand their empires into this region. Magdalena wants her hotel to be equidistant from each paved road. On the other hand, Vincenzo wants his hotel to be equidistant from the neighborhoods.
Consider the definitions of incenter and circumcenter, and centroid.
It has been previously seen that the incenter of a triangle is equidistant from its sides. Thereofre, a circle inscribed in the triangle and centered at the incenter can be drawn.
In a few of steps, it is possible to draw the inscribed circle or incircle of a triangle.
To follow another connection between circles and triangles, consider the circumcenter of a triangle. Recall that this point is equidistant from the vertices of the triangle. Therefore, a circle circumscribed at the triangle and centered at the circumcenter can be drawn.
The circumscribed circle or circumcircle of a triangle is the circle that passes through the three vertices of the triangle. The center of the circumscribed circle is the circumcenter of the triangle.
The circumscribed circle or circumcircle of a triangle can be drawn in a few of steps.
Inscribed and circumscribed circles can also be considered in real life!
LaShay built a triangular shaped farm and put gates in each corner.At night, she wants to monitor all three gates. Therefore, she will place a lamp post in her farm. Where should she place the lamp so that each of the three corners are illuminated? Define the region illuminated by the light.
Consider the definitions of the circles of a triangle.
With the topics seen in this lesson, the challenge presented at the beginning can be answered. The circle that is tangent to each side of the triangle is the inscribed circle of the triangle. The circle that passes through the three vertices of the triangle is the circumscribed circle of the triangle.
For example, consider a carpenter designing a triangular table with one leg. To determine the location of the leg, he will use the centroid of the table. Since the centroid is the center of mass, the table will be perfectly balanced.
First we should plot the points and graph the triangle.
The circumcenter is the point of intersection of the perpendicular bisectors of at least two sides of the triangle. Remember that a bisector cuts something in half. Therefore, we must find lines that are perpendicular to the sides at the side's midpoints.
A horizontal line is always perpendicular to a vertical line. Because SR is horizontal, any perpendicular line will be vertical. Similarly, we see that RT is vertical. Therefore, any perpendicular line is horizontal. Let's find the midpoint of each side using the Midpoint Formula.
Side | M(x_1+x_2/2,y_1+y_2/2) | Midpoint |
---|---|---|
SR | U(2+6/2,7+7/2) | U(4,7) |
RT | V(6+6/2,7+1/2) | V(6,4) |
Let's add these midpoints to our graph. We will also draw the perpendicular bisectors to locate their point of intersection. This is the circumcenter.
The circumcenter is located at (4,4). That means if we place the needle point of a compass at (4,4) then open it until it touches either S, R, or T, we can circumscribe the triangle.
P is the circumcenter of △ABC. Use the given information to find the length of PB.
From the exercise, we know that P is the circumcenter of △ ABC. By the Circumcenter Theorem, we know that the circumcenter of a triangle is equidistant from its vertices. Therefore, PA, PB, and PC all have the same length.
Since the three segments are congruent, we can equate the expressions we have been given for PA and PC and then solve for x by performing inverse operations until x has been isolated.
When we know the value of x, we can calculate either of the given expressions to find the length of PB. PA: 4( 10)+2 =42 units PC: 5( 10)-8 =42 units This means PB also equals 42 units.
In the figure below, PD=5x+8 and PE=7x+2. Find PF.
Examining the diagram, we see that PC bisects ∠ C and PB bisects ∠ B.
Therefore, these segments are angle bisectors and they intersect at the triangle's incenter. By the Incenter Theorem, the incenter of a triangle is equidistant from the sides of the triangle.
Since these three segments are congruent we equate expressions for PD and PE and solve for x.
Now that we know the value of x, we can substitute this into the expressions for PD or PE. If we find their lengths, we also find the length of PF. PD: 5( 3)+8 =23 units PE: 7( 3)+2 =23 units This means PF has a length of 23 units.
Which of the points A through C on the x-axis and D through F on the y-axis fall on the inscribed circle to △XYZ? It may be helpful to copy the triangle on graphing paper.
To inscribe a circle, we need to find the incenter of the triangle. This is the point of intersection of at least two of the triangle's angle bisectors.
There are four steps needed to construct an angle bisector. Let's remove the coordinate plane for now.
If we draw the angle bisector of at least two of the triangle's sides, we can find the incenter where these lines intersect.
Notice that the incenter is equidistant from every side of the triangle.
To inscribe the circle, we also have to find the circle's radius. This segment will be perpendicular to any of the sides of the triangle. To find it, we must again use our compass. There are four steps involved.
When we know where the inscribed circle will intersect one of the triangle's sides, we can open our compass to match this distance and then inscribe the circle.
As we can see, the circle intersects B and E.
Complete the statement with
The circumcenter of a right triangle is always on the hypotenuse.
If we increase the right angle, the triangle becomes obtuse. This also moves the circumcenter outside of the triangle.
Conversely, if we decrease the right angle, the triangle becomes acute. The circumcenter then moves inside of the triangle.
Therefore, the missing word is always.
Let's construct a triangle where the perpendicular bisector of one side intersects the opposite vertex of that side.
Let's recall the Perpendicular Bisector Theorem.
Perpendicular Bisector Theorem |- Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.
Since CD is the perpendicular bisector to AB, we know by the theorem that the distance from C to A and the distance from C to B are the same.
As we can see, △ ABC has two congruent sides. Therefore, it is an isosceles triangle and the word we are looking for is always.
Find the coordinates of the centroid of the triangle with the given vertices.
Let's first draw △ FGH in a coordinate plane.
The centroid of a triangle is the point of intersection of the triangle's medians. A median of a triangle is the segment drawn between the midpoint of a triangle's side and its opposite vertex. Therefore, we need to start by calculating the midpoints using the Midpoint Formula.
Side | Points | M(x_1+x_2/2,y_1+y_2/2) | Midpoint |
---|---|---|---|
FG | ( 4,5), ( 6,9) | I(4+ 6/2,5+ 9/2) | I(5,7) |
GH | ( 6,9), ( 8,1) | J(6+ 8/2,9+ 1/2) | J(7,5) |
FH | ( 4,5), ( 8,1) | K(4+ 8/2,5+ 1/2) | K(6,3) |
Let's add these midpoints to the diagram.
Finally, we will draw the medians of at least two sides. For good measure, we will draw all three.
The coordinates of the centroid are (6,5).
As in Part A, we will begin by drawing the triangle in a coordinate plane.
In order to draw the triangle's medians we must find the midpoint of each side. For this purpose we will use the Midpoint Formula.
Side | Points | M(x_1+x_2/2,y_1+y_2/2) | Midpoint |
---|---|---|---|
AC | ( 3,3), ( 6,7) | D(3+ 6/2,3+ 7/2) | D( 92,5) |
BC | ( 12,2), ( 6,7) | E(12+ 6/2,2+ 7/2) | E(9, 92) |
AB | ( 3,3), ( 12,2) | F(3+ 12/2,3+ 2/2) | F( 152, 52) |
Let's add these midpoints to our graph.
Let's draw the medians of at least two sides
The coordinates of the centroid are (7,4).
Which of the point(s), A through F, are on the circumference of the circumscribed circle to △XYZ?
To find the circumcenter, we must draw the perpendicular bisector of at least two of the triangle's sides and then locate their point of intersection.
To construct a perpendicular bisector there are two steps. For now, we will remove the coordinate plane leaving only the triangle.
If we draw the perpendicular bisector of at least two of the triangle's sides, we can locate the circumcenter where these lines intersect.
When we know the center of the circle, we can open our compass to match the distance between the cicrumcenter and one of the triangle's vertices and then draw the circle.
As we can see, B, D, and E are all on the circumference.