GEOMETRY FINAL REVIEW Theorem 1: If two chords intersect in a circle, the product of the lengths of the segments of one chord equal the product of the segments of the other. Intersecting Chords Rule: (segment piece)×(segment piece) = (segment piece)×(segment piece) Theorem 2: If two secant segments are drawn to a circle from the same external point, the product of the length of one secant segment and its external part is equal to the product of the length of the other secant segment and its external part. Secant-Secant Rule: (whole secant)×(external part) = (whole secant)×(external part) Theorem 3: If a secant segment and tangent segment are drawn to a circle from the same external point, the product of the length of the secant segment and its external part equals the square of the length of the tangent segment.
Transcript
GEOMETRY FINAL REVIEWTheorem 1: If two chords intersect in a circle, the product of the lengths of the segments of one
chord equal the product of the segments of the other.
Theorem 3: If a secant segment and tangent segment are drawn to a circle from the same external point, the product of the length of the secant segment and its external part
equals the square of the length of the tangent segment.
This theorem can also be stated as "the tangent being the mean proportional between the whole secant and its external part" (which yields the same final rule:
Equations of CirclesCircle with Center at
Origin (0,0)
where the center is (0,0) and the radius is r.
center (0,0)radius = 5
Circle with Center at Point (h,k)
where the center is (h,k) and the radius is r
center (4,4)radius = 5
Did you notice that the formula for the circle with the center at the originis just a special case of the formula for a circle at point (h,k)?When h = 0 and k = 0 (for the origin), the formula becomes
Where does the circle formulacome from?
Using the distance formula, the radius of the circle at the right is represented by
Equation of a Circle
Example 1: Write the equation of a circle with center at (3,-5) and a radius of 3.
Answer: Notice how the coordinate of -5 appears
as (y + 5)2
Example 2: Given the equation of a circle, , what are the coordinates of the center and the radius?
Answer:The center is (-3,6)
and the radius is
Notice how the answer for the radius is left in radical form. You should not estimate an answer into decimal form unless told to do so, or unless
working in a real world measurement situation. It is hard to measure a radical number of feet.
Slopes Equations of lines
The slope of a line is a rate of change and is represented by m.
Equations of line can take on several forms:
When a line passes through the points (x1, y1) and (x2, y2), the slope is
m = .
Lines that have a positive slope, rise from lower left to upper right. They go up hill.
Lines that have a negative slope, decline from upper left to lower right. They go down hill.
Lines that are horizontal have a slope of zero. (There is no "rise", creating a zero numerator.)
Lines that are vertical have no slope (undefined slope). (There is no "run", creating a zero denominator.)
Lines that are parallel have equal slopes.
Lines that are perpendicular have negative reciprocal slopes.(such as m = 2 with m = -1/2)
Slope Intercept Form:[used when you know, or can find, the slope, m, and the y-intercept, b.]
y = mx + b Point Slope Form:[used when you know, or can find, a point on the line (x1, y1), and the slope, m.]
Horizontal Line Form: y = 3 (or any number)Lines that are horizontal have a slope of zero. They have "run", but no "rise". The rise/run formula for slope always yields zero since rise = 0. y = mx + b y = 0x + 3 y = 3
Vertical Line Form: x = -2 (or any number)Lines that are vertical have no slope (it does not exist). They have "rise", but no "run". The rise/run formula for slope always has a zero denominator and is undefined.
The converse of a conditional statement is formed by interchanging the hypothesis and conclusion of the original statement, with the words "if" and "then" fixed.
Example:
Conditional: "If the space shuttle was launched, then a cloud of smoke was seen."
Converse: "If a cloud of smoke was seen, then the space shuttle was launched."
It is important to note that the converse of a true/false conditional statement is NOT necessarily true/false as the original statement.
The inverse of a conditional statement is formed by negating the hypothesis and negating the conclusion of the original statement.
Example:
Conditional: "If you grew up in Alaska, then you have seen snow."
Inverse: "If you did not grow up in Alaska, then you have not seen snow."
You may also use other words (in addition to NOT) to create the negation.
It is important to note that the inverse of a true/false conditional statement is NOT necessarily true/false as the original statement.
The contrapositive of a conditional statement is formed by negating both the hypothesis and the conclusion, and then interchanging the resulting negations. In other words, it does BOTH the jobs of the INVERSE and the CONVERSE.
Example:
Conditional: "If 9 is an odd number, then 9 is divisible by 2."
Contrapositive: "If 9 is not divisible by 2, then 9 is not an odd number."
An important fact to note is that:
If the original statement is TRUE, the contrapositive is TRUE, and vice versa;
If the original statement is FALSE, the contrapositive is FALSE, and vice versa.
In this case, they are said to be logically equivalent.
Medians: A median of a triangle is a segment joining any vertex to the midpoint of the opposite side. The medians of a triangle are concurrent. Notice that the point of concurrence is in the interior of the triangles.
Medians of an acute triangle. Medians of an obtuse triangle.
Archimedes showed that the point where the medians are concurrent is the center of gravity of a triangular shape of uniform thickness and density. This point where the medians are
concurrent is called the centroid of a triangle. If you cut a triangle out of cardboard and balance it on a pointed object, such as a pencil, the pencil will mark the location of the
triangle's centroid. The centroid divides the medians into a 2:1 ratio. The section of the median nearest the vertex
is twice as long as the section near the midpoint of the triangle's side.
When two or more transformations are combined to form a new transformation, the result is called a composition of transformations. In a composition, the first transformation produces an image upon
which the second transformation is then performed.
The symbol for a composition of transformations is an open circle.
The notation is read as a reflection in the x-axis following a translation of (x+3, y+4). Be careful!!! The process is done in reverse!!
You may see various notations which represent a composition of transformations:
could also be indicated by
Renaming a composition: It is possible that a composition of two transformations may be renamed by only one other transformational method. For example, the composition of a line reflection in the y-axis followed by a line reflection in the x-axis could be described as a single transformation of a reflection in the origin.
Methods for Proving (Showing) Triangles to be Congruent
SSS If three sides of one triangle are congruent to three sides of another triangle, the triangles are congruent.(For this method, the sum of the lengths of any two sides must be greater than the length of the third side, to guarantee a triangle exists.)
SAS If two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. (The included angle is the angle formed by the sides being used.)
ASA If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. (The included side is the side between the angles being used. It is the side where the rays of the angles would overlap.)
AAS If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. (The non-included side can be either of the two sides that are not between the two angles being used.)
HLRight
TrianglesOnly
If the hypotenuse and leg of one right triangle are congruent to the corresponding parts of another right triangle, the right triangles are congruent. (Either leg of the right triangle may be used as long as the corresponding legs are used.)
BE CAREFUL!!!Only the combinations listed above will give congruent triangles.
So, why do other combinations not work?
Methods that DO NOT Prove Triangles to be Congruent
AAA AAA works fine to show that triangles are the same SHAPE (similar), but does NOT work to also show they are the same size, thus congruent!
Consider the example at the right.
You can easily draw 2 equilateral triangles that are the same shape but are not congruent (the same size).
SSAor
ASS
SSA (or ASS) is humorously referred to as the "Donkey Theorem".
This is NOT a universal method to prove triangles congruent because it cannot guarantee that one unique
triangle will be drawn!! The SSA (or ASS) combination affords the possibility of creating zero,
one, or two triangles. Consider this diagram of triangle DEF. If for the second side, EF is equal to EG (the minimum distance needed to create a triangle), only one triangle can be drawn. However, if EF is greater than EG, two triangles can be drawn as shown by the dotted segment. Should EF be less than the minimum length needed to create a triangle, EG, no triangle can be drawn.
The possible "swing" of side can create two different triangles which causes our problem with this method. The first triangle, below, and the last triangle both show SSA, but they are
not congruent triangles.
The combination of SSA (or ASS) creates a unique triangle ONLY when working in a right triangle with the hypotenuse and a leg. This application is given the name HL (Hypotenuse-Leg) for Right Triangles to avoid confusion. You should not list SSA (or ASS) as a reason when writing a proof.
Once you prove your triangles are congruent, the "left-over" pieces that were not used in your method of proof, are also congruent.
Remember, congruent triangles have 6 sets of congruent pieces. We now have a "follow-up" theorem to be used AFTER the triangles are
known to be congruent:
Theorem: (CPCTC) Corresponding parts of congruent triangles are congruent.
There are five basic locus theorems (rules). Each theorem will be explained in detail in the following sections under this topic. Even though the
theorems sound confusing, the concepts are easy to understand.
Locus Theorem 1:
The locus of points at a fixed distance, d, from point P is a circle with the given point P as its
center and d as its radius.
Locus Theorem 2:
The locus of points at a fixed distance, d, from a line, l, is a pair of parallel lines d distance from l and on either
side of l.
Locus Theorem 3:
The locus of points equidistant from two points, P and Q, is the
perpendicular bisector of the line segment determined by the two
points.
Locus Theorem 4:
The locus of points equidistant from two parallel lines, l1 and l2 , is a line parallel to both l1 and l2 and midway between them.
Locus Theorem 5:
The locus of points equidistant from two intersecting lines, l1 and l2, is a pair of bisectors that bisect the angles
formed by l1 and l2 .
When attempting to solve a locus problem, there are certain steps that should be followed:
Steps to solving a locus problem:1. Draw a diagram showing the given lines and points.
2. Read carefully to determine the needed condition(s).
3. Locate one point that satisfies the needed condition and plot it on your diagram. Locate several additional points that satisfy the condition and plot them as well. Plot enough points so that a pattern (a shape, a path) is starting to appear.
4. Through these plotted points draw a dotted line to indicate the locus (or path) of the points.
5. Describe in words the geometric path that appears to be the locus.
6. If TWO conditions exist in your problem (a compound locus), repeat steps 2-4 above for the second condition ON THE SAME DIAGRAM. Count the number of points where the two loci intersect. (Where do the dotted lines cross?)
compound locus problem involves two, or possibly more, locus conditions occurring at the same time. The different conditions in a compound locus problem are generally separated by the word
"AND" or the words "AND ALSO".
Strategy for Solving Compound Locus ProblemsIf TWO locus conditions exist in a problem (a compound locus), prepare each condition separately ON THE SAME DIAGRAM. After the two conditions are drawn separately,
count the number of points where the two loci conditions intersect. (It may help to draw the locus locations as dotted lines. Then find where the dotted lines cross.)
Steps:1. Draw a diagram showing the given information in the problem.
2. Read carefully to determine one of the needed conditions. (Look for the possibility of the words "AND" or "AND ALSO" separating the conditions.)
3. Plot the first locus condition. If you do not see one of the locus theorems at work in the problem, locate one point that satisfies the needed condition and plot it on your diagram. Then locate several additional points that satisfy the condition and plot them as well. Plot enough points so that a pattern (a shape) is starting to appear, or until you remember the needed locus theorem for the problem.
4. Through these plotted points draw a dotted line to indicate the locus (or path) of the points.
5. Repeat steps 2-4 for the second locus condition.
6. Where the dotted lines intersect will be the points which satisfy both conditions. These points of intersection will be the answer to the compound locus problem.
Consider: A treasure is buried in your backyard. The picture below shows your backyard which contains a stump, a teepee, and a tree. The teepee is 8 feet from the stump and 18 feet from the tree. The treasure is equidistant from the teepee and the tree AND ALSO 6 feet from the stump. Locate all possible points of the buried treasure.
Description:
The red line represents the locus which is equidistant from the teepee and the tree (the perpendicular bisector of the segment). The blue circle represents the locus which is 6 feet from the stump. These two
loci intersect in two locations. The treasure could be buried at either "X" location. Start digging!!
Note: The red line and blue circle could have been drawn "dotted" to indicate that they are locus locations.
In mathematics, polygons are similar if their corresponding angles are congruent and the ratio of their corresponding sides are in proportion.
Once we know that triangles (or any polygons) are similar, we also know some additional facts about the figures.
1. Ratio of Perimeters, Altitudes, Medians, Diagonals, and Angle Bisectors
If two polygons are similar, their corresponding sides, altitudes, medians, diagonals, angle bisectors and perimeters are all in the same ratio.
Example:If the sides of two similar triangles are in the ratio 4:9, what is
the ratio of their perimeters?
Answer: 4:9
2. Ratio of Areas
If two polygons are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Example:If the sides of two similar triangles are in the ratio of 3:5, find the ratio of their areas.
Answer: 9:25
Method 3:
When working with diagonal segments, use the Distance Formula to determine the length.
The advantage of the Distance Formula is that you do not need to draw a picture to find the answer. All you need to know are the coordinates of the endpoints of the segment.
It doesn't matter which point you start with. Just start with the same point for reading both the x and y coordinates.
The Distance Formula is really just a coordinate geometry way of writing the Pythagorean Theorem. If you cannot remember the Distance Formula, you can always draw a graph and use the Pythagorean Theorem as was done in Method 2.
The Distance Formula can be used to find the lengths of all forms of line segments:
horizontal, vertical and diagonal.
Midpoint
If the line segments are diagonally positioned, more thought must be paid to the solution. When you are finding the coordinates of the midpoint of a segment, you are actually finding the average (mean) of the x-coordinates and the average (mean) of the y-coordinates.
This concept of finding the average of the coordinates can be written as a formula:
The Midpoint Formula:
The midpoint of a segment with endpoints (x1 , y1) and (x2 , y2) has
coordinates
NOTE:The Midpoint Formula works for
all line segments: vertical, horizontal or diagonal.
Find the midpoint of line segment .
A(-3,4)B(2,1)The midpoint will have coordinates
AnswerNOTE: Don't be surprised if your answer contains a fraction. Answers may be left in fractional form
or written as decimals.
"Mean Proportional" may also be referred to as a "Geometric Mean".
Remember the rule for working with proportions: the product of the means equals
the product of the extremes.
In a mean proportional problem, the "means" are the same values.
Definition:
The mean proportional of two positive numbers a and b is the positive number x
such that . When solving, .
Notice that the x value appears TWICE in the "means" positions.
Theorem: The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle.
Since these triangles are similar, we can establish proportions relating the
corresponding sides. Two valuable theorems can be formed using these
proportions.
Theorem: The altitude to the hypotenuse of a right triangle is the mean proportional between the segments into which it divides the hypotenuse.
Altitude Rule:
Theorem: Each leg of a right triangle is the mean proportional between the hypotenuse and the projection of the leg on the hypotenuse.
orLeg Rule:
Formulas for Working with Angles in Circles(Intercepted arcs are arcs "cut off" or "lying between" the sides of the specified angles.)
There are basically five circle formulas thatyou need to remember:
1. Central Angle: A central angle is an angle formed by two intersecting radii
such that its vertex is at the center of the circle.
Central Angle = Intercepted Arc
<AOB is a central angle. Its intercepted arc is the minor arc from A to B.
m<AOB = 80°
Theorem involving central angles:In a circle, or congruent circles, congruent central angles have congruent arcs.
2. Inscribed Angle:An inscribed angle is an angle with its vertex "on" the circle,
formed by two intersecting chords.
Inscribed Angle = Intercepted Arc
<ABC is an inscribed angle. Its intercepted arc is the minor arc from A to C.
m<ABC = 50°
Special situations involving inscribed angles:
An angle inscribed in asemi-circle is a right angle.
In a circle, inscribed circles that intercept the same arc are congruent.
A quadrilateral inscribed in a circle is called a cyclic quadrilateral.
The opposite angles in a cyclic quadrilateral are supplementary.
3. Tangent Chord Angle:An angle formed by an intersecting tangent and chord has its
vertex "on" the circle.
Tangent Chord Angle =
Intercepted Arc
<ABC is an angle formed by a tangent and chord.Its intercepted arc is the minor arc from A to B.
m<ABC = 60°
4. Angle Formed Inside of a Circle by Two Intersecting Chords:
When two chords intersect "inside" a circle, four angles are formed. At the point of intersection, two sets of vertical
angles can be seen in the corners of the X that is formed on the picture. Remember: vertical angles are equal.
Angle Formed Inside by Two Chords =
Sum of Intercepted Arcs
Once you have found ONE of these angles, you automatically know the sizes of the other three by
using your knowledge of vertical angles (being congruent) and adjacent angles forming a straight line
(measures adding to 180).
<BED is formed by two intersecting chords.
Its intercepted arcs are . [Note: the intercepted arcs belong to the set of
vertical angles.]
also, m<CEA = 120° (vetical angle)
m<BEC and m<DEA = 60° by straight line.
5. Angle Formed Outside of a Circle by the Intersection of:"Two Tangents" or "Two Secants" or "a Tangent and a Secant".
The formulas for all THREE of these situations are the same:
Angle Formed Outside = Difference of Intercepted Arcs (When subtracting, start with the larger arc.)
Two Tangents:<ABC is formed by two tangentsintersecting outside of circle O.
The intercepted arcs are minor arc and major arc . These two arcs together comprise the entire circle.
Special situation for this set up: It can be proven that <ABC and central <AOC are supplementary. Thus the angle formed by the two tangents and its first intercepted arc also add to 180º.
Two Secants:<ACE is formed by two secantsintersecting outside of circle O.
The intercepted arcs are minor arcs and .
a Tangent and a Secant:<ABD is formed by a tangent and a secant
intersecting outside of circle O.
The intercepted arcs are minor arcs and .
An arc is part of a circle's circumference.
Definition: In a circle, the degree measure of an arc is equal to the measure of the central angle that intercepts the arc.
Definition:
In a circle, the length of an arc is a portion of the circumference.
Remembering that the arc measure is the measure of the central angle, a definition can be formed as:
Example:
In circle O, the radius is 8, and the measure of minor arc is 110 degrees. Find the length of minor arc to the nearest integer.
Solution:
= 15.35889742 = 15
Understanding how an arc is measured makes the next theorems common sense.
Theorem: In the same circle, or congruent circles, congruent central angles have congruent arcs.
Theorem:(converse)
In the same circle, or congruent circles, congruent arcs have congruent central angles.
Remember: In the same circle, or congruent circles, congruent arcs have congruent chords. Knowing this theorem makes the next theorems seem straight forward.
Theorem: In the same circle, or congruent circles, congruent central angles have congruent chords.
Theorem:(converse)
In the same circle, or congruent circles, congruent chords have congruent central angles.
A quadratic equation is defined as an equation in which one or more of the terms is squared but raised to no higher power.
The general form is ax2 + bx + c = 0, where a, b and c are constants.
We learned in Algebra that linear quadratic systems can be solved algebraically and graphically. See the Algebra site if you need practice with algebraic solutions. In Geometry, we will concentrate on the
graphical solutions to these systems.
Linear - quadratic system:(where the quadratic is in one variable - only one variable is squared)
y = x - 2 (linear)y = x2 - 4x - 2 (quadratic - a parabola)
This familiar linear- quadratic system, where only one variable is squared in the quadratic, will be the graph of a parabola and a straight line. When a parabola and a straight line are graphed on the same set of axes, three situations are possible.
The equations will intersect in two locations. Two real solutions.
The equations will intersect in one location. One real solution.
The equations will not intersect.No real solutions.
Solve this problem manually (without graphing calculator): y = x2 - 4x - 2 y = x - 2
1. We will start with the graph of the quadratic equation. You can start with either equation.
y = x2- 4x- 2
2. We want to produce a representative graph of the parabola, one that will show the turning point. Find the equation of the axis of symmetry to guarantee that the turning point of the parabola will be graphed.
Axis of symmetry formula:
a = 1, b = -4, c = -2
Substituting:
x = 2
3. Choose this new x value as the center of a domain for graphing the parabola. Create a chart of values. Three values are usually tested above and below this x-value.
Substitute each value of x in the quadratic equation to find the corresponding values for y to complete the chart.
x y-1012345
x y-1012345
3-2-5-6-5-23
4. Plot these points on graph paper. Sketch the parabola.
5. Now, graph the linear equation on the same set of axes. You can use the same table of values and simply find the y values for the straight line. Or you can use the slope and y-intercept to graph the line.
Standard form for a line is: y = mx + bwhere m is the slope, and b is the y-intercept.
For y = x - 2, m = 1, and b = -2Draw a starting point at -2 on the y-axis. Use slope (rise over run) to find a second point by going up 1 and to the right 1, or down 1 and to
the left 1.
6. Using a straight-edge, draw the line.
7. Last, find the point(s) where the two graphs intersect (if they do intersect).
These graphs intersect at 2 points whose coordinates are: (0,-2) and (5,3)Solution Set: {(0,-2),(5,3)}
8. Let's check these answers to be sure we are correct. Place each point into BOTH equations.
Wasn't it nice how the points of intersection turned out to be nice integer values that were easily seen on the graph!
The problem we just solved worked out "nicely" since the intersection points were easily seen as integer values on the graph paper. Of course, this does not happen with all graphs. How could we tell, by looking at the graph, if a point of intersection was (2.3, 1.5), for example?
Answer: We could not tell "by looking". We would have to solve the system algebraically to find such an intersection point, or we would have to use our graphing calculator with the intersect option.
Linear - quadratic system:(where the quadratic is in two variables - both variables are squared)
Solve this problem manually (without graphing calculator): 1. We will start with the graph of the quadratic
equation, the circle. You can start with either equation.
The general form of a circle with center at the origin is x2 + y2 = r2, where r is the radius.
For this equation, x2 + y2 = 25, the center is at the origin, (0,0) and the radius is 5 (the square root of 25).
2. It is not necessary to make a chart to produce this graph. Since we know the center and the radius, we can easily draw the graph.
A compass can be used to make a more accurate looking graph.
3. Now, graph the linear equation on the same set of axes. Using the slope and y-intercept of the line will produce the graph quickly. Use a straight-edge to draw the line.
4y = 3x
The y-intercept is 0 and the slope is 3/4.
4. Last, find the point(s) where the two graphs intersect (if they do intersect).
The graphs intersect at points (-4,-3) and (4,3). Solution Set: {(-4,-3), (4,3)}
