Prerequisite 1 - Graphical Representation of Data
I) Vocabulary -
A) The Cartesian Plane - the rectangular coordinate plane that represents ordered pairs fo real numbers by points in a plane.
1) x - axis: the horizontal real number line
2) y - axis; the vertical real number line
3) origin - the point of intersection of the x-axis and the y-axis
4)Quadrants - the two axes divide the plane into four parts - the upper right quadrant is quadrant I, the upper left quadrant is quadrant II, the lower left quadrant is quadrant III, and the lower right quadrant is quadrant IV. They go counterclockwise from the positive side of the x-axis.
5) Ordered pair - each point in the plane corresponds to an ordered pair of real numbers (x, y)
called the coordinates of the point.
a) x-coordinate: represents the directed distance from the y-axis to the point
b) y-coordinate: represents the directed distance from the x-axis to the point.
B) Representing Data Graphically:
1) Scatterplot
2) bar graph
3) line graph
4) histogram
5) Interpreting a model based upon the data:
a) look at the data and know how to find the best-fit line
C) Distance Formula: d = √((x1 - x2)2 + (y1 - y2)2)
this is derived from the pythagorean theorem, where the difference of the x's is represented by "a" and the difference of the y's is represented by "b" and so
AB = a2 + b2 = c2
Example: Given points (-2, -5) and (-3, 4), what is the distance between the two points.
√((-2 - -3)2 + (-5 - 4)2) =
√((1)2 + (-9)2) =
√(1+81)=
√82
E) Midpoint Formula: the mean, or average, of the x-coordinates and the y-coordinates.therefore the midpoint formula is: ((x1+ x2) /2, (y1+ y2) /2)
example: Given A(-1, 7) and B(3, -3) what is their midpoint?((-1 + 3)/2, (7 + -3)/2) = (2/2, 4/2) = (1, 2)
F) Standard equation of a circle with radius r and center (h, k) is:
r2 = (x - h)2 + (y - k)2
OR
r = √((x - h)2 + (y - k)2)
Prerequisite 2: Graphs of Equations
I) Vocabulary:
A) Solution Point - for an equation in variables x and y, a point (a, b) is a solution point if the substitution of x = a and y = b satisfies the equation.
1) Graph of the equation - the set of all solution points of an equation.
B) How to Sketch the Graph of an Equation by Point Plotting
1) If possible, rewrite the equation so that one of the variables is isolated on one side of the equation.
2) Make a table of several solution points (usually 5 - 7)
3) Plot these points in the coordinate plane.
4) Connect the points with a smooth curve.
C) Using a Graphing Utility to Graph an Equation - to graph an equation involving x and y on a graphing utility, use the following procedure.
1) Rewrite the equation so that "y"is isolated on the left side.
2) Enter the equation into a graphing utility.
3) Determine a viewing window that shows all important features of the graph.
4) Graph the equation.
D) Throughout this course, you will learn that there are many ways to approach a problem.
1) a numerical approach: construct and use a table.
2) a graphical approach: draw and use a graph.
3) an algebraic approach: use the rules of algebra.
Prerequisite 3: Lines in the Plane
I) Vocabulary:
A) the slope of a line - represents the number of units a line rises or falls vertically for each unit of horizontal change from left to right.
Slope: (nonvertical line): Ratio of the vertical change (the rise, Δ y) to the horizontal change (the run, Δ x)
Formula: slope = m = (Δy/Δx) = (y2 - y1)÷ (x2 - x2)
II) Postulate:
A) In the coordinate plane, 2 nonvertical lines are parallel if and only if they have the same slope. Any two vertical or horizontal lines are parallel.
B) Slope- Intercept Form of the Equation of a Line: y = mx + b where m is the slope of the line and b is the y-intercept (where it crosses the y-axis so point (0,b)
C) Point-slope form of an equation of the line that passes through the point (x1, y1) and has a slope of m is
y - y1 = m(x - x1)
D) Summary of Equations of Lines
1. General form: Ax + By + C = 0
2. Vertical line: x = a
3. Horizontal line: y = b
4. Slope-intercept form: y = mx + b
5. Point-slope form: y - y1 = m(x - x1)
E) Parallel Lines - two distinct non-vertical lines are parallel if and only if their slopes are equal.
F) Perpendicular Lines - two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals of each other. If m1 = a/b then m2 = -b/a.
The product of the two slopes are -1. (a/b)(-b/a) = -1
m1 - -1/(m2)
Example: if m1 = 2/3 then
m2 = -3/2
Monday, August 13, 2007
Wednesday, August 8, 2007
Geometry Chapter 9.5 Trigonometric Ratios
Chapter 9.5 Trigonometric Ratios:
I) Vocabulary:
A) A Trigonometric Ratio - is a ratio of the lengths of two sides of a right triangle.
1) Sine - abbreviated sin
2) Cosine - abbreviated cos
3) Tangent - abbreviated tan
check out: http://www.tpub.com/math1/20b.htm
Let ΔABC be a right triangle. The sine, the cosine, and the tangent of the acute angle ∠A are defined as follows:
The side opposite angle A = side a
The side opposite angle B = side b
The side opposite angle C = side c = hypotenuse
Sin A = (side opposite angle A)/hypotenuse = a/c
Cos A = (side adjacent angle A)/hypotenuse = b/c
Tan A = (side opposite angle A)/(side adjacent angle A) = a/b
Example: Given ΔABC with AB = 13, AC = 5 and BC = 12 with right ∠C, find the 3 different ratios for angle A and angle B:
Sin A = 12/13
Cos A = 5/13
Tan A = 12/5
Sin B = 5/13
Cos B = 12/13
Tan B = 5/12
Do you notice a pattern?
The Sin A = Cos B and Cos A = Sin B plus the tangents are just reciprocals of each other. This is because the Sine and Cosine are cofunctions of each other.
Example: Find the sine, cosine and tangent of 30°, 45° and 60°.
Sin 30° = 1/2
Cos 30° = √(3)/2
Tan 30° = √(3)/3
Sin 45° = √(2)/2
Cos 45° = √(2)/2
Tan 45° = 1
Sin 60° = √(3)/2
Cos 60° = 1/2
Tan 60° = √3
B)Angle of Elevation - the angle that your line of sight makes with a line drawn horizontally.
Example: The angle of elevation from the base of a water slide to the top of a waterslide is about 13°. The slide extends horizontally about 58.2 meters. Find the vertical height of the slide.
1) Draw a diagram. Place the appropriate values where they belong.
2) Using the trigonometric functions, find the one you need.
a) We need to know the side opposite the angle and we know the side adjacent to the angle so this would be tangent.
b) Tan 13° = h/58.2
h = 58.2 (tan 13°)
h = 13.43652872 meters
Chapter 9.6 Solving Right Triangles:
I) Vocabulary:
A) Solve a right triangle - to determine the measures of all six parts, 3 sides and 3 angles.
Example: Given ΔABC, angle C is the right angle, angle A = 26°, AB = 4.5 inches, find AC, BC and angle B.
measure of angle A = 26°
measure of angle B = ?
measure of angle C = 90°
AB = c = 4.5 inches
BC = a = ?
AC = b = ?
A triangle has 180° so 26°+ 90° = 116°, 180° - 116° = 64° = measure of angle B
From the angle A, we know the hypotenuse or side "c" = 4.5, to find side "a", this is opposite to angle A so we would use sine function.
sin 26° = a/4.5
a = 1.972670161 inches
From the angle A, we know the hypotenuse and need to find side "b", this is adjacent to angle A so we would use the cosine function.
cos 26° = b/4.5
b = 4.044573208 inches
Therefore we have now solved the triangle:
measure of angle A = 26°
measure of angle B = 64°
measure of angle C =90°
a = 1.972670161 inches
b = 4.044573208 inches
c = 4.5 inches
I) Vocabulary:
A) A Trigonometric Ratio - is a ratio of the lengths of two sides of a right triangle.
1) Sine - abbreviated sin
2) Cosine - abbreviated cos
3) Tangent - abbreviated tan
check out: http://www.tpub.com/math1/20b.htm
Let ΔABC be a right triangle. The sine, the cosine, and the tangent of the acute angle ∠A are defined as follows:
The side opposite angle A = side a
The side opposite angle B = side b
The side opposite angle C = side c = hypotenuse
Sin A = (side opposite angle A)/hypotenuse = a/c
Cos A = (side adjacent angle A)/hypotenuse = b/c
Tan A = (side opposite angle A)/(side adjacent angle A) = a/b
Example: Given ΔABC with AB = 13, AC = 5 and BC = 12 with right ∠C, find the 3 different ratios for angle A and angle B:
Sin A = 12/13
Cos A = 5/13
Tan A = 12/5
Sin B = 5/13
Cos B = 12/13
Tan B = 5/12
Do you notice a pattern?
The Sin A = Cos B and Cos A = Sin B plus the tangents are just reciprocals of each other. This is because the Sine and Cosine are cofunctions of each other.
Example: Find the sine, cosine and tangent of 30°, 45° and 60°.
Sin 30° = 1/2
Cos 30° = √(3)/2
Tan 30° = √(3)/3
Sin 45° = √(2)/2
Cos 45° = √(2)/2
Tan 45° = 1
Sin 60° = √(3)/2
Cos 60° = 1/2
Tan 60° = √3
B)Angle of Elevation - the angle that your line of sight makes with a line drawn horizontally.
Example: The angle of elevation from the base of a water slide to the top of a waterslide is about 13°. The slide extends horizontally about 58.2 meters. Find the vertical height of the slide.
1) Draw a diagram. Place the appropriate values where they belong.
2) Using the trigonometric functions, find the one you need.
a) We need to know the side opposite the angle and we know the side adjacent to the angle so this would be tangent.
b) Tan 13° = h/58.2
h = 58.2 (tan 13°)
h = 13.43652872 meters
Chapter 9.6 Solving Right Triangles:
I) Vocabulary:
A) Solve a right triangle - to determine the measures of all six parts, 3 sides and 3 angles.
Example: Given ΔABC, angle C is the right angle, angle A = 26°, AB = 4.5 inches, find AC, BC and angle B.
measure of angle A = 26°
measure of angle B = ?
measure of angle C = 90°
AB = c = 4.5 inches
BC = a = ?
AC = b = ?
A triangle has 180° so 26°+ 90° = 116°, 180° - 116° = 64° = measure of angle B
From the angle A, we know the hypotenuse or side "c" = 4.5, to find side "a", this is opposite to angle A so we would use sine function.
sin 26° = a/4.5
a = 1.972670161 inches
From the angle A, we know the hypotenuse and need to find side "b", this is adjacent to angle A so we would use the cosine function.
cos 26° = b/4.5
b = 4.044573208 inches
Therefore we have now solved the triangle:
measure of angle A = 26°
measure of angle B = 64°
measure of angle C =90°
a = 1.972670161 inches
b = 4.044573208 inches
c = 4.5 inches
Tuesday, August 7, 2007
Geometry Chapter 11 Area of Polygons and Circles
Chapter 11.1 Angle Measures in Polygons:
I) lets look at the sum of polygon angle measures:
Polygon - Number of sides - Number of triangles can make from the polygon - sum of the measures of the interior angles:
A) Triangle - 3 sides - 1 triangle - 180°
B) Quadrilateral - 4 sides - 2 triangles - 360°
C) Pentagon - 5 sides - 3 triangles - 540°
D) Hexagon - 6 sides - 4 triangles - 720°
E) Heptagon - 7 sides - 5 triangles - 900°
F) Octagon - 8 sides - 6 triangles - 1080°
G) Nonagon - 9 sides - 7 triangles - 1260°
H) Decagon - 10 sides - 8 triangles - 1440°
I) Undecagon - 11 sides - 9 triangles - 1620°
J) Dodecagon - 12 sides - 10 triangles - 1800°
Do you see a pattern? The number of sides minus 2 is equal to the number of triangles so...
II) Theorem:
A) Polygon Interior Angles Theorem - the sum of the measures of the interior angles of a convex n-gon is (n - 2) (180°)
Example: What is the sum of the measures of the interior angles of a convex 20-gon?
(20 - 2)(180°) = 3240°
B) The measures of each interior angle of a regular n-gon is:
(1/n)(n-2)(180°)
Example: Given a regular hexagon, what is the measure of each angle?
(1/6)(6-2)(180°) = (1/6)(4)(180°) = 120°
C) Polygon Exterior Angles Theorem - the sum of the measures of the exterior angles of a convex polygon, one angle iat each vertex, is 360°.
Example: If the exterior angles of a polygon is x, 2x, 2x, 4x, and 3x, solve for x.
x + 2x + 2x + 4x + 3x = 360
12 x = 360
x = 30
Example: Given a regular polygon has 20 sides, what is the measure of an exterior angle and an interior angle?
360/20 = 18, so every exterior angle is 18°
180° - 18° = 162° , so every interior angle is 162°
(20 - 2)(180) = (162)(20)
3240 = 3240 so it checks
Example: Given the measrue of each interior angle of a regular n-gon is 120°, how many sides does the polygon have?
(1/n)(n-2)(180) = 120
(n-2)(180) = 120n
180n - 360 = 120n
60n = 360
n = 6
If the interior angle is 120°, then the exterior angle is 60°
360/60 = 6 sides, same answer!
Chapter 11.2 Areas of Regular Polygons:
I) Area of an Equilateral Triangles is:
A = ((√3)/4) s2
Example: Given an equilateral triangle with side length of 4 what is the area?
A = ((√3)/4)(42)
A = ((√3)/4)(16)
A = 4√3 square units
II) Vocabulary:
A) Center of the polygon - when you circumscribe a circle about a regular polygon, the center point of the circle is the center of the polygon.
B) Radius of the polygon - when you circumscribe a circle about a regular polygon, from the center point of the circle to the vertex of the polygon is the radius of the polygon.
C) Apothem of the polygon - the distance from the center point of the polygon perpendicular to any side of the polygon.
D) A Central angle of a regular polygon - is an angle whose vertex is the center and whose sides contain two consecutive vertices of the polygon. You can divide 360° by the number of sides to find the measure of each central angle of the polygon.
III) Theorem:
A) Area of a Regular Polygon - the area of a regular n-gon with side length "s" is half the product of the apothem "a" and the perimeter "P"
A = (1/2)(a)(P)
Chapter 11.3 Perimeters and Areas of Similar Figures
I) Theorems:
A) Areas of Similar Polygons- if two polygons are similar with the lengths of the corresponding sides in the ratio of a:b, then the ratio of their areas is a2:b2
Example: Given two similar hexagons with corresponding sides of 4 and 5, what is the ratio of their areas?
42:52
16:25
Chapter 11.4 Circumference and Arc Length
A) Circumference - of a circle is the distance around the circle.
C = 2π r = πd
B) Arc length corollary:
in a circle, the ratio of the length of a given arc to the circumference is equal to the ratio of the measure of the arc to 360º
Example: recall, the central angle is equal to it's intercepted arc
Given circle P with central angle APB = 30º and radius of 8 radius, what is the measure of the intercepted arc AB?
(measure of the arc AB)/(2π8) = (30/360)
(measure of the arc AB)/ (16π) = (1/6)
measure of the arc AB = 16π/6
measure of the arc AB = 8π/3 inches or about 8.37758041 inches
Chapter 11.5 Areas of Circles and Sectors
I) Vocabulary:
A) Sector of a Circle - is the region bounded by two radii of the circle and their intercepted arec.
II) Theorems:
A) Area of a Circle - the are of a circle is π times the square of the radius
A = πr2
Example: given a circle with radius of 9 cm, what is the area?
A = π(9)2 = 81π square cm.
B) Area of a Sector - the ratio of the area A of a sector of a circle to the area of the circle is equal to the ratio of the measure of the intercepted arc to 360º
Example: Given Circle C with points A and B on the circle, the major arc AB is 293º and the radius is 10 cm, what is the area of the sector?
(Area of sector)/(π 102) = 293º /360º
Area of sector/100π = 293/360
Area of sector = 1465π/18 which is about 255.6907354 square cm.
Chapter 11.6 Geometric Probability
I) Vocabulary
A) Probability is a number from 0 to 1 that represents the chance that an event will occur.
B) Geometric Probability - involves geometric measures such as length or area
II) Theorems:
A) Probability and length - Let line segment AB be a segment that contains the segment CD. If a point K on line segment AB is chosen at random, then the probability that it is on CD is as follows:
P(Point K is on line segment CD) = (length of CD)/(length of AB)
Example: Given AB = 5 and CD = 2, then the probability that point K is on CD is
5/2
B) Probability and Area - Let J be a region that contains region M. If a point K in J is chosen at random, then the probability that it is in region M is as follows:
P(Point K is in region M) = (Area of M)/(Area of J)
Example: You mow a rectangular lawn of 20 feet by 30 feet and it has a garden of 5 feet by 6 feet. What is the geometric probability that if someone tosses a ball that it will land in the garden?
(5)(6) = 30 square feet
(20)(30) = 600 square feet so
30/600 = 1/20
I) lets look at the sum of polygon angle measures:
Polygon - Number of sides - Number of triangles can make from the polygon - sum of the measures of the interior angles:
A) Triangle - 3 sides - 1 triangle - 180°
B) Quadrilateral - 4 sides - 2 triangles - 360°
C) Pentagon - 5 sides - 3 triangles - 540°
D) Hexagon - 6 sides - 4 triangles - 720°
E) Heptagon - 7 sides - 5 triangles - 900°
F) Octagon - 8 sides - 6 triangles - 1080°
G) Nonagon - 9 sides - 7 triangles - 1260°
H) Decagon - 10 sides - 8 triangles - 1440°
I) Undecagon - 11 sides - 9 triangles - 1620°
J) Dodecagon - 12 sides - 10 triangles - 1800°
Do you see a pattern? The number of sides minus 2 is equal to the number of triangles so...
II) Theorem:
A) Polygon Interior Angles Theorem - the sum of the measures of the interior angles of a convex n-gon is (n - 2) (180°)
Example: What is the sum of the measures of the interior angles of a convex 20-gon?
(20 - 2)(180°) = 3240°
B) The measures of each interior angle of a regular n-gon is:
(1/n)(n-2)(180°)
Example: Given a regular hexagon, what is the measure of each angle?
(1/6)(6-2)(180°) = (1/6)(4)(180°) = 120°
C) Polygon Exterior Angles Theorem - the sum of the measures of the exterior angles of a convex polygon, one angle iat each vertex, is 360°.
Example: If the exterior angles of a polygon is x, 2x, 2x, 4x, and 3x, solve for x.
x + 2x + 2x + 4x + 3x = 360
12 x = 360
x = 30
Example: Given a regular polygon has 20 sides, what is the measure of an exterior angle and an interior angle?
360/20 = 18, so every exterior angle is 18°
180° - 18° = 162° , so every interior angle is 162°
(20 - 2)(180) = (162)(20)
3240 = 3240 so it checks
Example: Given the measrue of each interior angle of a regular n-gon is 120°, how many sides does the polygon have?
(1/n)(n-2)(180) = 120
(n-2)(180) = 120n
180n - 360 = 120n
60n = 360
n = 6
If the interior angle is 120°, then the exterior angle is 60°
360/60 = 6 sides, same answer!
Chapter 11.2 Areas of Regular Polygons:
I) Area of an Equilateral Triangles is:
A = ((√3)/4) s2
Example: Given an equilateral triangle with side length of 4 what is the area?
A = ((√3)/4)(42)
A = ((√3)/4)(16)
A = 4√3 square units
II) Vocabulary:
A) Center of the polygon - when you circumscribe a circle about a regular polygon, the center point of the circle is the center of the polygon.
B) Radius of the polygon - when you circumscribe a circle about a regular polygon, from the center point of the circle to the vertex of the polygon is the radius of the polygon.
C) Apothem of the polygon - the distance from the center point of the polygon perpendicular to any side of the polygon.
D) A Central angle of a regular polygon - is an angle whose vertex is the center and whose sides contain two consecutive vertices of the polygon. You can divide 360° by the number of sides to find the measure of each central angle of the polygon.
III) Theorem:
A) Area of a Regular Polygon - the area of a regular n-gon with side length "s" is half the product of the apothem "a" and the perimeter "P"
A = (1/2)(a)(P)
Chapter 11.3 Perimeters and Areas of Similar Figures
I) Theorems:
A) Areas of Similar Polygons- if two polygons are similar with the lengths of the corresponding sides in the ratio of a:b, then the ratio of their areas is a2:b2
Example: Given two similar hexagons with corresponding sides of 4 and 5, what is the ratio of their areas?
42:52
16:25
Chapter 11.4 Circumference and Arc Length
A) Circumference - of a circle is the distance around the circle.
C = 2π r = πd
B) Arc length corollary:
in a circle, the ratio of the length of a given arc to the circumference is equal to the ratio of the measure of the arc to 360º
Example: recall, the central angle is equal to it's intercepted arc
Given circle P with central angle APB = 30º and radius of 8 radius, what is the measure of the intercepted arc AB?
(measure of the arc AB)/(2π8) = (30/360)
(measure of the arc AB)/ (16π) = (1/6)
measure of the arc AB = 16π/6
measure of the arc AB = 8π/3 inches or about 8.37758041 inches
Chapter 11.5 Areas of Circles and Sectors
I) Vocabulary:
A) Sector of a Circle - is the region bounded by two radii of the circle and their intercepted arec.
II) Theorems:
A) Area of a Circle - the are of a circle is π times the square of the radius
A = πr2
Example: given a circle with radius of 9 cm, what is the area?
A = π(9)2 = 81π square cm.
B) Area of a Sector - the ratio of the area A of a sector of a circle to the area of the circle is equal to the ratio of the measure of the intercepted arc to 360º
Example: Given Circle C with points A and B on the circle, the major arc AB is 293º and the radius is 10 cm, what is the area of the sector?
(Area of sector)/(π 102) = 293º /360º
Area of sector/100π = 293/360
Area of sector = 1465π/18 which is about 255.6907354 square cm.
Chapter 11.6 Geometric Probability
I) Vocabulary
A) Probability is a number from 0 to 1 that represents the chance that an event will occur.
B) Geometric Probability - involves geometric measures such as length or area
II) Theorems:
A) Probability and length - Let line segment AB be a segment that contains the segment CD. If a point K on line segment AB is chosen at random, then the probability that it is on CD is as follows:
P(Point K is on line segment CD) = (length of CD)/(length of AB)
Example: Given AB = 5 and CD = 2, then the probability that point K is on CD is
5/2
B) Probability and Area - Let J be a region that contains region M. If a point K in J is chosen at random, then the probability that it is in region M is as follows:
P(Point K is in region M) = (Area of M)/(Area of J)
Example: You mow a rectangular lawn of 20 feet by 30 feet and it has a garden of 5 feet by 6 feet. What is the geometric probability that if someone tosses a ball that it will land in the garden?
(5)(6) = 30 square feet
(20)(30) = 600 square feet so
30/600 = 1/20
Geometry 10.7 Locus of points
Chapter 10.7 Locus:
I)Vocabulary:
A) Locus - in a plane is the set of all points in a plane that satisfy a given condition or a set of given conditions. The word locus is derived from the Latin word for location. The plural of locus is loci, pronounced "low-sigh"
- can be described as the path of an object moving in a plane.
B) Finding a locus
1) Draw any figures that are given in the statement of the problem.
2) Locate several points that satisfy the given condition.
3) Continue drawing points until you can recognize a pattern.
4) Draw the locus and describe it in words.
Examples:
1) Given a point C, what is the locus of points 3 inches from C?
When you draw this you will get: the locus of points is a circle with center point C and a radius of 3 inches.
2) Given line AB, what is the locus of points 2 cm from line AB?
the locus of points 2 cm from line AB is 2 lines parallel to AB on either side of line AB 2 cm from AB
3) Given lines AB and CD, what is the locus of points equidistant from lines AB and CD?
the locus of points is a line parallel to both AB and CD that is half way between the lines AB and CD.
4) Given ∠BAC, what is the locus of points equidistant from ray AB and ray AC?
the locus of points is the angle bisect of ∠BAC.
5)Given point A and point B, what is the locus of points equidistant from point A and point B?
the locus of points is the perpendicular bisector of line segment AB.
C) Earthquakes - the epicenter of an earthquake is the point on the Earth's surface that is directly above the earthquake's origin. A seismograph measures ground motion during an earthquake. The seismograph measures the distance to the epicenter, but not he direction to the epicenter. To locate the epicenter, readings from three seismographs in different locations are needed.
1) locating an epicenter:
You are given readings from 3 different seismographs:
1) point A(-2, 2) and distance to the earthquake is 3 miles
2) point B(4, -1) and distance to the earthquake is 6 miles
3) point C(1, -5) and distance to the earthquake is 5 miles
Drawing these three circles, the point of interception is the epicenter. This would be (-2, -1)
I)Vocabulary:
A) Locus - in a plane is the set of all points in a plane that satisfy a given condition or a set of given conditions. The word locus is derived from the Latin word for location. The plural of locus is loci, pronounced "low-sigh"
- can be described as the path of an object moving in a plane.
B) Finding a locus
1) Draw any figures that are given in the statement of the problem.
2) Locate several points that satisfy the given condition.
3) Continue drawing points until you can recognize a pattern.
4) Draw the locus and describe it in words.
Examples:
1) Given a point C, what is the locus of points 3 inches from C?
When you draw this you will get: the locus of points is a circle with center point C and a radius of 3 inches.
2) Given line AB, what is the locus of points 2 cm from line AB?
the locus of points 2 cm from line AB is 2 lines parallel to AB on either side of line AB 2 cm from AB
3) Given lines AB and CD, what is the locus of points equidistant from lines AB and CD?
the locus of points is a line parallel to both AB and CD that is half way between the lines AB and CD.
4) Given ∠BAC, what is the locus of points equidistant from ray AB and ray AC?
the locus of points is the angle bisect of ∠BAC.
5)Given point A and point B, what is the locus of points equidistant from point A and point B?
the locus of points is the perpendicular bisector of line segment AB.
C) Earthquakes - the epicenter of an earthquake is the point on the Earth's surface that is directly above the earthquake's origin. A seismograph measures ground motion during an earthquake. The seismograph measures the distance to the epicenter, but not he direction to the epicenter. To locate the epicenter, readings from three seismographs in different locations are needed.
1) locating an epicenter:
You are given readings from 3 different seismographs:
1) point A(-2, 2) and distance to the earthquake is 3 miles
2) point B(4, -1) and distance to the earthquake is 6 miles
3) point C(1, -5) and distance to the earthquake is 5 miles
Drawing these three circles, the point of interception is the epicenter. This would be (-2, -1)
Geometry Chapter 10.6 Equations of Circles
Chapter 10.6 Equations of Circles:
I) Standard equation of a circle with radius r and center (h,k) is:
r2 = (x - h)2 + (y - k)2
OR
r = √((x - h)2 + (y - k)2)
Example: Given circle O with center point (5, 3) and radius of 6, what is the standard equation of circle O?
r2 = (x - h)2 + (y - k)2
62 = (x - 5)2 + (y - 3)2
36 = (x - 5)2 + (y - 3)2
Example: Given circle O with center point ( -2, 4) and radius of 9, what is the standard equation of circle O?
92 = (x - -2)2 + (y - 4)2
81 = (x + 2)2 + (y - 4)2
Example: Given the standard equation of circle O: (x + 12)2 + (y - 4)2 = 25, what is the center point and radius?
center point is (-12, 4) and radius of 5
I) Standard equation of a circle with radius r and center (h,k) is:
r2 = (x - h)2 + (y - k)2
OR
r = √((x - h)2 + (y - k)2)
Example: Given circle O with center point (5, 3) and radius of 6, what is the standard equation of circle O?
r2 = (x - h)2 + (y - k)2
62 = (x - 5)2 + (y - 3)2
36 = (x - 5)2 + (y - 3)2
Example: Given circle O with center point ( -2, 4) and radius of 9, what is the standard equation of circle O?
92 = (x - -2)2 + (y - 4)2
81 = (x + 2)2 + (y - 4)2
Example: Given the standard equation of circle O: (x + 12)2 + (y - 4)2 = 25, what is the center point and radius?
center point is (-12, 4) and radius of 5
Geometry Chapter 10.5 Segment Lengths in Circles
Chapter 10.5 Segment Lengths in Circles:
I) Vocabulary:
A) when two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord.
B) Tangent Segment - is tangent to a circle at an endpoint.
C) Secant Segment - is a segment that has an endpoint on a circle.
D) External Segment - is part of a secant segment from an external point to the circle. It is does not have any part of the segment in the interior part of the circle.
II) Theorem:
A) If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
Example: Given circle P with chord AB and chord CD that intersect at point E, AE = 9, ED = 12, CE = 15, EB = x, solve for x.
(9)(x) = (15)(12)
9x = 180
x = 20
B) If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = 9, PQ = 11, RS = 10 and ST = x, solve for x.
(RP)(RQ) = (RS)(RT)
9 (11+9) = 10 (x + 10)
(9)(20) = 10x + 100
180 = 10x + 100
80 = 10x
8 = x
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = x, PQ = 65, RS = 15, and ST = 35. Solve for x.
(RP)(RQ) = (RS)(RT)
x (x + 65) = (15)(15 + 35)
x2 + 65x = 750
x2 + 65x - 750 = 0
(x + 75)(x - 10) = 0
x = - 75, x = 10
Since a length cannot be a negative number, the only answer would then be 10 units.
C) If a secant segment and a tangent segment share an endpoint ooutside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = x, CD = 12, and EA = 8, solve for x.
(x)(x + 12) = 82
x2 + 12x = 64
x2 + 12x - 64 = 0
(x + 16)(x - 4) = 0
x = -16, x = 4
Since a length cannot be a negative number, the only answer would then be 4 units.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = 12, CD = 36, and EA = x, solve for x.
(12)(12 + 36) = x2
(12)(48) = x2
576 = x2
24 = x
I) Vocabulary:
A) when two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord.
B) Tangent Segment - is tangent to a circle at an endpoint.
C) Secant Segment - is a segment that has an endpoint on a circle.
D) External Segment - is part of a secant segment from an external point to the circle. It is does not have any part of the segment in the interior part of the circle.
II) Theorem:
A) If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
Example: Given circle P with chord AB and chord CD that intersect at point E, AE = 9, ED = 12, CE = 15, EB = x, solve for x.
(9)(x) = (15)(12)
9x = 180
x = 20
B) If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = 9, PQ = 11, RS = 10 and ST = x, solve for x.
(RP)(RQ) = (RS)(RT)
9 (11+9) = 10 (x + 10)
(9)(20) = 10x + 100
180 = 10x + 100
80 = 10x
8 = x
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = x, PQ = 65, RS = 15, and ST = 35. Solve for x.
(RP)(RQ) = (RS)(RT)
x (x + 65) = (15)(15 + 35)
x2 + 65x = 750
x2 + 65x - 750 = 0
(x + 75)(x - 10) = 0
x = - 75, x = 10
Since a length cannot be a negative number, the only answer would then be 10 units.
C) If a secant segment and a tangent segment share an endpoint ooutside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = x, CD = 12, and EA = 8, solve for x.
(x)(x + 12) = 82
x2 + 12x = 64
x2 + 12x - 64 = 0
(x + 16)(x - 4) = 0
x = -16, x = 4
Since a length cannot be a negative number, the only answer would then be 4 units.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = 12, CD = 36, and EA = x, solve for x.
(12)(12 + 36) = x2
(12)(48) = x2
576 = x2
24 = x
Wednesday, August 1, 2007
Geometry Chapter 10.3 Inscribed Angles and 10.4 Other Angle Relationships in Circles
Chapter 10.3 Inscribed Angles
I) Vocabulary
A) Inscribed Angles - an angle whose vertex is on a circle and whose sides contain chords of the circle.
B) Intercepted arc - the arc that lies in the interior of an inscribed angle and has endpoints on the angle.
C) If all of the vertices of a polygon lie on a circle, the polygon is inscribed in the circle and the circle is circumscribed about the polygon.
II) Theorems:
A) Measure of an inscribed angle - if an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc.
Example: if inscribed angle ABC of circle D is 20º, what is the measure of arc AC?
20º times 2 = 40º so the measure of arc AC is 40º.
B) If two inscribed angles of a circle intercept the same arc, then the angles are congruent.
Example: If angle A and angle B are both inscribed angles and both intercept arc CD, then angle A = angle B
C) If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of a n inscribed triangle is a diameter of the circle, then the triangle is a right triangle is a right triangle and the angle opposite the diameter is the right angle.
D) A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.
Example: Given quadrilateral ABCD is inscribed in circle E, angle A = 115º, angle B = angle D = y, and angle C = x, solve for x and y.
measure of angle A + measure of angle C = 180º
115 º+ x = 180 º
x = 65º
measure of angle B + measure of angle D = 180º
y + y = 180º
2y = 180º
y = 90º
Example: Given quadrilateral ABCD is inscribed in Circle E, angle A = 26y, angle B = 3x, angle C = 2x, and angle D = 21y, solve for both x and y.
measure of angle A + measure of angle C = 180º
26y + 2x = 180
measure of angle B + measure of angle D = 180º
3x + 21y = 180
Now solve both equations for x:
2x = -26y + 180
x = -13y + 90º
3x = -21y + 180
x = -7y + 60
Since x is equal to itself, by substitution:
-13y + 90 = -7y + 60
90 = 6y + 60
30 = 6y
y = 5
Now substitute y = 5,
x = (-7)(5) + 60 = -35 + 60 = 25
check to make sure both answers work!
Chapter 10.4 Other Angle Relationships in Circles
I) Theorems:
A) If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. m∠ 1 = (1/2)measure of arc AB.
Example: Given circle P with angle ACB that intercepts arc AB, the measure of arc AB = 50º, what is the measure of angle ACB?
m∠ACB = (1/2) measure of arc AB
m∠ACB = (1/2) (50)
m∠ACB = 25º
B) If two chords intersect in the interior of a circle, then the measure of each angle in one half the sum of the measures of the arcs intercepted by angle and its vertical angle.
m∠1 = (1/2)(measure arc CD + measure of arc AB)
Example: If Chord AB and chord CD intersect circle P at point E, the measure of arc AD = 25 degrees and the measure of arc CB = 75º what is the m∠AED?
measure of angle AED = (1/2)(25 + 75 )
measure of angle AED = (1/2)(100)
measure of angle AED = 50º
C) If a tangent and a secant, two tangents, or two secants intersect in the exterior of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs.
Example: Given Secant PAB and tangent PC of circle O, the measure of arc BC = 200º and the measure of arc AC = 30º, what is the measure of angle APC?
measure of angle APC = (1/2)(measure of arc BC - measure of arc AC)
measure of angle APC = (1/2)(200 - 30)
measure of angle APC = (1/2)(170)
measure of angle APC = 85º
Example: Given secant PAB and secant PCD of circle O, the measure of arc BD = 105º and measure of arc AC = 51º, what is the measure of angle APC ?
measure of angle APC = (1/2)(105 - 51 )
measure of angle APC = (1/2)(54)
measure of angle APC = 27º
Example: Given tangent PA and tangent PC, the measure of angle APC = 40º, what are the measures of the major arc AC and the minor arc AC?
m∠APC = (1/2)(major arc AC - minor arc AC)
if we let the minor arc AC = x, then the major arc AC = 360 - x
m∠APC = (1/2)((360 -x) - x)
m∠APC = (1/2)(360 - 2x)
40 = 180 - x
x = 220º
so measure of major arc AC = 220º and the measure of minor arc AC = 140º
I) Vocabulary
A) Inscribed Angles - an angle whose vertex is on a circle and whose sides contain chords of the circle.
B) Intercepted arc - the arc that lies in the interior of an inscribed angle and has endpoints on the angle.
C) If all of the vertices of a polygon lie on a circle, the polygon is inscribed in the circle and the circle is circumscribed about the polygon.
II) Theorems:
A) Measure of an inscribed angle - if an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc.
Example: if inscribed angle ABC of circle D is 20º, what is the measure of arc AC?
20º times 2 = 40º so the measure of arc AC is 40º.
B) If two inscribed angles of a circle intercept the same arc, then the angles are congruent.
Example: If angle A and angle B are both inscribed angles and both intercept arc CD, then angle A = angle B
C) If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of a n inscribed triangle is a diameter of the circle, then the triangle is a right triangle is a right triangle and the angle opposite the diameter is the right angle.
D) A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.
Example: Given quadrilateral ABCD is inscribed in circle E, angle A = 115º, angle B = angle D = y, and angle C = x, solve for x and y.
measure of angle A + measure of angle C = 180º
115 º+ x = 180 º
x = 65º
measure of angle B + measure of angle D = 180º
y + y = 180º
2y = 180º
y = 90º
Example: Given quadrilateral ABCD is inscribed in Circle E, angle A = 26y, angle B = 3x, angle C = 2x, and angle D = 21y, solve for both x and y.
measure of angle A + measure of angle C = 180º
26y + 2x = 180
measure of angle B + measure of angle D = 180º
3x + 21y = 180
Now solve both equations for x:
2x = -26y + 180
x = -13y + 90º
3x = -21y + 180
x = -7y + 60
Since x is equal to itself, by substitution:
-13y + 90 = -7y + 60
90 = 6y + 60
30 = 6y
y = 5
Now substitute y = 5,
x = (-7)(5) + 60 = -35 + 60 = 25
check to make sure both answers work!
Chapter 10.4 Other Angle Relationships in Circles
I) Theorems:
A) If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. m∠ 1 = (1/2)measure of arc AB.
Example: Given circle P with angle ACB that intercepts arc AB, the measure of arc AB = 50º, what is the measure of angle ACB?
m∠ACB = (1/2) measure of arc AB
m∠ACB = (1/2) (50)
m∠ACB = 25º
B) If two chords intersect in the interior of a circle, then the measure of each angle in one half the sum of the measures of the arcs intercepted by angle and its vertical angle.
m∠1 = (1/2)(measure arc CD + measure of arc AB)
Example: If Chord AB and chord CD intersect circle P at point E, the measure of arc AD = 25 degrees and the measure of arc CB = 75º what is the m∠AED?
measure of angle AED = (1/2)(25 + 75 )
measure of angle AED = (1/2)(100)
measure of angle AED = 50º
C) If a tangent and a secant, two tangents, or two secants intersect in the exterior of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs.
Example: Given Secant PAB and tangent PC of circle O, the measure of arc BC = 200º and the measure of arc AC = 30º, what is the measure of angle APC?
measure of angle APC = (1/2)(measure of arc BC - measure of arc AC)
measure of angle APC = (1/2)(200 - 30)
measure of angle APC = (1/2)(170)
measure of angle APC = 85º
Example: Given secant PAB and secant PCD of circle O, the measure of arc BD = 105º and measure of arc AC = 51º, what is the measure of angle APC ?
measure of angle APC = (1/2)(105 - 51 )
measure of angle APC = (1/2)(54)
measure of angle APC = 27º
Example: Given tangent PA and tangent PC, the measure of angle APC = 40º, what are the measures of the major arc AC and the minor arc AC?
m∠APC = (1/2)(major arc AC - minor arc AC)
if we let the minor arc AC = x, then the major arc AC = 360 - x
m∠APC = (1/2)((360 -x) - x)
m∠APC = (1/2)(360 - 2x)
40 = 180 - x
x = 220º
so measure of major arc AC = 220º and the measure of minor arc AC = 140º
Geometry Chapter 10.2 Arcs and Chords,
Chapter 10.2 Arcs and Chords:
I)Vocabulary:
A) Central angle - in a plane, an angle whose vertex is the center of a circle.
B) Minor Arc - if the measure of a central angle APB is less than 180º, then A and B and the points of circle P in the interior of angle APB form a minor arc of the circle.
C) Major Arc - The points A and B and the points of circle P in the exterior of angle APB for a major arc. OR if the measure of a central angle APB is greater than 180º and less then 360º, then A and B and the points of circle P in the interior of ∠APB form a major arc of the circle.
D) Semicircle - if the endpoints of an arc are the endpoints of a diameter.
E) To name a minor arc, use the endpoints of the arc.
Example: given a minor arc on circle C with endpoints A and B, its name is arc AB and notation is the letters AB with a small arc over the two letters.
F) To name a major arc, use the endpoints and a point on the circle between the endpoints.
Example: given a major arc on circle C with endpoints A and B and a point D between A and B, its name is arc ADB.
G) Measure of a minor arc or major arc - is defined to be the measure of its central angle .
Example: measure of minor arc AB = measure of angleACB
II) Theorems:
1) Arc Addition Postulate or Arc Partition Postlate - the measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs.
Example: Arc ABC = arc AB + arc BC
2) Congruent arcs - two arcs of the same circle or of congruent circles that have the same measure.
3) In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.
4) If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
5) If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
6) In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
I)Vocabulary:
A) Central angle - in a plane, an angle whose vertex is the center of a circle.
B) Minor Arc - if the measure of a central angle APB is less than 180º, then A and B and the points of circle P in the interior of angle APB form a minor arc of the circle.
C) Major Arc - The points A and B and the points of circle P in the exterior of angle APB for a major arc. OR if the measure of a central angle APB is greater than 180º and less then 360º, then A and B and the points of circle P in the interior of ∠APB form a major arc of the circle.
D) Semicircle - if the endpoints of an arc are the endpoints of a diameter.
E) To name a minor arc, use the endpoints of the arc.
Example: given a minor arc on circle C with endpoints A and B, its name is arc AB and notation is the letters AB with a small arc over the two letters.
F) To name a major arc, use the endpoints and a point on the circle between the endpoints.
Example: given a major arc on circle C with endpoints A and B and a point D between A and B, its name is arc ADB.
G) Measure of a minor arc or major arc - is defined to be the measure of its central angle .
Example: measure of minor arc AB = measure of angleACB
II) Theorems:
1) Arc Addition Postulate or Arc Partition Postlate - the measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs.
Example: Arc ABC = arc AB + arc BC
2) Congruent arcs - two arcs of the same circle or of congruent circles that have the same measure.
3) In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.
4) If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
5) If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
6) In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
Geometry Chapter 10.1 Tangents to Circles
Chapter 10.1 Tangents to Circles
I) Vocabulary:
A) Circle - is the set of all pints in a plane that are equidistant from a given point, called the center of the circle.
B) Radius - the distance from the center to a point on the circle. Two circles are congruent if they have the same radius.
C) Diameter - the distance across the circle, through its center. The diameter is twice the radius.
D) Chord - is a segment whose endpoints are points on the circle.
E) Diameter - is a chord that passes through the center of the circle.
F) Secant - is a line that intersects a circle in two points.
G) Tangent - is a line in the plane of a circle that intersects the circle in exactly one point.
H) Tangent Circles - coplanar circles that intersect in one point.
I) Concentric - coplanar circles that have a common center.
J) Common tangents - a line or segment that is tangent to two coplanar circles.
1) Common internal tangent - intersects the segment that joins the centers of the two circles.
2) Common external tangent - does not intersect the segment that joins the centers of the two circles.
K) Interior of a circle - consists of the points that are inside the circle.
L) Exterior of a circle - consists of the points that are outside the circle.
M) Point of tangency - the point at which a tangent line intersects the circle to which it is tangent.
N) Theorems:
1) If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.
Example: Given line AB is tangent to circle C with point A on circle C, AC is a radius of 5 units, AB is 12 units, what is the length of line segment CB?
Since line AB is a tangent line, we know that AC and AB are perpendicular so angle CAB is a right angle so triangle CAB is a right triangle. Therefore we can use pythagorean theorem to find CB.
52 + 122 = (CB)2
25 + 144 = (CB)2
169 = (CB)2
13 = CB
2) In a plane, if a line is perpendicular to a radius of a circle at its endpoint on the circle, then the line is tangent to the circle.
Example: Given radius DE = 11 units on circle D, there is an external point F, line segment DF = 61 and line segment EF = 60, is line EF a tangent line to circle D?
112 + 602 = 121 + 3600 = 3721
612 = 3721
Therefore, since the two smaller sides squared are equal to the longer side squared, this is the converse of pythagorean theorem so we know that angle DEF is a right angle, so therefore line EF is perpendicular to line segment DE so therefore EF is a tangent line to circle D.
3) If two segments from the same exterior point are tangent to a circle, then they are congruent.
Example: Given tangent SR = 2x + 7 and tangent SB = 5x - 8, where both tangents intersect at point S off of circle C with points R and B are on circle C, solve for x.
2x + 7 = 5x - 8
7 = 3x - 8
15 = 3x
5 = x
I) Vocabulary:
A) Circle - is the set of all pints in a plane that are equidistant from a given point, called the center of the circle.
B) Radius - the distance from the center to a point on the circle. Two circles are congruent if they have the same radius.
C) Diameter - the distance across the circle, through its center. The diameter is twice the radius.
D) Chord - is a segment whose endpoints are points on the circle.
E) Diameter - is a chord that passes through the center of the circle.
F) Secant - is a line that intersects a circle in two points.
G) Tangent - is a line in the plane of a circle that intersects the circle in exactly one point.
H) Tangent Circles - coplanar circles that intersect in one point.
I) Concentric - coplanar circles that have a common center.
J) Common tangents - a line or segment that is tangent to two coplanar circles.
1) Common internal tangent - intersects the segment that joins the centers of the two circles.
2) Common external tangent - does not intersect the segment that joins the centers of the two circles.
K) Interior of a circle - consists of the points that are inside the circle.
L) Exterior of a circle - consists of the points that are outside the circle.
M) Point of tangency - the point at which a tangent line intersects the circle to which it is tangent.
N) Theorems:
1) If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.
Example: Given line AB is tangent to circle C with point A on circle C, AC is a radius of 5 units, AB is 12 units, what is the length of line segment CB?
Since line AB is a tangent line, we know that AC and AB are perpendicular so angle CAB is a right angle so triangle CAB is a right triangle. Therefore we can use pythagorean theorem to find CB.
52 + 122 = (CB)2
25 + 144 = (CB)2
169 = (CB)2
13 = CB
2) In a plane, if a line is perpendicular to a radius of a circle at its endpoint on the circle, then the line is tangent to the circle.
Example: Given radius DE = 11 units on circle D, there is an external point F, line segment DF = 61 and line segment EF = 60, is line EF a tangent line to circle D?
112 + 602 = 121 + 3600 = 3721
612 = 3721
Therefore, since the two smaller sides squared are equal to the longer side squared, this is the converse of pythagorean theorem so we know that angle DEF is a right angle, so therefore line EF is perpendicular to line segment DE so therefore EF is a tangent line to circle D.
3) If two segments from the same exterior point are tangent to a circle, then they are congruent.
Example: Given tangent SR = 2x + 7 and tangent SB = 5x - 8, where both tangents intersect at point S off of circle C with points R and B are on circle C, solve for x.
2x + 7 = 5x - 8
7 = 3x - 8
15 = 3x
5 = x
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