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Transcript
```William Low
Complement and Synthesis
Geometry Formulae
Perimeter and Area of Plane Figures:

Perimeter = Sum of the lengths of all the sides.
bh
 Area of a triangle:
A=
.
2
 Area of a parallelogram: A = b  h .
( B  b)  h
 Area of a trapezoid:
A=
.
2
Dd
 Area of a rhombus:
A=
.
2
Solids:
Solid
Formula for the
Formula for the total
lateral area
area
Formula for the Volume
CUBE
A=4s2
A=6s2
RIGHT PRISM
A = 2 h (l + w)
A=2(hl+hw+lw)
CYLINDER
A=2πrh
A=2πr(h+r)
CONE
A=πrs
A=πr(s+r)
V= s3
or
V = (area of base)(height)
V=l×w×h
or
V = (area of base)(height)
V=πr2h
or
V = (area of base)(height)
1
V= πr2h
3
or
1
V = (area of base)(height)
3
Analytic Geometry Formulas:

Ax1  By1  C
Distance between a point and a line : d =
A2  B 2
or
d=
mx1  y1  b
m2  1
m = slope of the line; b = the y-intercept of the line; ( x1 , y1) = the cords of the point.

Point of division of a line segment:
bx1  ax2
,
ab
nx  mx2
x= 1
,
nm
x=
x = x1 

Distance between two points :
d=
a
 ( x 2  x1 ) ,
ab
by1  ay 2
or
ab
ny  my2
y= 1
or
nm
y=
y = y1 
a
 ( y 2  y1 )
ab
( x 2  x1 ) 2  ( y 2  y1 ) 2
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William Low
Angles:
1. Adjacent angles whose external sides are in a straight line are supplementary.
2. Vertically opposite angles are congruent.
3. If a transversal intersects two parallel lines , then:
i. the alternate interior angles are congruent.
ii. the alternate exterior angles are congruent.
iii. the corresponding angles are congruent.
4. If two corresponding (or alternate interior or alternate exterior) angles are congruent, then they are
formed by two parallel lines and a transversal.
Triangles:
The sum of the measures of the interior angles of a triangle is 180°.
In any triangle, the longest side is opposite the largest angle.
In any isosceles triangle, the angles opposite the congruent sides are congruent.
In any equilateral triangle, each angle measures 60°.
In any isosceles triangle, the perpendicular bisector of the side adjacent to the congruent angles is
the bisector of the angle opposite this side as well as the median and the altitude to this side.
10. In any right triangle, the acute angles are complementary.
11. In any isosceles right triangle, each acute angle measures 45°.
12. In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of
the lengths of the other two sides (Pythagorean Theorem).
13. A triangle is right angled if the square of the length of one of its sides is equal to the sum of the
squares of the lengths of the other two sides.
14. In a right triangle, the length of the side opposite a 30° angle is equal to half the length of the
hypotenuse.
15. Two triangles whose corresponding sides are congruent must be congruent.
16. If two sides and the contained angle of one triangle are congruent to the corresponding two sides
and the contained angle of another triangle, then the triangles must be congruent.
17. If two angles and the contained side of one triangle are congruent to the corresponding two angles
and the contained side of another triangle, then the triangles must be congruent.
18. If two angles of one triangle are congruent to the two corresponding angles of another triangle, the
triangles must be similar.
19. If the lengths of the corresponding sides of two triangles are in proportion, then the triangles must
be similar.
20. If the lengths of two sides of one triangle are proportional to the lengths of the two corresponding
sides of another triangle and the contained angles are congruent, then the triangles must be similar.
21. In a right triangle, the sine of an acute angle is equal to the ratio obtained by dividing the length of
the side opposite this angle by the length of the hypotenuse :
5.
6.
7.
8.
9.
sin A =
a
,
c
where a is the length of the side opposite angle A and c is the length of the
hypotenuse.
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22. In a right triangle, the cosine of an acute angle is equal to the ratio obtained by dividing the length
of the side adjacent to this angle by the length of the hypotenuse :
cos A =
b
,
c
where b is the length of the side adjacent to angle A and c is the length of the
hypotenuse.
23. In a right triangle, the tangent of an acute angle is equal to the ratio obtained by dividing the length
of the side opposite this angle by the length of the side adjacent to it :
tan A =
a
,
b
where a is the length of the side opposite angle A and b is the length of the side
24. The lengths of the sides of any triangle are proportional to the sines of the angles opposite these
sides (Law of Sines).
sin A sin B sin C


a
b
c
25. The square of the length of a side of any triangle is equal to the sum of the squares of the lengths
of the other two side minus twice the product of the lengths of the other two sides multiplied by
the cosine of the contained angle (Law of Cosines) :
a 2  b 2  c 2  2bc cos A
b 2  a 2  c 2  2ac cos B
c 2  a 2  b 2  2ac cos C
26. The opposite angles of a parallelogram are congruent.
27. The opposite sides of a parallelogram are congruent.
28. The diagonals of a parallelogram bisect each other.
29. The diagonals of a rectangle are congruent.
30. The diagonals of a rhombus are perpendicular to each other.
Circles:
31. All the diameters of a circle are congruent.
32. In a circle, the measure of a diameter is equal to twice the measure of the radius.
33. The axes of symmetry of a circle contain its centre.
34. The ratio of the circumference of a circle to its diameter is a constant known as π :
C=πd
or
C=2πr,
where C is the circumference, d is the diameter and r is the radius.
35. The area of a circle is equal to π r 2: A = π r 2 where A is the area and r is the radius.
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Principles Specific to this Course
Isometries and Congruent Figures:
36. An isometry preserves co linearity, parallelism, the order of points, distances , and the measures of
angles. In addition, translations and rotations preserve the orientation of the plane.
37. Any translation will transform a straight line into another line parallel to it.
38. Plane figures or solids are congruent if and only if there is an isometry that maps one figure onto
the other.
39. In congruent plane figures or solids, the measure of the following elements are equal :
a).
the corresponding segments and angles
b).
the perimeters
c).
the areas
d).
the volumes
40. Any point on the perpendicular bisector of a segment is equidistant from the two endpoints of this
segment.
41. Any point on the bisector of an angle is equidistant from the sides of this angle.
42. In any right triangle, the length of the median to the hypotenuse is equal to half the length of the
hypotenuse.
43. The three perpendicular bisectors of the sides of a triangle are concurrent in a point that is
equidistant from the three vertices.
44. The diagonals from one vertex of a convex polygon form n – 2 triangles, where n is the number
of sides in that polygon.
45. The sum of the measures of the interior angles of a polygon is 180° ( n – 2 ), where n is the
number of sides in the polygon.
46. In a convex polygon, the sum of the measures of the exterior angles, one at each vertex, is 360°.
Similarity transformations and Similar Figures:
47. Any similarity transformation preserves collinearity, parallelism, the order of points, the
orientation of the plane, the measures of angles, and the ratio of the distances.
48. Any dilatation will transform a straight line into another line parallel to it.
49. Plane figures or solids are similar if, and only if, there is a similarity transformation that maps one
figure onto the other.
50. In similar plane figures or solids :
a).
the ratio of the lengths of the corresponding segments is equal to the scale factor
b).
the ratio of the measures of the corresponding angles is 1.
c).
the ratio of the areas is equal to the square of the scale factor.
d).
the ratio of the volumes is equal to the cube of the scale factor
51. Plane figures or solids with a scale factor of 1 are congruent.
52. Any straight line that intersects two sides of a triangle and is parallel to the third side forms a
smaller triangle that is similar to the larger triangle.
53. Transversals intersected by parallel lines are divided into segments of proportional lengths.
54. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and
its length is equal to one-half the length of the third side.
55. The three medians of a triangle are concurrent in a point that is two-thirds the distance from each
vertex to the mid-point of the opposite side.
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Unit 1
Graphically Solving a System of Equations of Different Degrees
Objective:
To solve graphically a system of two equations in two variables, one of the first
degree, the other of the second degree; also to determine the number of solutions and
the respective positions of the two curves with respect to each other.
The standard equation for an equation of the first degree:
y=mx+b
To graphically represent an equation of degree 0 or degree 1:
1. Transform the equation to the form y = m x + b.
2. Make a table of values with at least three coordinates.
3. Place the points on a Cartesian graph.
4. Join these points to draw the line.
The standard equation for an equation of the second degree:
y=ax2+bx+c
b 
,
The vertex is given by 
 , where the discriminant Δ = b 2 – 4 a c
2
a
4
a


The intercept on the y–axis is the point (0 , c)
The zeros are x =
b 
 b  b 2  4ac
, or x =
2a
2a
To graphically represent an equation of degree 2:
1. Transform the equation to the form y = a x 2 + b x + c
2. Calculate the coordinates of the vertex, the y-intercept and the zeros and complete a table of
values.
3. Place the points on a Cartesian graph.
4. Join these points to draw the parabola.
Procedure to solve a system of equations:
1. Represent each of the two equations graphically.
2. Determine the points of intersection, if any.
3. Present the solutions of the system as couples (ordered pairs).
Note : The graphical solution of a system of equations with two variables where one is a linear
equation and the other a quadratic equation , three situations are present:
1. If the line is non-vertical secant to the parabola there will be two solutions to the system.
2. If the line is a tangent to the parabola there will be one solution to the system.
3. If the line is disjoint to the parabola the system will have no solutions.
Use of a graphic calculator to find the solutions to a system of equations:
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Unit 2
Algebraic Solutions of Systems of Equations of Different Degrees
Objective:
To solve algebraically , by the method of comparison or the method of substitution , a
system of two equations in two variables. One equation is of the 1 st degree and the
second equation is of the 2nd degree,
Comparison Procedure:
1. Isolate the same variable in the two equations.
2. Compare the two equations obtained.
3. Solve the resulting single variable.
4. Substitute the obtained value in one of the original equations.
5. Verify the solutions in each of the original equations.
6. Present the solutions in the form of a couple (ordered pair).
Substitution Procedure:
1. Isolate one of the variables in one of the equations.
2. Substitute the expression obtained in the other equation of the system.
3. Solve the resulting single variable.
4. Substitute the obtained value in one of the original equations.
5. Verify the solutions in each of the original equations.
6. Present the solutions in the form of a couple (ordered pair).
Unit 3
Solution of Actual Problems Using a System of Equations in Two Variables
Objective:
To translate a practical situation (word problem) into a system of two equations in
two variables of which one equation is of the 1st degree and the second is of the 2nd
degree , and to solve this system of equations.
Procedure:
1. Carefully read the question to ascertain what are the required unknowns of the problem.
2. Identify these unknowns by the variables x & y.
3. Present the problem mathematically as a system of equations.
4. Solve this system of equations graphically , by the method of comparison or by the method of
substitution..
Unit 4
Operations on Polynomial Functions
Objective:
To calculate the sum, the differenc, or the product of two polynomial functions of
different degree and described as equations, graphically, or a table of values.
Adding or subtracting two polynomial functions f (x) and g (x) is like adding or subtracting
polynomials:
(f + g)(x) = f (x) + g (x)
(f – g)(x) = f (x) – g (x)
Multiplying two polynomial functions f (x) and g (x) is like multiplying polynomials:
(f • g)(x) = f (x) • g (x)
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Examples1, 2, 3 and Exercise 4.1 page 4.2
Examples 4 and 5 page 4.5
Exercise 4.2 page 4.13
To carry out an operation on two polynomial functions, from a table of values, it is sufficient to carry out
the operation on the images of x.
To carry out an operation on the graphic representations of two polynomial functions , it is sufficient to
carry out the operation on the images of x values.
Examples 8 and 9 page 4.15
Example 10 page 4.19
Exercise 4.3 page 4.21
Use of a graphic calculator to find the sum, difference, product solutions to a system of equations:
Unit 5
Characteristics of the Graphic Forms
Objective:
To represent graphically the sum, difference, product of two polynomial functions of
degree 2 or less, from their individual graphs or from their equations. Also to
describe certain characteristics of these graphs and to recognize in the primary
functions, certain important points.



The slope of a line is positive if the line rises from left to right.
The slope of a line is negative if the line lowers from left to right.
The slope of a line is zero if the line is horizontal.
To graph the sum, difference, or product of two polynomial functions using their equations:
1. Perform the required operation on the equations.
2. Graph the result of the operation.
or
1. Graph each function.
2. Perform the operation using the graph of the function.
Example 1 page 5.2
Exercise 5.1 page 5.6
To find certain characteristics of the function representing the sum, difference or product we first look at
the initial functions and certain points.
 If we have two constant functions, the sum is also a constant function that is parallel to the other
two.
(Example 2)
 If we have a constant function (horizontal line) and a first degree function (oblique line) of the form
y = m x + b, the sum of these two functions is a line parallel to the oblique line i.e. with the same
slope.
(Example 3)
 If we have a constant function and a parabola (quadratic function), the sum of these functions is a
parabola that has undergone a vertical translation. It opens in the same direction as the initial parabola
and has the same axis of symmetry.
(Example 4)
 If we have two oblique lines with different slopes (m1 ≠ m2), the sum of these lines is also an oblique
line with a slope that is different from the slopes of the initial two lines.
(Example 5)
 If we have two oblique lines with slopes m1 and −m2 respectively, the sum is a horizontal line with
slope m = 0.
(Example 6)
 The sum of an oblique line and a parabola is a parabola that opens in the same direction as the initial
parabola.
(Example 7)
C:/My Documents/My Work/Adult Ed/Math 436 Notes/Complement & Synthesis
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William Low
 The sum of two parabolas where a1 = – a2 and b1 ≠ b2 is an oblique line.
(Example 8)
 The sum of two parabolas where a1 = – a2 and b1 = – b2 is a horizontal line.
(Example 9)
 The sum of two parabolas where a1 ≠ – a2 is a parabola
(Example 10)
If f (x) is a constant function and:
 g (x) is also a constant function, then (f + g)(x) and (f – g)(x) is also a constant function.
 g (x) is a function of the form y = m x + b , then (f + g)(x) will be parallel to g (x) and (f – g)(x) will
be perpendicular to g (x).
 g (x) is a function of the form y = a x 2 + b x + c , then (f + g)(x) will be a parabola translated on the
axis of symmetry with the same orientation as g (x) and (f – g)(x) will be a parabola with the same
axis of symmetry but opposite orientation as g (x).
If f (x) is a function of the form y = m1 x + b1 and:
 g(x) is a constant function, then (f + g)(x) and (f − g)(x) will be lines parallel to f (x).
 g(x) is a function of the form y = m2 x + b2 , where m1 ≠ − m2 , for (f + g)(x) or m1 ≠ m2 for
(f − g)(x), both will produce an oblique line.
 g(x) is a function of the form y = m2 x + b2 , where m1 = − m2 , for (f + g)(x) or m1 = m2 for
(f − g)(x), both will produce a horizontal line.
 g(x) is a function of the form y = a x 2 + b x + c , then (f + g)(x) is represented by a parabola
conserving the orientation of g (x) and (f – g)(x) is represented by a parabola with the inverse
orientation of g (x).
Given f(x), a function of the form y = a1 x2 + b1 x + c1 represented by a parabola.
 If g(x) is a function of the form y = a2 x2 + b2 x + c2 , where a1 = – a2 and b1= – b2 , then (f + g)(x) is a
constant function with a slope of zero.
 If g(x) is a function of the form y = a2 x2 + b2 x + c2 , where a1 = – a2 and b1 ≠ b2 , then (f + g)(x) is
represented by an oblique line.
 If g(x) is a function of the form y = a2 x2 + b2 x + c2 , where a1 ≠ – a2, then (f + g)(x) is represented by a
parabola.
Exercise 5.2 page 5.15
 The difference between two lines with equal slopes is a horizontal line.
(Example 11)
 The difference between a line and a parabola is a parabola that opens in the opposite direction as the
initial parabola because it now has the opposite sign.
(Example 12)
 The difference between two parabolas, where a1 = a2 is an oblique line.
(Example 13)
If the composite function is the result of a product of two functions:
 Two constant functions will give a constant function.
 A constant function and a linear function will give a linear function.
 Two linear functions will give a parabola oriented upwards if both linear functions have slopes in the
same direction.
 Two linear functions will give a parabola oriented downwards if the linear functions have slopes in the
opposite direction.
 Two linear functions will give a parabola with the axis of symmetry being the abscissa of the
intersection of the two functions.
Exercise 5.3 page 5.24
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Given two initial functions, determine the characteristics of the function (f • g)(x)
1.
Find the zeros of the initial two functions.
2.
Determine how the functions behave in the intervals before, and after and at the zero.
3.
Determine whether the function representing the product has a maximum or a minimum. If so,
determine the coordinates of this point.
4.
Determine in which interval the function representing the product is increasing or decreasing.
5.
Determine which type of graph corresponds to the function representing the product.
To determine certain characteristics of the function representing a product:
1.
Find the zeros of the initial two functions.
2.
Determine the sign of the functions by completing a table, if necessary..
3.
Determine whether the function representing the product has a maximum or a minimum. If so,
determine the coordinates of this point.
4.
Determine in which interval the function representing the product is increasing or decreasing.
5.
Determine which type of graph corresponds to the function representing the product.
N.B. The product of two functions is a function of a degree equal to the sum of the degrees of the two
initial functions.
TABLE
x<a
x=a
a<x<b
x=b
x>b
Sign of f(x)
Sign of g(x)
Sign of (f • g)(x)
Exercise 5.4 Page 5.35
Practice Exercises page 5.35
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Unit 6 Distance Between a Point and a Line
Objective: To determine the distance between a point and a line.
Recall equations associated with straight lines :
 the slope of the line and the y-intercept from the standard equation y = m x + b.
y  y1
 the slope of the line in terms of the coordinates of two points on the line m = 2
.
x2  x1
 the equation of the line given the slope and the coordinates of one point on the line m =
y  y1
.
x  x1
1
or
m2 m1 = – 1.
m1
 the length of a line between two points given the coordinates of these points :
 the slope of a line perpendicular to a given line m2  

d  ( x 2  x1 ) 2  y 2  y1 ) 2
 the length of a vertical line
 the length of a horizontal line

if the line is oblique.
d v = | x 2 – x 1| .
d h = | y 2 – y 1| .
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Example 1
Determine the distance between point P(2 , 2) and the line y = – 3x +4. Then find the point of intersection
of these two lines and calculate the distance between the two points.
Procedure to determine the distance between a point and a line:
1. Determine the slope of the given line using the coordinates of two points on the line.
2. Determine the equation of the given line using the slope and the coordinates of one point on the
line.
3. Determine slope of the line segment from the point to the line. This line will be perpendicular to
the given line.
4. Using this slope and the coordinates of the point , determine the equation of the line segment
between the point and the line.
5. Find the point of intersection between the two lines by solving this system of equations.
6. Calculate the length of line segment from the point to the line by using the distance formula..
Alternate procedure to determine the distance between a point and a line:
1.
Write the equation in the form y = m x + b or in the form A x + B y + C = 0.
2.
Determine values of m , b if in the form y = m x + b or the values of A , B , C if in the form
A x + B y + C and the values of the coordinates of the point ( x 1 , y 1).
3.
Place these values in the appropriate formula
d=
 y1  b  ax1
a2 1
or
Ax1  By1  C
A2  B 2
4. Carry out the calculations.
Exercise 6.1 page 6.5
Unit 7
Calculation of the Perimeter and the Area of a Polygon
Objective:
To determine the measures of one or several sides and the equation of one or several
altitudes of a polygon. Also to calculate the perimeter or the area of this polygon.
Procedure:
1. Calculate the coordinates of the vertices of the polygon.
2. Calculate the distance between the two vertices appropriate to the area. Use the distance formula
if the line is oblique or the alternate formulae if the line is vertical or horizontal.
Procedure:
To determine the equation of one or several altitudes , knowing the equations of the lines :
1. Find the slope of the base of the polygon by isolating the variable y (y = m x + b).
2. Determine the slope of the line perpendicular to this base.
3. Calculate the coordinates of an appropriate vertex.
4. Find the equation of the altitude of the polygon.
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Note:
 The perimeter of a polygon is the sum of the measures of the lengths of the sides of the polygon.
 The area of a polygon is the measure which expresses the expanse of a limited surface.
 Area of a square :
A = s2
 Area of a rectangle :
A=b×h
Dd
 Area of a rhombus :
A=
2
 Area of a parallelogram :
A=b×h
B  b   h
 Area of a trapezoid :
A=
2
1
 Area of a triangle :
A = b  h 
2
Procedure:
To calculate the Perimeter of a Polygon :
1. Calculate the coordinates of the vertices, if necessary.
2. Determine the length of each side.
3. Add the lengths of all the sides of the polygon.
Procedure:
To Calculate the Area of a polygon :
1. Determine the appropriate formula.
2. Determine the lengths required in the formula.
3. Perform the calculations.
Unit 8
Geometric Deductions
Objective:
Given the coordinates of the vertices, determine the equation of particular straight
lines (altitudes , medians , perpendicular bisectors etc.) of a triangle or of a
 To find the coordinates of the point that divides a line segment in a particular ratio, given the endpoints
of that line segment, let (x 1 , y 1) be the first endpoint (or starting point) and (x 2 , y 2) be the other
a
 bx  ax2 by1  ay 2 
endpoint (or destination) then apply the formula  1
is the given ratio.
,
 where
b
ba 
 ba
 The mid-point of a line segment is the point that dived the line segment in two equal parts. The
x  x2 y1  y 2
coordinates of the mid-point are M  1
.
,
2
2
 The altitude is a straight line, normally drawn from a vertex, perpendicular to a side of a triangle or of a
quadrilateral that is opposite the vertex.
 The median is a line segment that connects a vertex to the middle of the opposite side.
 The perpendicular bisector of a line segment is a line perpendicular to the line segment at its mid-point.
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Unit 9
Analytic Geometry and Proofs
Objective:
To describe the steps to a solution or to complete missing parts to the steps.

What are the properties of a
parallelogram
rectangle
square
rhombus
trapezoid
isosceles trapezoid
isosceles triangle
equilateral triangle
right triangle
scalene triangle
To solve a problem:
 Determine the important information.
 Formulate the hypotheses.
 Determine the conclusion to be proved.
 Use a line of deductive reasoning in steps.
 Justify each step in the proof using analytical geometry.
Hypothesis: A tentative assumption made in order to draw out and test its logical or empirical
consequences; an assumption or concession made for sake of argument; an interpretation of
a practical situation or condition taken as the ground for action.
Deduction: The deriving of a conclusion by reasoning; inference in which the conclusion about
particulars follows necessarily from general or universal premises;
compare induction – a conclusion reached by logical deduction.
Example:
Show that the four points A (–1 , 4 ) , B ( 7 , 5 ) , C ( 5 , 0 ) , D (–3 , –1 ) are the vertices of
a parallelogram.

What is given? What is the important data? Four points represented by coordinate pairs.

What hypothesis could be formulated? If these points represent the four vertices of a parallelogram,
then lines connecting these vertices should result in the two opposite sides being parallel.

What is needed to be proved? Opposite sides of the quadrilateral are parallel.

How can this proof be obtained? How can the hypothesis be proved?
The points can be represented on a Cartesian graph and the slopes of each line calculated. If the lines
represent the sides of a parallelogram then the slopes of opposite sides should be equal.

What methods could be used to support the applicable reasoning?
This particular hypothesis can be proved using the formula for the slope of a straight line given two
points on the line.
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Example:
Given the three vertices A (–1 , –3 ) , B ( 6 , 1) , and C ( 2 , –5 ) of a triangle, prove that it
is a right triangle.
Given?
Hypotheses?
Conclusion to prove?
How?
Three vertices.
A right triangle has a right angle.
That the given vertices form a right angle.
Plot on a Cartesian plane , find slopes , show perpendicularity
y  y1
1
m2  
m 2
m1
x 2  x1
Support for the reasoning? :
Problems to Prove:
1. Use analytic geometry to prove the following statement :
In any right triangle, the length of the median to the hypotenuse is equal to half the length of the
hypotenuse
Hypothesis:
 In triangle ABC , the angle A is a right angle.
 The line segment AM is the median to the line segment BC.
To Prove:
m AM 
1
m BC
2
Method:
i. Prove that A is a right angle by showing that the line segment BA is perpendicular to the line segment
y  y1
y  y1
AC by finding the slope of each line using the formula m1 = 2
and m2 = 2
for each of the
x2  x1
x2  x1
1
lines respectively and showing that m2  
.
m1
ii. Find the coordinates (x , y) of M using the formula for finding the midpoint
x  x2
y  y2
and y  1
.
x 1
2
2
iii. Prove that AM is the median to BC by showing that BM = MC using the distance formula






d  ( x 2  x1 ) 2  y 2  y1 ) 2 to find the length of each line.
iv. Find the distance between the points A (0 , 0) and M (x , y) using the distance formula
d  ( x 2  x1 ) 2  y 2  y1 ) 2
v. Now find the length of the hypotenuse BC by again using the distance formula
1
d  ( x 2  x1 ) 2  y 2  y1 ) 2 and show that m AM  m BC .
2
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2. Prove the following statement:
In the right triangle ABC, the altitude drawn from the vertex A determines two similar triangles.
Hypothesis:
 In triangle ABC , the angle A is a right angle.
 The line segment AH is the altitude drawn from the vertex A.
To Prove:
Triangle AHC is similar to triangle AHB.
Method:
i. Angle AHC = Angle AHB both being equal to 90° since AH is given as an altitude.
ii. Angle CAH is complementary to angle HAB since their sum is angle CAB, is given as a right angle.
iii. Angle ABH is complementary to angle HAB since angle AHB is a right angle and the sum of the
interior angles of a triangle = 180°.
iv. Angles CAH and ABH are therefore congruent since they are each the complement to the same angle
HAB.
v. Triangle AHC is therefore similar to triangle AHB since the have two corresponding angles that are
congruent.
3. Use analytic geometry to prove the following statement:
The line segment that joins the midpoints of two sides of a triangle,
is parallel to the third side.
Hypothesis:
 PQR is a triangle.
 M is the midpoint of side PQ.
 N is the midpoint of side QR.
To Prove:
The line segment MN is parallel to side PR.
Method:
i. Find the coordinates (x , y) of M using the formula for finding the midpoint
x  x2
y  y2
and y  1
.
x 1
2
2
ii. Find the coordinates (x , y) of N using the formula for finding the midpoint
x  x2
y  y2
and y  1
.
x 1
2
2
y  y1
iii. Find the slope of the line segment MN using the slope formula m = 2
.
x2  x1
y  y1
iv. Find the slope of the line segment PR again using the slope formula m = 2
.
x2  x1
v. Show that the slope of MN = the slope of PR, thus proving that the line segment MN is parallel to side
PR.
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4. Prove the following statement:
Any point on the bisector of an interior angle of a triangle is equidistant from the sides of that
triangle.
Hypothesis:


Ray AP is the bisector of the angle BAC.
Point P is a point on the bisector AP.
Construction:
The perpendicular distances are drawn from point P on the bisector to rays AB
and AC.
To Prove
:
Point P is equidistant from the sides of the angle BAC , i.e. mBP  mCP.
Method:
i. Angle BAP = angle CAP since AP is given as a bisector of angle BAC.
ii. Angle ABP = angle ACP = 90° since the line segments BP and CP were constructed as perpendicular
distances.
iii. Angle BPA = angle CPA since we have shown two angles of triangle ABP = the corresponding two
angles of triangle ACP therefore the third angle in each triangle must be equal.
iv. Line segment AP = line segment AP since it is the common side of both triangles ABP and ACP.
v. Triangles ABP and ACP are congruent since two angles and the contained side of one triangle are
congruent to the corresponding two angles and the contained side of the other triangle. This is a
sufficient condition for congruency.
vi. Thus we have proves that mBP  mCP.
3. A quadrilateral is formed by joining the following points:
A (3, 5), B (5, – 5), C (2, – 10), D (– 8, – 6)
Using the formulas and principles related to the slopes and Measures of line segments, show that this
4. A quadrilateral is formed by joining the following points:
A (– 1, 2) , B (3, 3) , C (6, – 0.5) , D (0, – 2)
Using the formulas and principles related to the slopes and measures of line segments, show that this
Unit 10
Characteristics of Isometries and Similar Figures
An isometric transformation is a geometrical transformation that conserves congruency.
Translation, rotation, and reflection are Isometries.
A translation is a displacement of an object in a constant direction in a plane.
A rotation is a displacement in a figures position determined by an angle of rotation in a given direction
about a point called the centre of rotation.
A reflection is a mirror image about a line called the axis of reflection.
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Isometric Properties
Any isometry:
1. Transforms one line into another line.
2. Conserves the length of line segments.
3. Conserves the angles of the geometric figure.
4. Conserves the order of the points.
5. Conserves the lines of one figure parallel to another.
6. Conserves the lines of one figure perpendicular to another.
7. Conserves the ratio of lengths.
These properties are said to be invariants of the isometry
A dilatation is a geometric transformation which enlarges or reduces a figure while preserving the
measures of its angles and keeping the ratio of the lengths of its sides constant.
Properties of a dilatation:
Any dilatation preserves:
1.
The angle measures.
2.
The order of points.
3.
The parallelism.
4.
The ratio of lengths.
These properties are said to be the invariants of the dilatation.
A transformation of similitude is a geometric transformation that conserves similarity for all sides for all
sides while preserving the measures of its angles.
Isometric figures have equivalent angles, equivalent corresponding sides and keep the same form.
Similar figures have their angles equivalent, their corresponding sides proportional and keep the same
form.
Congruent figures have congruent corresponding angles and sides, and have the same shape. Similar
figures have congruent corresponding angles, proportional corresponding sides and the same shape.
Characteristics of Isometric Figures:
 All corresponding segments are congruent.
 All corresponding angles are congruent.
 The parameters are equal.
 The areas are equal.
 The volumes are equal.
Congruent triangles have three fundamental properties:
1. The measure of an angle and the lengths of the two sides which form this angle are all the
information needed to draw one and only one triangle.
S-A-S Property: Two triangles are congruent if two sides and the contained angle of one
triangle are congruent to the corresponding two sides and contained angle of the other
triangle. The notation S-A-S means an angle between two sides.
2. The measures of two angles and the length of the side contained between these two angles are
all the information needed two draw one and only one triangle.
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A-S-A Property: Two triangles are congruent if two angles and the included side in one
triangle are congruent to the corresponding two angles and included side of the other
triangle. The notation A-S-A means a side between two angles.
3. The lengths of the three sides of a triangle are all the information needed two draw exactly one
triangle.
S-S-S Property: Two triangles are congruent if the three sides of one triangle are congruent to
the corresponding three sides of the other. The notation S-S-S means three congruent sides.
Congruent Triangles: Two triangles are congruent if they are the exact same size and shape i.e. if they
are identical (SAS, ASA, SSS).
The bisector of an angle is the half-line that divides the angle into two congruent angles.
A median is the line segment that joins one of the vertices of a triangle to the midpoint of the opposite
side.
Properties of congruent figures:

All corresponding segments are congruent.

All corresponding angles are congruent.

The perimeters are equal.

The areas are equal.
Similarity of Triangles:
Two triangles are similar if there exists a similarity between them. In two similar triangles,
corresponding angles are congruent and corresponding sides are proportional.
A similarity is a composition of transformations that changes a given figure into another similar figure.
Indication of similarity: ∆ A B C − ∆ A′ B′ C′
Similar triangles have three similar properties:
1. If the measures of two angles of two triangles are known and the corresponding angles are
congruent , it can be concluded that the two triangles are similar.
A-A Property: Two Triangles are similar if they have two congruent corresponding angles. The
notation A-A means two congruent angles.
2. If the lengths of the three sides of two triangles are known, and if the lengths of the
corresponding sides are proportional, it can be concluded that these triangles are similar.
S-S-S Property: Two triangles are similar if the three corresponding sides are proportional.
The notation S-S-S means three proportional sides.
3. In two triangles, if two pairs of corresponding sides are proportional and if the contained
angles are congruent, the triangles are similar.
S-A-S Property: Two triangles are similar if they have a congruent angle contained between
two corresponding proportional sides. The notation S-A-S means a congruent angle between
two corresponding proportional sides.
Similar Triangles: Triangles are similar if their corresponding angles are equal (if two pairs of angles are
equal [AA], if all three sides are proportional, if there are two proportional sides and the angle in between
them is the same. Corresponding sides of similar triangles are proportional.
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Properties of similar figures:

All corresponding angles are congruent.

All corresponding sides are proportional.

All corresponding segments are proportional.

The ratio of the perimeters is equal to the ratio of the lengths of the corresponding sides.

The ratio of the areas is equal to the square of the ratio of the lengths of the corresponding sides.
Properties of congruent solids:

All corresponding segments are congruent.

All the areas are equal.

All the volumes are equal.

The ratio of the lengths of the corresponding segment is 1.
Properties of similar solids:

All corresponding segments are proportional.

The ratio of the areas is equal to the square of the ratio of the lengths of the sides.

The ratio of the volumes is equal to the cube of the ratio of the lengths of the sides.
length of side of l arg er structure
length of side of smaller structure
area of l arg er structure
area of smaller structure
k
 k2
volume of l arg er structure
volume of smaller structure
 k3
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Unit 11
Equivalent Figures and Solids
Objective:
To recognize equivalent plane figures as figures with the same area. Also to recognize
equivalent solids as solids with the same volume.
Two figures are equivalent if they have the same area.
bh
 Area of a triangle:
A=
.
2
 Area of a rectangle:
A=b×h

Area of a parallelogram:
A =bh .

Area of a square:
A=s2

Area of a trapezoid:
A=

Area of a rhombus:

Area of a circle:
( B  b)  h
.
2
Dd
A=
.
2
A =  r2
Equivalent solids have the same volume.

Volume of a cube:

Volume of a rectangular prism: V = l × w × h

Volume of a cylinder:
V =  r2 × h

Volume of a cone:
V=
V=s3
 r2 h
3
Unit 12
Deducing Unknown Measurements or Ratios
Objective:
To calculate the unknown measurements or ratios in congruent , similar or equivalent
figures and justify the steps in your calculation.
The corresponding sides of similar figures are proportional.
The corresponding segments of congruent figures are congruent.
The areas of equivalent figures are equal.
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Unit 13
Proofs Involving Plane Figures
Objective: To state the minimum conditions required for two triangles to be congruent or similar.
Also to complete a proof that requires the application of the concepts of congruence and
similarity and in which certain steps or justifications are missing.
Congruent triangles have three fundamental properties:
1. S-A-S Property: Two triangles are congruent if two sides and the contained angle of one triangle
are congruent to the corresponding two sides and contained angle of the other triangle. The
notation S-A-S means an angle between two sides.
2. A-S-A Property: Two triangles are congruent if two angles and the included side in one triangle
are congruent to the corresponding two angles and included side of the other triangle. The
notation A-S-A means a side between two angles.
3. S-S-S Property: Two triangles are congruent if the three sides of one triangle are congruent to the
corresponding three sides of the other. The notation S-S-S means three congruent sides.
Congruent Triangles: Two triangles are congruent if they are the exact same size and shape i.e. if they
are identical (SAS, ASA, SSS).
Similar triangles have three similar properties:
1. If the measures of two angles of two triangles are known and the corresponding angles are
congruent , it can be concluded that the two triangles are similar.
A-A Property: Two Triangles are similar if they have two congruent corresponding angles. The
notation A-A means two congruent angles.
2. If the lengths of the three sides of two triangles are known, and if the lengths of the
corresponding sides are proportional, it can be concluded that these triangles are similar.
S-S-S Property: Two triangles are similar if the three corresponding sides are proportional.
The notation S-S-S means three proportional sides.
3. In two triangles, if two pairs of corresponding sides are proportional and if the contained
angles are congruent, the triangles are similar.
S-A-S Property: Two triangles are similar if they have a congruent angle contained between
two corresponding proportional sides. The notation S-A-S means a congruent angle between
two corresponding proportional sides.
Similar Triangles: Triangles are similar if their corresponding angles are equal (if two pairs of angles are
equal [AA], if all three sides are proportional, if there are two proportional sides and the angle in between
them is the same. Corresponding sides of similar triangles are proportional.
Two angles are vertically opposite if:
1.
they have the same vertex;
2.
the sides of one triangle are extensions of the sides of the other triangle.
Vertically opposite angles are always equal measure:
Two angles are alternate-exterior angles if they are situated:
1.
on either side of a transversal;
2.
outside the other two lines without being adjacent.
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Alternate-exterior angles formed by a transversal and two parallel lines are always equal in measure and
therefore congruent.
Two angles are alternate-interior angles if they are situated:
1.
on either side of a transversal;
2.
inside the other two lines without being adjacent.
Alternate-interior angles formed by a transversal and two parallel lines are always equal in measure and
therefore congruent.
Two angles are corresponding:
1.
if they are situated on the same side as a transversal;
2.
3.
if one is situated inside the other two lines and the other is situated outside the other two lines.
Corresponding angles formed by a transversal and two parallel lines are always equal in measure and
therefore congruent.
Unit 14
Solving Geometry Problems
Objective:
To solve problems using the concepts of congruence, similarity or equivalence, by
applying a rigorous approach that involves deductive reasoning.
Ina right triangle, the length of the side opposite the 30° angle is equal to half the length of the
hypotenuse.
In a right angled triangle:
sin A =
side opposite to angle A a

hypotenuse
c
cos A =
side adjacent to angle A b

hypotenuse
c
tan A = side opposite to angel A = a
side adjacent to angle A b
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