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Mathematics 20
Module 3
Lesson 20 Review of Lessons 14-19
Mathematics 20
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Lesson 20
Mathematics 20
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Lesson 20
Review of Lessons 14-19
Introduction
This lesson will be a review of the concepts that you have learned in Lessons 14-19.
Lessons 14 and 15 were the consumer math section where you have seen the important
link of mathematics to facing the financial challenges of everyday life. In lessons 16 to 19
you have used your mathematical skills to explore the world of geometry.
Each section will be set up so that it is a review of a lesson. A number of practice
exercises will be provided with each review. Work through the practice exercises so that
you feel comfortable with the concepts in that particular lesson. If you find that you are
having difficulties, go back to the actual lesson and review the material there.
It is so important to understand the material and be able to apply it to a number of
different situations.
The assignment for this lesson will be a review of the entire course. It will be very similar
to the final examination that you will receive at the conclusion of this course. Every
aspect of the course will be covered. Once again, if there is some area that you are having
problems with, go back to those sections and review the material.
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Lesson 20
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Lesson 20
Objectives
After completing this lesson you will be able to
•
solve problems involving the mathematical concepts in Lessons 14 to 19.
•
apply the mathematical concepts from Lessons 14 to 19 to solve
real-world problems.
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Lesson 20
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Lesson 20
20.1 Review of Lesson 14
Some important concepts to remember in this section are:
•
there are many advantages and disadvantages of credit. Be informed!
•
the Simple Interest Formula is:
I = prt
where:
•
•
•
•
I is the interest.
p is the principle amount.
r is the interest rate.
t is the time in number of days.
•
Interest on purchases is charged only if the balance owing is not paid in full by the
due date. If the balance is not paid in full, interest will be charged on the unpaid
balance and on any purchases that were made during the month, up to and
including the date of the billing statement.
•
the Monthly Payment Formula is:
Monthly Payment Formula



r

12
M=p 
1
1 
12 t

r 


1 +

12 


Where:
•
•
•
•
Mathematics 20









M is the amount of the monthly payment.
p is the principal or the amount of the loan.
r is the annual interest rate.
9
= 0 .09 = r
Example: 9 % =
100
 r

is the monthly interest rate 

 12

t is the term in years.
12t is the term in months 
335
Lesson 20
Exercise 20.1
1.
Calculate the cost of credit of each of these items.
Item
Cash Price
Monthly Payment
Number of Months
a) Mountain Bike
$425.90
$51.80
10
b) CD Player
$215.33
$28.48
8
c) Golf Clubs
$639.85
$66.43
12
d) Cordless Telephone
$148.50
$40.00
4
e) Gold Bracelet
$234.99
$42.99
6
f)
2.
Which item has the lowest percentage increase in the final cost?
Determine the cost of credit for each option and decide which option is the best.
Cash Price
Option A
Option B
a) $1490.55
$260.00 per month
for 6 months
$156.50 per month
for 12 months
b) $883.67
$101.00 per month
for 9 months
$300.00 per month
for 3 months
c) $540.99
$70.59 per month
for 10 months
$100.00 per month
for 6 months
d) $5189.46
$250.00 per month
$200.00 per month
for 2 years
for 3 years
e)
Mathematics 20
What can you say, in general, about the options for
paying back a loan?
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Lesson 20
3.
4.
Calculate the amount of interest that has accrued by the statement date which is
the first of the month, and determine the balance that will be carried forward to the
next statement.
Balance on Last
Statement
Payment
New
Purchases
Cash
Advances
Daily Interest
Rate
Due
Date
a) $130.80
$25.00
(June 3)
$61.99
(June 13)
-
0.05%
July 1
b) $644.07
$125.00
(Dec. 23)
$35.82
(Dec. 7)
$100.93
(Dec. 22)
-
0.055%
Jan. 1
c) $431.00
$100.00
(Jan. 24)
$452.01
(Jan.10)
$50.00
(Jan.19)
0.04%
Feb. 1
d) $98.62
$60.00
(Mar. 28)
$29.99
(Mar. 6)
$54.00
(Mar. 14)
$60.00
(Mar. 11)
$40.00
(Mar. 20)
0.045%
Apr. 1
Vern wants to buy a boat that costs $7120.00. The interest rate at the bank is
1
10 %. Determine the monthly payments if Vern borrows the money over 4 years.
2
How much would Vern save if he borrowed the money over 2 years?
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Lesson 20
5.
Cindy and Lin are buying a new home. The amount of the mortgage will be
$81 500 and will be paid over 20 years at an interest rate of 7%. What will their
monthly payments be?
6.
Sajamin borrows $6350.00 to buy a used car. The interest rate is 12% and Sajamin
will pay for the car over 4 years. What are monthly payments?
7.
Paul's monthly payments on a new stereo system are $130.00 each month. When
Paul bought the system, the interest rate was 9% and the length of the term was 3
years. What was the original cost of the stereo system?
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Lesson 20
20.2 Review of Lesson 15
``
Some important concepts to remember in this section are:
•
the final cost of a taxable item is:
Final Cost = Original Cost + 6% PST x Original Cost + 5% GST x Original Cost
•
The formula for mill rate is:
Mill Rate =
•
Budgeted Expenses
Budgeted Expenses
=
 1000
Assessment
Assessment
1000
The formula for property tax is:
Assessment
Property Tax = Mill Rate 
1000
Exercise 20.2
1.
Fill in the following table.
Item
Cost
Coat
$76.22
Hockey Skates
GST
PST
Final Cost
$14.31
Snowmobile
$671.16
Calculator
$63.50
2.
The town of Cupar has a total budget of $187 971 assessed value of $2 725 220.
What is its municipal mill rate?
3.
In Cupar, the Howards own a home that is worth $8 620 and the property is worth
$1 750. What are their municipal property taxes?
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Lesson 20
4.
The Howards taxes are due June 30. The discount that the Howards receive for
paying their taxes is listed in the table.
a)
b)
March 30
April 30
May 30
June 30
5%
4%
2%
0%
What are their municipal taxes if they pay by March?
What are their municipal taxes if they pay by May?
5.
Sid and Mary Lou own a farm in the R.M. of Sherwood. Their taxable assessment is
$12 860. The municipal tax is 24.00 mills. The School tax is 78.8 mills. How much
will Sid and Mary Lou pay in taxes? Separate the amount for municipal and school
taxes and then find the total.
6.
Sid and Mary Lou’s taxes are due on Dec. 30, but they plan to pay their taxes by
September 30th. According to the table for discounts in question #4, how much
would they pay? What would be the amount of money that they would save?
(March 30 is the same as September 30)
7.
A penalty of 0.75 % per month will be added to taxes that are not paid on time.
How much would Sid and Mary Lou pay if they didn't pay until May of the following
year?
8.
Calculate the property taxes. In each of the following cases the mill rate is given for
that particular municipality. Take into consideration the date the taxes would be
paid and apply the discount or penalty according to the table in question 4. Taxes
are due June 30 with a penalty of 0.75% per month late fee.
Assessed
Land
Value
Mill
Rates
Building
Municipal
Tax
Library
Rate
Public
School
a) $6 000
$7 670
41.05
---
69.46
February
b) $13 000
$24 716
52.86
8.24
72.33
July
c) $9 576
$10 265
53.10
---
74.55
May
d) $12 837
$31 930
48.62
8.89
79.02
September
e) $4 126
$8 980
49.00
---
Mathematics 20
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73.99
Separate
School
Payment
Date
March
Lesson 20
20.3 Review of Lesson 16
Some important concepts to remember in this section are:
•
the word congruent means to have the same measure.
•
congruent segments are segments which have the same measure.
•
congruent angles are angles which have the same measure.
•
two triangles are congruent if the corresponding angles and sides of the two
triangles are congruent.
•
the Congruence Postulates are used to determine if two triangles are congruent.
•
Side-Side-Side (SSS) Congruence Postulate
If three sides of one triangle are congruent, respectively, to three sides of a second
triangle, then the two triangles are congruent.
•
Side-Angle-Side (SAS) Congruence Postulate
If two sides and the included angle of one triangle are congruent, respectively, to
two sides and the included angle of a second triangle, then the two triangles are
congruent.
•
Angle-Side-Angle (ASA) Congruence Postulate
If two angles and the included side of one triangle are congruent, respectively, to
two angles and the included side of a second triangle, then the two triangles are
congruent.
•
Angle-Angle-Side (AAS) Congruence Postulate
If two angles and the non-included side of one triangle are congruent, respectively,
to two angles and the non-included side of a second triangle, then the two triangles
are congruent.
•
Hypotenuse-Leg (HL) Theorem
If the hypotenuse and a leg of a right triangle are congruent, respectively, to the
hypotenuse and a leg of a second right triangle, then the two right triangles are
congruent.
•
triangles can be constructed through both informal and formal methods. The
formal method of construction is done with a straight edge and compass.
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Lesson 20
Exercise 20.3
1.
2.
For each of the pairs of triangles, like markings indicate the congruent parts.
Name the congruence postulate, if any, that will prove the triangles congruent.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
List the corresponding parts of the two triangles if FUN  DAY .
Mathematics 20
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Lesson 20
RST with SL  TK and RST  RTS .
RSL  RTK
3.
Given:
Prove:
4.
Given the following three segments, construct FGH . What congruence postulate
determines that you can do this?
5.
Given the following two angles and segment construct WXY . What congruence
postulate determines that you can do this?
20.4 Review of Lesson 17
Some important concepts to remember in this section are:
•
there are many different reasons to use when proving two segments or angles are
congruent and these are listed in the appendix.
•
when proving two triangles are congruent, always mark the congruent parts on the
two triangles first. Then determine which congruence postulate you are going to
use.
•
the last statement in the two-column proof should be the same as what the question
asked you to prove.
•
once you have proven that two triangles are congruent, you can also say that the
corresponding parts are also congruent.
•
when constructing a right triangle it is necessary to construct a perpendicular line
from one of the legs.
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Lesson 20
Exercise 20.4
1.
Mark the congruent parts on the triangles from the information that is given.
a.
Given:
Isosceles JKL with
RJ bisecting KJL
b.
Given:
Figure ABCD with
AB  AD , BC  DC , AB  CD
c.
Given:
Figure with RS TU ,
V is the midpoint of ST .
d.
Given:
ABC with AR  BS ,
RSC is isosceles.
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Lesson 20
e.
Given:
X
XYZ with altitudes
YN to XZ ; ZM to XY
and XM  XN .
M
1
2
Y
f.
2.
Given:
3
4
N
Z
AM is the  bisector to BC .
Using the information given in #1, prove the following.
a. Prove:
JKR  JLR
b. Prove:
ADB  CBD
c. Prove:
RSV  UTV
d. Prove:
ACR  BCS
e. Prove:
YN  ZM
f. Prove:
ABC is isosceles
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Lesson 20
3.
Construct a right triangle with leg lengths, AB and CD.
20.5 Review of Lesson 18
Some important concepts to remember in this section are:
•
two polygons are similar if their vertices can be paired so that:
•
corresponding angles are congruent.
•
corresponding sides are in proportion (lengths have the same ratio).
•
the scale factor is the ratio of the lengths of any two sides of the similar polygon.
•
if two solids are similar with a scale factor of a : b , then corresponding areas have a
ratio of a 2 : b2 , and corresponding volumes have a ratio of a 3 : b3 .
Exercise 20.5
1.
Hexagon ABCDEF is similar to hexagon UVWXYZ. List the corresponding sides
and vertices.
Mathematics 20
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Lesson 20
2.
In each of the following pairs of polygons determine the lengths of the missing sides
or the measures of the missing angles.
a.
b.
c.
d.
20
Mathematics 20
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Lesson 20
3.
Find the length of the missing side in the following similar right triangles.
a.
b.
c.
d.
Mathematics 20
348
Lesson 20
4.
5.
Determine the proportion of the area and volume for the following similar solid
polygons.
a)
The length of the radius on cylinder A is 4 cm and on cylinder B is 8 cm.
b)
The width of box A is 9 cm and of box B is 12 cm.
c)
The height of pyramid B is 13.2 cm and pyramid A is 4.4 cm.
From the ratios in question 4, determine the value of the
i) area
ii) volume of the solid B.
a)
Cylinder A • area
• volume
-
65.36 cm2
150.72 cm3
b)
Box A
• area
• volume
-
102 cm2
54 cm3
c)
Pyramid A
• area of base • volume
-
Mathematics 20
62 m2
294.36 m3
349
Lesson 20
20.6 Review of Lesson 19
Some important concepts to remember in this section are:
•
it is necessary to understand the vocabulary of circles. This was discussed in
Section 19.1.
•
a central angle of a circle is an angle in the plane of the circle with its vertex
coinciding with the centre of the circle.
•
a central angle forms both a major arc and a minor arc with the circle.
•
a circle can be both inscribed within a triangle and circumscribed around the
triangle through the method of a formal construction.
•
a tangent can be constructed to a circle from a point outside the circle.
•
a number of theorems have been developed about circles.
Exercise 20.6
1.
Given circle with centre O and tangents AP and BP .
a.
If AP = 8 and OP = 10, what is the radius of the circle?
b.
If AP = 8 and OP = 10, then BP = _______________.
c.
If the measure of angle APB is 56°, how large is angle AOB?
Mathematics 20
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Lesson 20
2.
3.
For the circle with centre O and CO perpendicular to AB, if the
, then:
, and
a.
m AC =
i.
m OCE
=
b.
m CB =
j.
m CEO
=
c.
m BD =
k.
m CFO
=
d.
m AD =
l.
m BFE
=
e.
m AE =
m.
m OFE
=
f.
m COA
=
n.
m AOD
=
g.
m BOE
=
o.
m OAD
=
h.
m COE
=
In the circle with centre O chords AB and CD intersect at P. (See Assignment 19)
a.
If CP = 4, PD = 6, and PB = 3, then AP = __________.
b.
If BP = 4, AP = 15, and CP = 6, then PD = __________.
Mathematics 20
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Lesson 20
4.
Chords AB and CD intersect at P. If CP = 4, PD = 10, PB = x, and AP = x + 3, find
the value of x.
Mathematics 20
352
Lesson 20
Answers to Exercises
Exercise 20.1
1.
a.
b.
c.
d.
e.
f.
$518.00
$227.84
$797.16
$160.00
$257.94
Golf clubs
2.
a.
Option A
Option B
$1 560.00
$1 878.00
b.
Option A
Option B
$909.00
$900.00
Best
c.
Option A
Option B
$705.90
$600.00
Best
d.
Option A
Option B
$6 000.00
$7 200.00
e.
Generally, less money will have to be paid back if
the loan is paid over a shorter period of time.
3.
a.
b.
c.
d.
$169.99
$667.30
$842.48
$225.52
4.
$182.30/month for 4 years
$330.20/month for 2 years.
Best
Best
Savings of $825.60 if paid in 2 years instead of 4 years.
Mathematics 20
5.
$631.88
6.
$167.22
7.
$4 088.09
353
Lesson 20
Exercise 20.2
1.
Item
Cost
GST
PST
Final Cost
Coat
$76.22
$4.57
$3.81
$84.60
Hockey
Skates
$238.50
$14.31
$4.93
$264.74
Snowmobile
$13 423.20
$805.39
$671.16
$14 899.75
Calculator
$57.21
$3.43
$2.86
$63.50
Exercise 20.3
Mathematics 20
2.
69 mills
3.
$715.53
4.
a.
b.
5.
School Tax
- $1 013.37
Municipal Tax - $ 308.64
Total
- $1 322.01
6.
$1 255.91
$66.10
7.
$1 371.59
8.
a.
b.
c.
d.
e.
$1 589.98
$5 070.19
$2 482.05
$6 249.56
$1 531.31
1.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
ASA
SSS
none
none
ASA
SAS
ASA
none
none
none
SSS
none
679.75
701.22
354
Lesson 20
2.
3.
Exercise 20.4
FU  DA
FN  DY
UN  AY
F  D
U  A
N  Y
1.
2.
3.
Statement
SL  TK
RSL  RTK
R  R
4.
RSL  RTK
4.
Refer to 16.4 in lesson 16.
SSS Postulate
5.
Refer to 16.4 in lesson 16
ASA Postulate
1.
a.
Reason
1.
Given
2.
Given
3.
Congruence is an
equivalence relation on
the set of triangles
(reflexivity)
4.
AAS Postulate
b.
Mathematics 20
355
Lesson 20
c.
d.
e.
f.
Mathematics 20
356
Lesson 20
2.
a.
1.
2.
Statement
JKL is isosceles
JK  JL
3.
4.
5.
6.
RJ bisects KJL
KJR  LJR
JR  JR
JKR  JLR
Reason
1.
Given
2.
Definition of isosceles
triangle
3.
Given
4.
Definition of bisect
5.
Reflexive property
6.
SAS Postulate
3.
4.
5.
Statement
AB  AD , BC  DC
BAD and DCB are
right angles
AB  CD
BD  DB
ADB  CBD
Reason
1.
Given
2.
Perpendicular lines meet
to form right angles
3.
Given
4.
Reflexive property
5.
HL Theorem
1.
2.
Statement
RS TU
S  T
3.
V is the midpoint of
Reason
1.
Given
2.
If two lines are parallel,
the alternate interior
angles are congruent
3.
Given
4.
5.
ST
SV  TV
SVR  TVU
4.
5.
6.
RSV  UTV
6.
b.
1.
2.
c.
Mathematics 20
357
Definition of midpoint
Vertical angles are
congruent.
ASA Postulate
Lesson 20
d.
Statement
AR  BS
RSC is isosceles
CR  CS
2  3
Reason
1.
Given
2.
Given
3.
4.
Angles opposite
5.
Angles that form a linear
pair are supplementary
6.
1 & 2 are
supplementary
3 & 4 are
supplementary
1  4
6.
7.
ACR  BCS
7.
Supplements of
congruent angles are
congruent
SAS Postulate
Statement
Altitudes YN to XZ
and ZM to XY
YN  XZ ; ZM  YX
XMZ & XNY are
right angles
XM  XN
X  X
XMZ  XNY
YN  ZM
Reason
1.
Given
Reason
1.
Given
4.
Statement
AM is the  bisector
to BC
AM  BC
AMB and AMC
are right angles
AMB  AMC
5.
6.
7.
8.
9.
BM  CM
AM  AM
AMB  AMC
AB  AC
ABC is isosceles
5.
6.
7.
8.
9.
1.
2.
3.
4.
5.
e.
1.
2.
3.
4.
5.
6.
7.
f.
1.
2.
3.
Mathematics 20
358
2.
3.
4.
5.
6.
7.
2.
3.
4.
Definition of altitude
Perpendicular lines meet
to form right angles
Given
Reflexive property
ASA Postulate
CPCTC
Definition of  bisector
Perpendicular lines meet
to form right angles
Any two right angles are
congruent.
Definition of  bisector
Reflexive property
SAS Postulate
CPCTC
Definition of isosceles
Lesson 20
Exercise 20.5
3.
Hint: Draw the perpendicular bisector to form a right
angle.
1.
A ~ U
B ~ V
X ~ W
D ~ X
E ~ Y
F ~ Z
2.
a.
E = 58°
F = 60°
D = 62°
A = 62°
c = 7.5
d=6
b.
x = 11.25
c.
P  88 
Q  140 
N  123 
w = 6.6
x = 11
y = 13.6
z = 14.6
d.
y = 9.6
x = 7.2
a.
b.
c.
d.
x = 46.4 m
x = 10.1 m
x = 14.4 m
x = 70.0 m
3.
Mathematics 20
AB ~ UV
BC ~ VW
CD ~ WX
DE ~ XY
EF ~ YZ
FA ~ ZU
359
Lesson 20
4.
a.
Area =
16 1

64 4
Volume =
1
8
b.
Area =
9
16
Volume =
27
64
c.
Area =
1
9
Volume =
1
27
5.
Exercise 20.6
Mathematics 20
a.
i.
ii.
Area B = 261.4 cm2
Volume B = 1205.76 cm3
b.
i.
ii.
Area B = 181.3 cm2
Volume B = 128 cm3
c.
i.
ii.
Area B = 558 m2
Volume B = 7947.72 m3
1.
a.
b.
c.
6
8
124° (Hint: PAO  PBO by the SSS Postulate)
2.
a.
b.
c.
d.
e.
f.
g.
h.
90°
90°
50°
130°
150°
90°
30°
120°
3.
a.
b.
AP = 8
PD = 10
4.
x=5
i.
j.
k.
l.
m.
n.
o.
360
30°
30°
60°
30°
120°
130°
25°
Lesson 20
Mathematics 20
Module 3
Assignment 20
Mathematics 20
361
Lesson 20
Mathematics 20
362
Lesson 20
Optional insert: Assignment #20 frontal sheet here.
Mathematics 20
363
Lesson 20
Mathematics 20
364
Lesson 20
Assignment 20
Values
(40)
A.
Multiple Choice: Select the best answer for each of the following and place
a check () beside it.
1.

The value of 7  5  14  2 4
____
____
____
____
2.
3.
Mathematics 20
a.
b.
c.
d.

3
is ***.
 16
47
36
 83
If x  25 x  3 y   3 x  2 the simplified equation with y isolated is ***.
____
a.
____
____
b.
c.
____
d.
3  5x
3
y  1  5x
y  1  2x
1 6x
y
3
y
The graph of y  x  2 is ***.
____
a.
____
c.
____
b.
____
d.
365
Lesson 20
4.
The simplified form of
____
____
____
____
5.
6.
____
a.
____
b.
____
c.
____
d.
3
2
x> 1
x>
3
3
, x>
2
2
x <  1, x > 1
x< 
The equivalent form of
____
a.
____
b.
____
c.
d.
2
is ***.
3
6
3
2
3
6
3
2
3
The expression  x  2   9 is equivalent to ***.
2
____
____
____
____
Mathematics 20
62
6 2 2
104  42
30  32
The solution to x  3 x  1  5 is ***.
____
7.
a.
b.
c.
d.
54  50  24  18 is ***.
a.
b.
c.
d.
x  3 x  3 
x  11 x  7 
x 2  13
x  5 x  1
366
Lesson 20
8.
When x 2  4 x  5 is divided by x + 1, the quotient is ***.
____
____
____
____
9.
10.
11.
Mathematics 20
a.
b.
c.
d.
x 1
x  5 x  1
x 5
0
The equivalent simplifed form of
____
a.
____
____
____
b.
c.
d.
a  b0 4 a 4
2 a 2 b2 a  bb 2
1
ab
1
2
ab
The equivalent form of a  3   a  1 
1
____
a.
____
b.
____
c.
____
d.
If f  x  
____
a.
____
b.
____
____
c.
d.
is ***.
1
is ***.
4
a  2a  3
2
2
a  2a  3
 a  3  a 1
2a  4
2
a  2a  3
2
2x  1  5
, then f(2) is ***.
 x  2 ( x  3)
7
7
5
0
not defined
367
Lesson 20
12.
The interest for one month on a $800 amount at an interest rate of 8%
is ***.
____
____
____
____
13.
a.
b.
c.
d.
A  D
AB = DE
AC  DF
AB  DE
a.
b.
c
d.
ABC  FED
AB  EF
BAC  EDF
B   D
From the diagram it is possible to
conclude that ***.
____
____
____
____
Mathematics 20
$251.00
$3 514.00
$140.00
$2 430.00
If ABC  DEF , then it is also true that ***.
____
____
____
____
16.
a.
b.
c.
d.
If ABC  DEF , then because corresponding parts of congruent
triangles are congruent, the immediate conclusion is ***.
____
____
____
____
15.
$64.00
$8.00
$5.33
$5.21
The mill rate in a town is 251 mills. A home and lot assessed at
$14 000 would have a property tax of ***.
____
____
____
____
14.
a.
b.
c.
d.
a.
b.
c.
d.
ABC  EDF
ABC  DEF
B  E
CB  EF
368
Lesson 20
17.
If ABC is an equilateral  then the
measure of 1 is ***.
____
____
____
____
18.
19.
____
a.
3
____
b.
3
____
c.
4
____
d.
4
1
2
1
2
If AB = 8 then AE is equal to ***.
a.
b.
c.
d.
2
3
4
6
A picture frame in the shape of a regular 10 sided polygon is to be
constructed. At what angle should the frame pieces be cut?
____
____
____
____
Mathematics 20
15
75
105
115
In the given diagram, the length of DC is equal to ***.
____
____
____
____
20.
a.
b.
c.
d.
a.
b.
c.
d.
36°
60°
72°
144°
369
Lesson 20
Mathematics 20
370
Lesson 20
Part B can be answered in the space provided. You also have the option to
do the remaining questions in this assignment on separate lined paper. If
you choose this option, please complete all of the questions on the separate
paper. Evaluation of your solution to each problem will be based on the
following.
(40)
B.
•
A correct mathematical method for solving the problem is shown.
•
The final answer is accurate and a check of the answer is shown where
asked for by the question.
•
The solution is written in a style that is clear, logical, well organized,
uses proper terms, and states a conclusion.
1.
A rectangular box has sides whose lengths are  x  2  cm,  x  2  cm,
and 2 x  1 cm.
Write the polynomial in standard form which represents the volume of
the box.
What is the volume when x is 10 cm?
Mathematics 20
371
Lesson 20
2.
Calculate the length of the line segment from points  3,  2  to point
4, 1 on the coordinate plane.
3.
Divide x 3  x 2  13 x  14 by x  2 .
4.
Simplify
Mathematics 20
x
 x  3 
 x  3  x  9 
2
9
2
3
6
2
1
.
372
Lesson 20
5.
6.
Mathematics 20
Simplify
2
5
 2
.
b  9 b  8 b  15
2
The velocity of an object at any time t in seconds is given by
vt   3 t 2  47 t  30
Find the time at which the object is motionless.
373
Lesson 20
7.
Solve the equation x   3 x  4 for all admissible roots.
8.
Sketch the graph of y  3  x  5   3 and analyze it by discussing
domain, range, x and y intercepts, axis of symmetry, vertex, concavity
and maximum minimum points.
Mathematics 20
2
374
Lesson 20
y
x
Mathematics 20
375
Lesson 20
Answer Part C on separate lined paper. Please include any tables or graphs
that you are required to do with the assignment.
(5)
C.
1.
Suppose that it is equally likely that a dart could land anywhere on
the given figure. What is the probability that it will land in the circle?
(10)
(5)
2.
3.
a.
Given that DE is parallel to
BC , prove that
ADE ~ ABC .
b.
If BC = 2 find the length of DE.
c.
What is the ratio of the area of
ADE to the area of ABC ?
Prove APO  BPO where A and B are points of tangency.
Prove AOC  BOC (5 mark Bonus)
_____
(100)
Mathematics 20
376
Lesson 20
Appendix
Reasons for Proofs
1.
Given
2.
Definition of perpendicular lines
•
Two lines that meet or intersect to form right angles.
3.
Definition of a right angle
•
An angle with a measure of 90  .
4.
Definition of a bisector of an angle
•
The bisector of ABC is a ray BD in the interior of ABC such that
ABD  DBC .
5.
Definition of the midpoint of a segment
•
The point that divides the segment into two congruent segments.
6.
Definition of a segment bisector
•
A line, segment, ray, or plane that intersect the segment at its midpoint.
7.
Definition of a perpendicular bisector of a segment
•
A line, ray, or segment that is perpendicular to the segment at its midpoint.
8.
Definition of an altitude of a triangle
•
The segment from a vertex of a triangle and perpendicular to the line
containing the opposite side.
9.
Definition of a median of a triangle
•
A segment from a vertex of a triangle to the midpoint of the opposite side.
10.
Definition of complementary angles
•
Two angles whose measures have the sum of 90  .
Mathematics 20
377
Lesson 20
11.
Definition of supplementary angles
•
Two angles whose measures have the sum of 180  .
12.
Properties of Equality
a)
Addition Property
If a  b and c  d , then a  c  b  d .
b)
Subtraction Property
If a  b and c  d , then a  c  b  d .
c)
Multiplication Property
If a  b , then ca  cb .
d)
Division Property
If a  b and c  0 , then
a b
 .
c c
e)
Reflexive Property
aa
f)
Symmetric Property
If a  b , then b  a .
g)
Transitive Property
If a  b and b  c , then a  c .
h)
Substitution Property
If a  b , then either a or b may be substituted for the other in any equation or
inequality.
13.
The sum of the measures of the angles of a triangle is 180 .
14.
Vertically opposite angles are congruent.
15.
Parallel Lines
a)
If two parallel lines are cut by a transversal, then the alternate interior
angles are congruent.
b)
If two parallel lines are cut by a transversal, then the corresponding angles
are congruent.
c)
If two parallel lines are cut by a transversal, then the same side interior
angles are supplementary.
Mathematics 20
378
Lesson 20
16.
Congruence Postulates
a)
b)
c)
d)
e)
f)
SSS
SAS
ASA
AAS
HL
LL
Mathematics 20
379
Lesson 20
Mathematics 20
380
Lesson 20