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b)
Statement
S(n) = 180°(n – 2)
S(8) = 180°(8 – 2)
S(8) = 180°(6)
S(8) = 1080°
1080º
= 135°
8
x + 135° = 180°
x = 45°
Justification
A regular octagon has
eight sides, so n is 8.
Measure of one interior
angle.
Let x be the measure of
an exterior angle.
Supplementary angles
Cha pt e r R e vie w, pa ge 1 0 6
c)
Statement
AHG = 135° and
FGH = 135°
HAF = 45° and
GFA = 45°
HAB = 135°
HAB = HAF + FAB
FAB = HAB – HAF
FAB = 135° – 45°
FAB = 90°
Similarly, ABE = 90°
AF || BE
Let S represent the sum of the angles indicated.
S = 2(a + b + c + d + e + f)
S = 2(360°)
S = 720°
The sum of the indicated angles is 720°.
Alternative solution:
The angles are external angles of the central
hexagon. Since there are two external angles at
each vertex, the sum of the angles is double the
sum of the external angles of a polygon, or
2(360°) = 720°.
Justification
Regular octagon
Exterior angle of
regular octagon
Regular octagon
Addition and
substitution
Solve for FAB.
Symmetry
Supplementary
angles
6. e.g., Label the indicated angles a, b, c, d, e, and f. As
a result, the measures of the supplementary angles are
as shown.
The interior angles in a hexagon sum to 720°.
Therefore,
720° = (180° – a) + (180° – b) + (180° – c)
+ (180° – d) + (180° – e) + (180° – f)
720° = 1080° – (a + b + c + d + e + f)
360° = a + b + c + d + e + f
By opposite angles, each angle that is opposite to a, b, c,
d, e, and f is equal to that angle, as shown.
1. e.g., The side bars coming up to the handle are
parallel and the handle is a transversal.
2. a) Corresponding pairs of angles: a, e; b, g; c, f;
d, h
b) e.g., No. The lines are not parallel, so
corresponding pairs cannot be equal.
c) I can see eight pairs of supplementary angles.
a, b; a, c; c, d; b, d; e, f; e, g; f, h; g, h
d) The other pairs of equal angles are vertically
opposite angles: a, d; b, c; e, h; f, g
3.
Statement
a = 35°
b + a = 180°
b = 145°
4. Let b be the 180° angle, a be the supplementary
angle to b, and c be the 40° angle.
Statement
a+b
a
c
a
AB
= 180°
= 40°
= 40°
=c
|| CD
Justification
Supplementary angles
Substitution
Given
Transitive principle
Corresponding angles
are equal.
5. a)
Statement
b = 76°
c = 76°
a + b = 180°
a = 104°
Foundations of Mathematics 11 Solutions Manual
Justification
Corresponding angles
Supplementary angles
Justification
Corresponding angles
Vertically opposite
angles
Supplementary angles
2-15
b)
9.
Statement
2a + 3a
5a
a
3a
3a
2a
2a
b
b
c + 2a
c + 72°
c
= 180°
= 180°
= 36°
= 3(36°)
= 108°
= 2(36°)
= 72°
= 3a
= 108°
= 180°
= 180°
= 108°
Justification
Interior angles on the
same side of a
transversal
Substitution
Vertically opposite
angles
Supplementary angles
Substitution
b) Measure ABF and BFH. Measure DBA and
BFE. Both pairs should be equal. So then the lines are
parallel.
7.
ST = TR
Justification
Given angles
Alternate interior angles
Transitive property
Base angles of TRS are
equal, so TRS is isosceles.
Property of isosceles triangle
8. a)
Statement
x = 40°
z = 45°
x + y + z = 180°
40° + y + 45° = 180°
y = 95°
OPL
NO
Substitution
6. e.g.,
a)
Statement
QRS = TRS
QRS = RST
TRS = RST
Statement
LO = LP
Justification
Alternate interior
angles
Alternate interior
angles
Supplementary angles
Substitution
OQN
Justification
Given
LOP is isosceles.
Property of isosceles
= POL
triangle
= NQ
Given
NOQ is isosceles.
Property of isosceles
= NOQ
triangle
Sum of interior
= PLO + POL + OPL
angles in
= 180° – (POL + OPL)
triangle
= 180° – 2POL
180°
PLO
PLO
Sum of interior
180° = QNO + NOQ + OQN
angles in
QNO = 180° – (NOQ + OQN)
triangle
QNO = 180° – 2NOQ
PLO and QNO are
PLO + QNO = 180° – 90°
the two acute angles
PLO + QNO = 90°
in right triangle LMN.
Sum of
180° – 2POL + 180° – 2NOQ = 90°
interior
360° – 2(POL + NOQ) = 90°
–2(POL + NOQ) = –270° angles of
triangle is
POL + NOQ = 135°
180°.
Substitution
POQ + POL + NOQ = 180° Supplementary
POQ + 135° = 180° angles
Substitution
POQ = 45°
10. a)
Statement
S(n) = 180°(n – 2)
S(15) = 180°(15 –
2)
S(15) = 180°(13)
S(15) = 2340°
Justification
For a 15-sided polygon,
n is 15.
b) The sum of the exterior angles of any polygon is
360°. The measure of one exterior angle is 360°
divided by the number of sides.
360°
= 24°
15
The measure of an exterior angle of a 15-sided
polygon is 24°.
b)
Statement
y + 68° = 180°
y = 112°
x = 68°
z + 72° + 68° = 180°
z = 40°
2-16
Justification
Supplementary angles
Corresponding angles
Sum of interior angles
in triangle
Chapter 2: Properties of Angles and Triangles
11.
Statement
S(n) = 180°(n – 2)
S(5) = 180°(5 – 2)
S(5) = 180°(3)
S(5) = 540°
540º
= 108°
5
AB = BC
BAC
Justification
A regular pentagon
has five sides, so n
is 5.
Measure of one
interior angle
Given
ABC is isosceles.
Property of isosceles
= BCA
triangle
Sum of
= BAC + BCA +
interior
angles in
= BAC + BCA + 108°
triangle
= BAC + BCA
Substitution
= 2BAC
= BAC
Property of
= BCA + ACD
equality
= BCD – BCA
180°
ABC
180°
72°
72°
36°
BCD
ACD
ACD = 108° – 36°
ACD = 72°
180° = ACD + CDE
Sum of interior
angles on same
side of transversal
AC || ED
Cha pt e r Ta s k, pa ge 10 7
e.g.,
A. I decided to create a logo for my local animal shelter,
to raise awareness of homeless animals. I want to
convey the message that dogs are good for you and give
you love. If you’re looking for love, come to the animal
shelter!
B. I chose a stylized heart, using parallel lines as thin
stripes to form the heart, because I think this design is
more interesting than a solid heart. I wanted to include a
doghouse as well, to show that animals live here. Again,
I wanted my design to be a bit different, to catch the eye,
so I chose a pentagon. I lined up the heart and the
pentagon so that one edge of each is parallel to the
other.
C. I started my logo by drawing a horizontal line. Then I
used a protractor to draw a line at an angle of 26° from
the horizontal. I drew a vertical line at the point of
intersection of these two lines. The angle between the
vertical line and the line drawn at a 26° angle should
have a measure of 64°. I measured it to be sure.
Next, I drew the H. I drew a line above the H, parallel to
the line at the base of the H, by measuring a 64° angle to
the vertical line (corresponding angles). I completed the
words and then drew another vertical line at the end of
the a.
Foundations of Mathematics 11 Solutions Manual
Next, I printed out an outline of a heart. I placed
the heart on top of a piece of graph paper and
used the graph paper lines as a guide to draw the
parallel lines for the stripes. When I finished, I
checked my work. I used a straightedge as a
transversal, and I made sure that each line was
lined up with the straightedge. Then I rotated the
heart into the position I wanted, under the words.
Finally, I used dynamic geometry software to draw
a regular pentagon. I determined that the sum of
the measures of the interior angles would be 540°.
S(n) = 180°(n – 2)
S(5) = 180°(5 – 2)
S(5) = 540°
This meant that the measure of each interior angle
540°
would be
= 108° .
5
I adjusted the vertices until all my angles were
108°. I knew that the sides would all have the
same measure.
I printed the pentagon. I added extra lines, as
shown, for the roof.
D. To make the final form of my logo, I erased the
lines and measures in my sketch. I drew a dog,
sized it to fit inside the doghouse, and glued it onto
the house. Then I scanned the completed logo and
touched it up.
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