31
SECTION 3: Triangle Geometry B A D A C B A D C A Height CD is 24 cm. The expression is m xy y 2 BC or any other equivalent expression. 1 2 3 4 5 6 7 8 9 10 11 12
SECTION 3: Triangle Geometry
B
A
D
A
C
B
A
D
C
A
Height CD is 24 cm.
The expression is m xyy 2BC or any other equivalent expression.
1
2
3
4
5
6
7
8
9
10
11
12
The length of cable AB is 5.6 metres.
Calculation of the measure of radius OD
OC m OF m = )OD (m2
400 = 25 16 = )OD (m2
m 20 = OD m
In a right triangle, the measure of one side of the
right angle is a mean proportional between the
length of the hypotenuse and the segment on the
hypotenuse adjacent to that leg.
Calculation of the area of the circular sector BOD
Area : r n2 2
360
20 106
360370
( )
Answer: The area of sector BOD is 370 m2.
Statements Justifications
1 CD m
BD m =
AD m
CD m
1 The altitude to the hypotenuse of a right
triangle is the mean proportional between
the segments of the hypotenuse.
CD m
26 =
19
CD m
)CD (m2 = 26 19 = 494
CD m = 22.23 494
2 BC m = )CD (m + )BD (m22
2 Pythagorean theorem.
= 494 + 262
13
14
15
= 494 + 676 In a right triangle, the square of the
hypotenuse is equal to the sum of the
squares of the other sides.
= 34.21 170 1
Answer: The length of cable needed to replace the 1st guy wire is 34.21 m. Accept a result
between 34.19 m and 34.21 m.
Measure of segment AC
)AB (m + )BC (m = AC m22
30 + 40 = AC m 22
m 50 = AC m
Pythagorean theorem
Measure of segment BE In a right triangle, the product of the measures
of the legs is equal to the product of the
measures of the hypotenuse and the altitude
drawn to the hypotenuse.
BC m AB m =
40 30 =
BE m =
BE m AC m
BE m 50
m 24
Measure of segment AE
)BE (m )AB (m = AE m22
24 30 = AE m 22
m 18 = AE m
Pythagorean theorem
16
Area of triangle ABE
Area = 2
BEmAEm
Area = 2
2418
Area = 216 m2
Answer: The area of the piece of land is 216 m2.
Measure of BC
9 = 12 15 = BC m 22
Measure of CE
m CE m AB = m AC m BC
m CE 15 =
12 9
m CE =
15
9 12 = 7.2
Measure of AE
6.92.712AEm22
Perimeter of triangle ACE : 12 + 7.2 + 9.6 = 28.8
Answer: The perimeter of triangle ACE is 28.8 cm.
A
B
C
E
17
Measure the roof’s base, AC
222
11ACm
41.1ACm
Measure of one side of square
1.41 2 0.2 = 1.01
Height BH of roof
BC m AB m = BH m AC m
1 1 BH m 1.41
BHm 0.71
Full height of kennel
1.01 + 0.71 = 1.72
Answer : The full height of the kennel is 1.72 m.
Find the hypotenuse AC
Answer: The height of the attachment point is 7.4 metres.
0.2 m0.2 mA
B
CH
18
19
m AC
= )BC (m + )AB (m22
15.81 m
m BD
= AC m
BC m AB m
In a right triangle the product of the legs equals
the square of the hypotenuse.
15.81
9 13
7.4 m
where m BD 7.4 m
A
B
C
D
Find the height AH
The height is the mean proportional between
the two segments into which the height divides
the hypotenuse.
HB m CH m = )AH (m2
24 = 6 4 = )AH (m2
24 = AH m
To find the angle of elevation AEH of the ladder.
EH m
AH m =AEH tan
3.266 1.5
4.899
1.5
24 = AEHtan
m AEH 72,976
Answer: m AEH 73
In right triangle ADC, we apply the Pythagorean
relation.
)DC (m + )AD (m = )AC (m222
20 122 2 2
mDCc h
16 = DC m
A
B CD
E
a) ADC AED Right angles.
b) DAC DAE Angles common to both triangles
c) ADE ~ ACD Two triangles are similar if they have two corresponding angles
congruent.
In two similar triangles, the corresponding sides are proportional.
20
21
AC m
AD m =
DC m
DE m
20
12 =
16
DE m
9.6 = 20
12 16 = DE m
Answer: 9.6 cm.
Length of segment BD
sin 28 = 10
BDm BDm 4.695 cm
Length of segment DH
Using the Pythagorean Theorem
222
HDmBHmBDm
(4.695)2 = 4
2 + 2HDm
HDm 2.458 cm
Length of segment HC
51.1HCm
05.6HCm4
46.2
4
HCm
46.2
DHm
BHm
HCm
DHm
The length of the altitude to the hypotenuse of a right triangle is
the geometric mean between the lengths of the segments into
which the altitude divides the hypotenuse.
Answer: The length of segment HC is 1.51 cm.
22
cm27ABm (Diameter is double the radius.)
cm32ADm (Subtraction)
Using altitude to the hypotenuse theorem
cm592.9CDm
92CDm
423CDm
2
2
Area of triangle ABC
=
2
592.927
= 129.5 cm2
Answer: The area of triangle ABC is 129.5 cm2.
Measure of segment BD
DC m AD m = )BD (m2
= 16 9 = 144
BD m = 12
In a right triangle, the altitude from the
hypotenuse is the proportional mean between
the two segments it determines on the
hypotenuse.
Measure of segment BC
)DC (m + )BD (m = BC m22
15 = 9 + 12 = 22 Pythagorean theorem
23
24
A
B
CD
E
F
Measure of segment DE
DE m BC m = DC m BD m
DE m 15 = 9 16
7.2 = 15 144 = DE m
In a right-angled triangle, the product of the
measures of the legs is equal to the product of
the measures of the hypotenuse and the
altitude drawn to the hypotenuse.
Measure of segment BE
)DE (m )BD (m = BE m22
6.92.712 22
Pythagorean theorem
Perimeter of rectangle FBED
P = DE m2 + BE m2
= 2 9.6 + 2 7.2 = 19.2 + 14.4 = 33.6
Answer: Rounded to the nearest tenth, the perimeter of rectangle FBED is 33.6 m.
Length of DC
222
ACmADmDCm
50.972124.7860DCm 222
m60.9850.9721DCm
Length of AE
DC m AE m = AC m AD m
98.60 AE m 78.24 60
m 47.61 AE m
Answer: To the nearest hundredth, the length of segment AE is 47.61 m.
25
Calculate the length of segment AB , given triangle ABC is isosceles and therefore AM= 12:
dm15
225
912
BMAMAB
22
22
Calculate the length of segment PM
BMAMABPM
(In a right triangle, the length of the hypotenuse multiplied by the length of the altitude to the
hypotenuse is equal to the product of the lengths of the sides of the right angle.)
PM • 15 = 12 • 9
PM = 7.2 dm
Answer: Segments PM and QM each measures 7.2 dm.
Height of the cone
ABm
BDm
BCm
ABm
BCmBDmABm2
39ABm2
cm196.527ABm
Area of the base
The radius of the base is 9 cm.
Area of the base = 92 = 81 cm
2
Volume of the cone
Volume = 3
2781 440.753 cm
3
Answer: The volume of the cone to the nearest cm3 is 441 cm
3.
26
27
MNR is a right triangle because m MNR = 90o
m MNR = 2
180 = 90
o
Angle MNR is an inscribed angle and therefore is
half the measure of its intercepted arc.
22 125MR m
13MR m
Pythagorean Theorem applied to right triangle MNR
222 )MR m()NR m()MN m(
)MR m()MP m()MN(m 2
13)MP m(52
cm923.1MP m
In a right triangle, each side is the geometric mean
between the hypotenuse and that side’s projection on
the hypotenuse.
22 923.15NP m
615.4NP m
Pythagorean Theorem applied to right triangle MNP
)NPMP( .
MP mMR mPR m
923.113PR m
077.11PR m
NR mPS mPR mPN m
4.615 11.077 PS m 12
In a right triangle, the length of the hypotenuse
multiplied by the length of the altitude to the
hypotenuse is equal to the product of the lengths of
the sides of the right angle.
28
M
N
P R
O
S
12
077.11615.4PS m
PS m 4.26
Answer: Line segment PS is 4.26 cm in length.
Accept an answer within the interval [4.1, 4.3].
B
B
A
B
B
D
D
B
C
C
C
29
30
31
32
33
34
35
36
37
38
39
A
C
A
A
C
C
B
C
C
C
B
D
A
C
40
41
42
43
44
45
46
47
48
49
50
51
52
53
D
B
A
D
D
D
C
C
A
C
C
C
D
B
54
55
56
57
58
59
60
61
62
63
64
65
66
67
A
B
D
A
D
B
B
If three sides of one triangle are congruent to three sides of another triangle, the two triangles are
congruent (or SSS).
Segment AD measures 20 m.
The height AC of the mainsail is 8 m.
Because of similar triangles
2.3
1
8
x where x = 2.5
The distance d can be deduced
8 2.5 = 5.5
Answer: The distance needed between the tree and the house is 5.5 metres.
Step 2 should read as follows:
CPA – DPB because they are vertically opposite angles.
68
69
70
71
72
73
74
75
76
77
78
79
Step 2 DAB BCD
because the opposite angles of a parallelogram are congruent.
Step 3 ABD CDB
because they are alternate interior angles formed by a transversal intersecting
two parallel segments CD//AB .
Step 1 DFE JFM because vertically opposite angles are congruent.
Step 3 DEF ~ JMF because the lengths of two sides of one triangle are
proportional to the lengths of the corresponding sides of
the other triangle and the contained angles are congruent.
Statements
Justifications
ABC ~ ADE AA : A is common
m ABC = m ADE = 90
ADm
ABm
DEm
BCm The corresponding sides of similar triangles are proportional.
8.20
8.0
DEm
5.0 By substitution
13DEm Property of proportions
80
81
82 SCHOOL
AB
C
D
E
F G H
0.8 m
1.0 m1.5 m
20 m
Note : The most important justification is the one that states that the corresponding sides of
similar triangles are proportional.
Height of the school
EHm = DHmDEm
= 13 + 1
= 14
Answer: The height of the school is 14 m.
Statements Justifications
1. AD DC 1. The hypothesis states that D is the midpoint of segment
AC.
2. BD BD 2. Reflexive property.
3. ADB CDB 3. The hypothesis states that ACDB .
4. ABD CBD 4. If two sides and the contained angle of one triangle are
congruent to two sides and the contained angle of another
triangle, then the triangles are congruent (or S-A-S).
Step 1 m ABC + m CBD = 180
m ABC + 120 = 180
m ABC = 60
because adjacent angles whose external sides are in
a straight line are supplementary.
Step 2 m BCA = m QEP = 40 because alternate exterior angles are congruent
when formed by a transversal intersecting two
parallel lines.
83
84
AD
C
B
Value of segment BE:
222
AEmABmBEm
222
4050BEm
900BEm2
30BEm
Solve the proportion: DCm
BEm
BCmABm
ABm
150
30
BCm50
50
200BCm
Answer: The length of the pond is 200 metres.
Statements
Justifications
1. EFI ~ EGH The two triangles have two corresponding angles
congruent :
E is common;
EFI EGH because when two parallel line are cut by a
transversal, the corresponding angles are congruent.
85
86 x + 1
4x + 13x
xA B
C D
E
F
G H
I
2. m EF = 2 units
EHm
EIm
EGm
EFm
25
1
4
x
x
x
x
5x2 + 2x = 4x
2 + 4x
x2 2x = 0
x(x 2) = 0
x1 = 0 (to be rejected)
x2 = 2 units = m EF
In similar triangles the corresponding sides are
proportional.
By substitution
Step 3 ABC ~ EDC
because 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 are similar.
Step 4 ABC EDC
because the corresponding angles of similar figures are congruent.
Value of x
Segments AI, CH, and EF are parallel, as are segment AC, IE and HF because the opposite sides
of a parallelogram are parallel.
x
CHFmACH m
because alternate interior angles formed by parallel lines
and a transversal are congruent.
40
1604
180203
180
CHFmEFH m
x
x
xx
because two consecutive angles in a parallelogram are
supplementary.
87
88
Measure of angle BDI:
61BDI m
19402BDI m
192
BGHmBDI m
x
because corresponding angles formed by parallel lines and
a transversal are congruent.
Answer: The numerical measure of angle BDI is 61.
Since BC//DF , m AFD = m ACB
Since AB//EF , m DAF = m EFC
Since two angles of one triangle are congruent to the two corresponding angles of the other
triangle, triangles DAF and EFC are similar and their corresponding segments are proportional in
length.
Similarity ratio
Since BC//DF and AB//EF , polygon BDFE is a parallelogram.
Hence, cm15BDmEFm .
1
3
cm5
cm15
ADm
EFmk
Perimeter of DAF
FAmDFmADm
5 + 10 + 7.5
22.5 cm
Perimeter of EFC
1
3
DAFofPerimeter
EFCofPerimeter
1
3
cm5.22
EFCofPerimeter
Answer: The perimeter of EFC = 67.5 cm.
89
Step 3 MBA ~ MNO because if two angles of one triangle and the
two corresponding angles of another
triangle are congruent, then the two
triangles are similar.
Step 4
cm4.22OMm
OMm
cm14
cm16
cm10
OMm
AMm
NOm
BAm
because the lengths of the corresponding
sides of two similar triangles are
proportional.
STATEMENTS JUSTIFICATIONS
1. Segments AD and BC are congruent.
The opposite sides of a rectangle are
congruent.
2. Segments AB and DC are congruent.
The opposite sides of a rectangle are
congruent.
3. Diagonal AC is common to both
triangles ABC and CDA.
By construction
4. Triangles ABC and CDA are
congruent.
If three sides of one triangle are congruent
to three sides of another triangle, then the
triangles are congruent (or SSS).
90
91
A
B
C
D
Measure of angle BPC
m BPC = m ADP = 65
because they are corresponding angles and PB//AD since ABPD is a parallelogram.
Value of x
The sum of the measures of the interior angles of triangle BPC is equal to 180. Hence
3x + 2x + 65 = 180
5x = 115
x = 23
Measure of angle PCB
m PCB = 2x = 2 23 = 46
Measure of angle BAP
m BAP = m PCB = 46 because the opposite angles of a parallelogram are isometric.
Answer The measure of angle BAP is 46.
Length of segment CF:
Transversals intersected by parallel lines are divided into segments of proportional lengths.
ADm
DGm
ACm
CFm
30
45
26
CFm
CFm = 39 cm
Length of segment AF:
CFmACmAFm
AFm = 26 + 39 = 65 cm
92
93
Length of segment EF:
AEF ABC because they are corresponding angles formed by a transversal intersecting two
parallel segments. For the same reason, AFE ACB. Hence, AEF ~ ABC and the lengths
of the corresponding segments are proportional.
ACm
AFm
BCm
EFm
26
65
14
EFm
EFm = 35 cm
Answer: The length of segment EF is 35 cm.
Step 2 AVS CWS
ATS CUS
because the corresponding angles of
congruent figures are congruent.
Step 3
BC//AD
DC//AB
because congruent alternate interior angles
are formed by parallel lines and a
transversal.
Example 1
Statements Justifications
1 ADE ~ ABC 1 2 are similar if they have 2 congruent angles.
since m A = m A A is common
m AED = mACD Right angles
2
BCm
DEm
AC m
AEm
9
4
3
x
x
2 If 2 angles are similar then the corresponding
sides are proportional.
94
95
3 9x = 4x + 12
5x = 12
x = 2.4
3 In a proportion, the product of the means is equal
to the product of the extremes.
Answer: The distance AE is 2.4 m.
Example 2
Statements Justifications
1 m A = m A 1 A is common to triangles ADE and ABC.
2 m AED = m ACB = 90 2 ED and BC are heights.
3 tan A =
AEm
4 =
3AEm
9
3 By definition,
θadjacentsideofmeasure
θoppositesideofmeasureθtan
4
x =
9
3x +
4 4x + 12 =
5x =
x =
9x
12
2.4
4 In a proportion, the product of the means is equal
to the product of the extremes.
Answer: The distance AE is 2.4 m.
Step 4 ABD BCE because 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.
Step 4 BEAD because the corresponding sides of
congruent figures are congruent.
96
Statements
Justifications
1. AC CE Point C is midpoint of segment AE.
2. BC CD Point C is midpoint of segment BD.
3. ACB DCE Vertically opposite angles are congruent.
4. ABC CDE
If two sides and the contained angle of one triangle are
congruent to two sides and the contained angle of another
triangle, then the triangles are congruent (or SAS).
measure of ,AB hypotenuse of the triangle
m68ABm 22
m10AB m
measure of BD : d
m10
m10
m9
m 6
d ABC and ADE are similar
d = 5 m
Answer: 5 m
97
98
A B
ED
C
A
B
d
D
C E8 m
6 m
9 m
m AB : d
Triangle ADC and triangle BEC are similar.
d
m1800
m1800
m1400
m600
Answer: d = 2400 m
Other acceptable method : The student uses trigonometry to solve the problem.
Measure of segment BC:
2BC m = 22
ACmAB m
2BC m = (18)2 + (24)
2
2BC m = 900
BC m = 30
Measure of segment MC:
MC m = 2
BC m
MC m = 2
30
MC m = 15
99
100 Statements Justifications
MON QOP Vertical angles are congruent.
MNO QPO Given (All right angles are congruent.)
MNO ~ QPO AA
101
M
P
N O
Q
Measure of segment MD:
ABm
MD m =
ACm
MC m
18
MDm =
24
15
MDm = 11.25
Answer: The measure of segment MD is 11.25 cm.
Show that triangles RCT and ABC are similar.
STATEMENT
JUSTIFICATION
1. CRT A When a transversal cuts two parallel lines,
corresponding angles are congruent.
2. RCT ACB Reflexivity
3. RCT ~ ACB Two triangles are similar if they have two
corresponding angles congruent (AA).
4.
CBm
CTm
ACm
RCm
In similar figures, the measures of corresponding
sides are proportional.
5.
4
1
4
1
xx
x
x = 2
Substitution
Answer: The value of x is 2 units.
102
C
1x 1
R T
x + 35
BA
STATEMENT JUSTIFICATION
1. m ADB = 90
m ABC = 90
Therefore,
m ADB = m ABC
By definition of altitude BD
Given that triangle ABC has a right angle at B.
2. m BAD = m CAB Angle common to triangles BAC and BAD.
3. ADB ~ ABC Two triangles are similar if they have two
corresponding angles congruent.
Length of side BC
ACm
ADm =
BCm
DEm
18
12 =
x
5
x = 7.5
Area of triangle ABC
A = 2
bh
A = 2
185.7
A = 67.5
Answer: The area of triangle ABC is 67.5 m2.
103
104
B
CDA
A
B
CD
E
5 m
12 m
18 m
Statements Justifications
1. m DC = 6 cm
m AC = 2 cm
4
3 8 cm = 6 cm
8 cm 6 cm = 2 cm
2.
m EC = 9 cm
m BC = 3 cm
4
3 12 cm = 9 cm
12 cm 9 cm = 3 cm
3.
ECm
BCm
DCm
ACm
9
3
6
2
4.
ACB DCE
Vertical angles are congruent.
5.
ABC ~ DEC
Two triangles are similar if the angles included between
corresponding proportional sides are congruent (S.A.S.
Similarity Theorem).
Note : Steps 1 and 2 are optional.
105 m DA = 8 cm
m EB = 12 cm
m EC =
4
EBm3
m DC =
4
DAm3
BA
C
DE
Statements Justifications
1. ABC ~ FDE Two pairs of corresponding angles are congruent :
ACB FED given in the hypothesis,
B D because the opposite angles of a parallelogram
are congruent.
2. m DF = 8.8 cm
CAm
EFm
BAm
DFm
45
18
22
DFm
hence, m DF = 8.8 cm
In similar triangles, the corresponding sides are
proportional.
By substitution
Measure of angle JLK
Two vertically opposite angles are congruent.
Hence:
75
DLHmJLKm
Value of x
The sum of the measures of the interior angles of a triangle is 180.
Hence:
21
1055
1807523
180JLKmLKJmKJLm
x
x
xx
106
107
A
B C
DE
F
Measure of angle FKB
Corresponding angles formed by two parallel lines and a transversal are congruent.
Hence:
63
213
3
KJLmFKBm
x
Answer: The numerical measure of angle FKB is 63.
