Download 1 7.1 Triangle Application Theorems (pg.298-301)

Geometry for Enjoyment and Challenge - Text Solutions
1
Ruth Doherty
7.1 Triangle Application Theorems (pg.298-301)
2.
7
Given: m∠1 = 130◦
m∠7 = 70◦
Prove:
Find the measures of ∠2, ∠3, ∠4,
∠5, ∠6
6
5
3
2
1
4
Solution:
m∠1 + m∠2 = 180 (Definition of Supplementary)
m∠2 = 50
m∠7 + m∠6 = 180 (Definition of Supplementary)
m∠6 = 110
m∠7 = m∠5 (Vertical Angles are congruent, so their measures are equal)
m∠5 = 70
m∠2 + m∠3 + m∠5 = 180 (Sum of the measures of the interior angles of a triangle)
m∠3 + 50 + 70 = 180
m∠3 = 60
m∠4 = m∠5 + m∠2 (Exterior Angle Equality)
m∠4 = 120
3.
C
Given: m∠CAB = 80◦
m∠CBA = 60◦
AE and BD are altitudes
Find:
m∠C and m∠AF B
E
D
F
A
B
Solution:
m∠CAB + m∠CBA + m∠C = 180 (Sum of the measures of the interior angles of a triangle)
80 + 60 + m∠C = 180
m∠C = 40◦
m∠ADB = 90 (Altitudes are perpendicular to the side of the triangle, ⊥⇒
m∠BEA = 90
,
= 90◦ )
m∠DBA + m∠ADB + m∠CAB = 180 (Sum of the measures of the interior angles of a
m∠DBA + 90 + 80 = 180
triangle)
m∠DBA = 10
m∠EAB + m∠AEB + m∠CBA = 180 (Sum of the measures of the interior angles of a
m∠DBA + 90 + 60 = 180
triangle)
m∠DBA = 30
m∠AF B = 180 − m∠DBA − m∠EAB (Sum of the measures of the interior angles of a 4)
m∠AF B = 140◦
4.
G
In the diagram as marked, if m∠G = 50, find
m∠M
M
H
J
Solution:
Let m∠M HJ = x and m∠M JH = y, so m∠M HJ + m∠M JH = x + y
2x + 2y + 50 = 180 (Sum of the measures of the interior angles of a triangle)
2(x + y) = 130
x + y = 65
m∠M +m∠M HJ +m∠M JH = 180 (Sum of the measures of the interior angles of a triangle)
m∠M = 180 − (x + y)
m∠M = 180 − 65
m∠M = 115
5. The measure of the three angles of a triangle are in the ratio of 4 : 5 : 6. Find the measure
of each.
Solution:
Let the measures of the three angles be x, y, and z.
So because the sum of the measures of the interior angles is 180, x + y + z = 180.
Now, x = 4s, y = 5s, z = 6s because we want the angles to be in a ratio of 4 : 5 : 6.
4s + 5s + 6s = 180 by substitution.
15s = 180
s = 12
x = 48, y = 60 and z = 72
6.
O
Given: m∠ORS = 4x + 6
m∠P = x + 24
m∠O = 2x + 4
Find: m∠O
S
R
Solution:
By the Exterior Angle Equality:
m∠ORS
4x + 6
4x + 6
x
m∠O = 2(22) + 4 = 48
=
=
=
=
m∠O + m∠P
(x + 24) + (2x + 4)
3x + 28
22
P
9. Tell whether each statement is true Always, Sometimes or Never.
Solution:
(a) The acute angles of a right triangle are complementary. Always
(b) The supplement of one of the angles of a triangle is equal in measure to the sum of the
other two angles of the triangle. Always
(c) A triangle contains two obtuse angles. Never
(d) If one of the angles of an isosceles triangle is 60◦ , the triangle is equilateral. Always
(e) If the sides of one triangle are doubled to form another triangle, each angle of the second
triangle is twice as large as the corresponding angle of the first triangle. Never
12. In 4DEF , the sum of the measures of ∠D and ∠E is 110. The sum of the measures of
∠E and ∠F is 150. Find the sum of the measures of ∠D and ∠F .
Solution:
A
D
E
150
110
F
C
By the Exterior Angle Equality, we know:
∠E + ∠EDF = ∠ADF and ∠E + ∠DF E = ∠DF C.
So, m∠DF C = 110◦ and m∠ADF = 150◦ , then because ∠DF C$∠EDF , m∠EDF = 30.
∠DF C$∠EF D as well, so m∠EF D = 70. Thus, m∠D + m∠F = 100
17.
Given: EF GH is a rectangle.
F H = 20
J, K, M , and O are midpooints
E
J
K
a) Find the perimeter of JKM O
b) What is the most descriptive name
for JKM O?
F
H
O
M
G
Solution:
(a) J is the midpoint of EH and K is the midpoint of EF . Then by the midline theorem,
KJ||F H and KJ = 12 F H. By the same reasoning, OM ||F H and M O = 21 F H. So
KJ = 10 and M O = 10. EF GH is a rectangle so the diagonals are congruent, which
means EG = 20. By the midline theorem, JO||EG and JO = 12 EG, KM ||EG and
KM = 12 EG. Therefore KM = 10 and JO = 10. So the perimeter of KJM O is 40.
(b) The sides of JKM O are all congruent. The opposite sides are also parallel, so JKM O is
a rhombus.
2
7.2 Two Proof-Oriented Triangle Theorems (p.304-306)
13.
Given: OHJM is an isosceles trapezoid,
with bases HJ and OM .
∠HP J ∼
= ∠JKH
M
O
P
K
Prove:
(a) 4HRJ is isosceles
R
(b) HP ∼
= JK
(c) R is equidistant from O and M .
H
J
Solution:
(a) It is given that ∠HP J ∼
= ∠JKH. OHJM is an isosceles trapezoid which means that the
bases angles are congruent, ∠P HJ ∼
= ∠KJH. By the reflexive property, HJ ∼
= HJ. So
∼
∼
⇒ ,
4JHP = 4HJK by AAS. Then ∠KHJ = ∠P JH by CPCTC. Therefore, by
4HRJ is isosceles.
(b) Also by CPCTC, P H ∼
= KJ.
(c) 4HRJ is isosceles so RH ∼
= RJ. And by CPCTC, HK ∼
= P J. Therefore by the subRK.
Because
OHJM
is
an
isosceles
trapezoid, the legs are
traction property, P R ∼
=
∼
∼
congruent so OH = M J. Thus by the subtraction property, OP = M K. ∠HP R$∠OP R
and ∠JKR$∠M KR, because they form straight angles. Given ∠HP R ∼
= ∠JKR, by the
congruent supplement theorem, ∠OP R ∼
= ∠M KR. Then 4OP R ∼
= 4M KR by SAS.
Which means OR ∼
= M R by CPCTC.
14.
Z
Given: AC||XY
AB||CY
∠ZAC ∼
= ∠XAB
C
A
Prove: ∠X ∼
= ∠Z
X
Solution:
Statements
Reasons
1. AC||XY
1. Given
2. AB||CY
3. ∠ZAC ∼
= ∠XAB
2. Given
4. ∠CY B ∼
= ∠ABX
∼
5. ∠CY B = ∠ZCA
4. P → CAC
6. ∠CY B ∼
= ∠ZCA
∼
7. ∠X = ∠ZAC
6. Transitive Property
8. ∠X ∼
= ∠XAB
∼
9. ∠Z = ∠XAB
8. Transitive Property
10. ∠X ∼
= ∠Z
10. Transitive Property
3. Given
5. P → CAC
7. P → CAC
9. No Choice Theorem
B
Y