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LESSON
5-3
Challenge
Isosceles Triangles and Roof Trusses
The wooden or metal framework that supports a roof is called a
roof truss. The simplest type of roof truss has the shape of an
isosceles triangle, as depicted by UABC at right. In the diagram,
the legs of the triangle, AB and CB, represent sloping beams that
are called rafters. The base, AC, represents the tie beam that
“ties together” the rafters. However, large roofs require trusses
with designs that are more complex than this.
For example, a king-post truss is pictured at right.
The king post, KZ, is a median of UJKL, and it provides
support for the rafters. Additional support for the rafters
comes from struts ZX and ZY , which are medians of
UJKZ and ULKZ, respectively. The outer triangle,
UJKL, is an isosceles triangle with base JL.
1. Refer to the diagram of the king-post truss. Write a flowchart proof to show that
‘JKZ > ‘LKZ.
2. Explain why it must be true that ‘XJZ > ‘XZJ. (Hint: How is ZX related to 'JKL?)
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3. Given that m‘KJZ 338, find the measures of all the other angles between the rafters,
beams, and struts of the king-post truss. Label the angle measures directly on the figure.
In the queen-post truss pictured at right, congruent queen posts NV
and TW are positioned so that they are perpendicular to tie beam
PR and so that PV # VW # WR. Struts MV , SW , and NT intersect
the rafters so that PM # MN # NQ and RS # ST # TQ. The
outer triangle, UPQR, is an isosceles triangle with base PR.
4. Using the information about the queen-post truss given above, prove each statement
on a separate sheet of paper. Use any form of proof that you want.
a. UQNT is isosceles.
b. UMPV is isosceles.
c. UVMN is isosceles.
5. Given that m‘NQT 1108, find the measures of all the other angles between the rafters,
beams, and struts of the queen-post truss. Label the angle measures directly on the figure.
© Houghton Mifflin Harcourt Publishing Company
142
Holt McDougal Analytic Geometry
Practice C
Reteach
1.
1. Possible answer: From the definition of a
parallelogram, DC is congruent to AB
and DC is parallel to AB . By the
Alternate Interior Angles Theorem, ‘BAC
is congruent to ‘DCA and ‘CDB is
congruent to ‘ABD. Therefore UABE is
congruent to UCDE by ASA. By CPCTC,
DE is congruent to BE and AE is
congruent to CE . Congruent segments
have equal lengths, so the diagonals
bisect each other.
2.
Statements
2. Possible answer: From the definition of a
rhombus, IH is congruent to FG , IF is
congruent to GH , and IH is parallel to
FG . By Alternate Interior Angles
Theorem, ‘GFH is congruent to ‘IHF
and ‘FGI is congruent to ‘HIG.
Therefore UFGJ is congruent to UHIJ by
ASA. By CPCTC, FJ is congruent to HJ
and GJ is congruent to IJ . So UFJI is
congruent to UGHJ by SSS. But UHIJ is
also congruent to UFIJ by SSS. And so
all four triangles are congruent by the
Transitive Property of Congruence. By
CPCTC and the Segment Addition
Postulate, FH is congruent to GI . By
CPCTC and the Linear Pair Theorem,
‘FJI, ‘GJF, ‘HJG, and ‘IJH are right
angles. So FH and GI are
perpendicular. By CPCTC, ‘GFH, ‘IFH,
‘GHF, and ‘IHF are congruent, so FH
bisects ‘IFG and ‘IHG. Similar
reasoning shows that GI bisects ‘FGH
and ‘FIH.
Reasons
1. UUXW and UUVW
are rt. Us.
1. Given
2. UX # UV
2. a. Given
3. UW # UW
3. b. Reflex. Prop. of #
4. c. UUXW # UUVW
4. d. HL
5. ‘X # ‘V
5. e. CPCTC
3. QR
WX
13 , RS
XY
7, SQ
YW
34 . So UQRS # UWXY by SSS, and
‘RSQ # ‘XYW by CPCTC.
4. AB
JK
5, BC
KL
10,
CA LJ
53. So UABC # UJKL
by SSS, and ‘CAB # ‘LJK by CPCTC.
5. MN
TU
3 5 , NP
UV
2 5 , PM
VT
65 . So UMNP # UTUV by SSS,
and ‘PMN # ‘VTU by CPCTC.
Challenge
3. The diagonals of a rectangle bisect each
other.
1.
4. The diagonals of a square are congruent
perpendicular bisectors that bisect the
vertex angles of the square.
5. The diagonals are congruent.
2. Explanations may vary.
© Houghton Mifflin Harcourt Publishing Company
A29
Holt McDougal Analytic Geometry
Practice B
3.
1. Possible answer: It is given that HI is
congruent to HJ , so ‘I must be
congruent to ‘J by the Isosceles Triangle
Theorem. ‘IKH and ‘JKH are both right
angles by the definition of perpendicular
lines, and all right angles are congruent.
Thus by AAS, UHKI is congruent to
UHKJ. IK is congruent to KJ by
CPCTC, so HK bisects IJ by the
definition of segment bisector.
4. Proofs will vary.
Problem Solving
1. 60 in2; Since the triangles are #, they
have the same measures. So, the
triangles also have the same areas.
2. 82 m; UUVW # UXYW by SAS, so
UV # XY by CPCTC. Therefore UV
82 m.
2. 58.1 ft
3. ‘P # ‘R because they are both rt. ‘s .
PQ # RQ because PQ RQ 65 ft.
‘NQP # ‘SQR because vert. ‘s are #.
Therefore UNPQ # USRQ by ASA. By
CPCTC, NP # SR . So SR NP 40 ft.
4. A
4.
XY
2
3. 45°
5. 36 or 9
4
3
6. 76°
7.
8. 10
9. 30°
10. 89
Practice C
1. Possible answer: UABC is an isosceles
triangle with vertices A(0, b), B(a, 0), and
C(a, 0). D is the midpoint of BC , so D
has coordinates (0,0). The slope of AD
b0 b
is
, so the slope is undefined. A
00 0
5. G
6. D
Reading Strategies
1. Possible answer: to abbreviate or make
a statement simpler to understand or
remember
line with an undefined slope is a vertical
line. The slope of BC is
00
0
0 . A line with a zero
a ( a ) 2a
2. Answers will vary. Students may mention
FBI, IRS, RSVP, FAQ, or others that are
popular in text messaging.
3. ‘B # ‘M; ‘A # ‘L; CB # NM ; AB # LM
slope is a horizontal line. Because AD is
vertical and BC is horizontal, AD A BC .
5-4 ISOSCELES AND EQUILATERAL
TRIANGLES
Practice A
1. ZY
2. XY ; XZ
3. ‘Z; ‘Y
4. ‘X
5. equiangular
6. opposite
7. angles
8. equilateral
9. 693 ft
11. 6.3
13. 4
1
yd
2
10. 50q
12. 60q
14. 65q
15. 8
© Houghton Mifflin Harcourt Publishing Company
A30
Holt McDougal Analytic Geometry