Download Chapter 4, continued

Chapter 4,
continued
4. J; If a triangle has sides that are 5 centimeters and
3 centimeters long, and the side that is 5 centimeters long
forms a 288 angle with the third side, there is not enough
information to assert that this triangle is unique. Also
needed is the length of the third side or the measure of
one of the other angles.
5. The triangle that is marked with the 21 unit side length is
equiangular, so it is equilateral. The triangle marked with
the 2x 2 3 unit side length is isosceles by the Converse
of Base Angles Theorem. The two triangles share a
common side, which is congruent to the 2x 2 3 unit side.
2x 2 3 5 21
}
} }
}
}
}
15. True; XY > RS, YZ > ST, and XZ > RT, so
nXYZ > nRST by SSS.
}
}
16. Not true; AC 5 5 and DB 5 4, so AC À DB.
Therefore n ABC À nDCB.
17. True; Ž QSR > Ž TSU by the Vertical Angles Theorem.
} }
} }
Because QS > TS and RS > US, nQRS > nTUS
by SAS.
18. Not true; The triangle vertices are in the incorrect order.
}
} }
DE > HG and EF > GF, so nDEF > nHGF by HL.
} }
19. DE > GH, Ž D > Ž G, Ž F > Ž J
} }
20. DF > GJ, Ž F > Ž J, Ž D > Ž G
}
21. Show nACD and nBED are congruent by AAS, which
2x 5 24
}
}
makes AD congruent to BD. nABD is then an isosceles
triangle, which makes Ž1 and Ž2 congruent.
x 5 12
22. Show nFKH > nFGH by HL. So Ž1 > Ž2 because
Chapter 4 Review (pp. 282–285)
corresponding parts of congruent triangles are congruent.
1. A triangle with three congruent angles is
23. Show nQSV > nQTV by SSS. So Ž QSV > Ž QTV
called equiangular.
2. In an isosceles triangle, base angles are opposite the
congruent sides while the congruent sides form the
vertex angle.
3. An isosceles triangle has at least two congruent sides
while a scalene triangle has no congruent sides.
because corresponding parts of congruent triangles
are congruent. Using vertical angles and the Transitive
Property, you get Ž1 > Ž2.
24. ŽL > Ž N, so x 5 65.
25. n WXY is equilateral;
1 }32 x 1 30 28 5 608
4. Sample answer:
3
2
} x 5 30
2
1
}
x 5 20
}
26. TU > VU;
3
Ž R and Ž N
}
} }
}
Corresponding sides: PQ and LM, PR and LN,
}
}
QR and MN
(2x 2 25)8 5 x8 1 208
6.
8x8 5 2x8 1 908
6x 5 90
4x 5 36
x 5 15
x59
(9(9) 1 9)8 5 908
9. mŽB 5 1808 2 508 2 708 5 608
}
27. (x, y) l (x 2 1, y 1 5)
}
10. AB > UT, so AB 5 15 m.
11. Ž T > ŽB, so mŽ T 5 608.
Q(2, 21) l (1, 4)
13. (2x 1 4)8 5 1808 2 1208 2 208
2x 5 36
x 5 18
14. 5x8 5 1808 2 358 2 908
1
Q
21
x
R
S
28. (x, y) l (x, 2y)
y
Q(2, 21) l (2, 1)
R(5, 22) l (5, 2)
S(2, 23) l (2, 3)
1
21
Q
x
R
12. Ž V > ŽA, so mŽ V 5 508.
2x 1 4 5 40
y
(9x 1 9)8 5 5x8 1 458
8.
8(15)8 5 1208
}
x51
S(2, 23) l (1, 2)
(2(45) 2 25)8 5 658
7.
8x 5 8
R(5, 22) l (4, 3)
x 5 45
S
29. (x, y) l (2x, 2y)
y
Q(2, 21) l (22, 1)
R(5, 22) l (25, 2)
S(2, 23) l (22, 3)
1
21
Q
5x 5 55
x 5 11
114
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
7x 1 5 5 13 2 x
5. Corresponding angles: Ž P and Ž L, Ž Q and Ž M,
x
R
S
Chapter 4,
continued
2. 7 2 ca21
Chapter 4 Test (p. 286)
2ca28
1. The triangle is equilateral and acute (or equiangular).
cq8
2. The triangle is scalene and right.
3. The triangle is isosceles and obtuse.
22
4. x8 5 1808 2 308 2 808
0
2
12
14
12
14
29
x 5 30
212
6. x8 5 1808 2 558 2 508
7. 3x 2 5 5 10
(15x 1 y)8 5 908
3x 5 15
15(5) 1 y 5 90
x55
28
y 5 15
8. n ABC > nEDC can be proven by SAS because
} } }
AC > EC, BC > DC, and Ž ACB > Ž ECD by the
Vertical Angles Congruence Theorem.
0
2
}
MN > PQ, NP > QM, and MP > PM by the
Reflexive Property.
} }
11. Given: n ABC is isosceles with base AC. BD bisects Ž B.
Prove: n ABD > nCBD
10
22
0
2
1
6. } z < 2
4
z<8
22
0
2
4
6
8
10
7. 5k 1 1q211
Reasons
5kq212
1. nABC is isosceles. 1. Given
12
2. Ž BAD > Ž BCD
} }
3. BD > BD
kq2}
5
2. Base Angles Theorem
12
25
3. Reflexive Property of
Congruence
}
24
4. Given
5. Ž ABD > Ž CBD
5. Definition of angle bisector
6. n ABD > nCBD
6. AAS Congruence Theorem
}
22
}
}
b. nPQR > nSTU by AAS if Ž P > Ž S or if
Ž R > Ž U.
22 < r
r > 22
24
22
0
2
4
6
8
9. 6x 1 7 < 2x 2 3
4x < 210
13. The figure transformation is a reflection.
10
14. The triangle transformation is a reflection.
x < 2}
4
15. The triangle transformation is a translation.
5
x < 2}2
Chapter 4 Algebra Review (p. 287)
5
22
1. x 2 6 > 24
24
2
2
14.4 > 27.2r
}
12. a. nPQR > nSTU by HL if PQ > ST or if QR > TU.
x>2
0
8. 13.6 > 20.8 2 7.2r
4. BD bisects Ž B.
1
8
y < 21
24
10. nMNP > nPQM can be proven by SSS because
Statements
6
y12<1
triangles are equilateral, so all angles are congruent by
the Corollary to the Base Angles Theorem.
}
4
5. 3( y 1 2) < 3
9. nFGH > nJKL can be proven by ASA because both
}
0
ta10
22
}
} }
24
5
4. } t 1 8a
2
}ta
x 5 75
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
10
xa29
3x 5 90
0
8
29qx
5. 2x8 1 x8 5 1808 2 908
21
6
3. 254q6x
x 5 70
}
4
22
0
2
10. 2v 1 12a 2 2v
va23
25
24
23
22
21
0
1
Geometry
Worked-Out Solution Key
115
Chapter 4,
continued
11. 4(n 1 5)q 2 n
20. {3x 2 15{ 5 0
4n 1 20q5 2 n
3x 2 15 5 0
5nq215
3x 5 15
nq23
x55
24
22
0
The solution is 5.
2
21. {6x 2 2{ 5 4
12. 5y 1 3q2( y 2 9)
6x 2 2 5 4
5y 1 3q2y 2 18
3yq221
yq27
6x 2 2 5 24
or
6x 5 6
6x 5 22
x51
x 5 2}3
1
1
26
24
22
0
The solutions are 2}3 and 1.
2
22. {8x 1 1{ 5 17
13. {x 2 5{ 5 3
x2553
8x 1 1 5 17
x 2 5 5 23
or
x58
x52
The solutions are 2 and 8.
x52
x 5 2}4
9
The solutions are 2}4 and 2.
x 1 6 5 22
or
x 5 24
23. {9 2 2x{ 5 19
x 5 28
9 2 2x 5 19
The solutions are 28 and 24.
05x
24. {0.5x 2 4{ 5 2
0.5x 2 4 5 2
16. {2 2 x{ 5 0.5
or
2 2 x 5 20.5
1.5 5 x
2.5 5 x
3x 2 1 5 28
3x 5 9
3x 5 27
x53
x 5 2}
7
3
7
The solutions are 2}3 and 3.
4x 1 5 5 27
or
4x 5 2
4x 5 212
1
x 5 }2
x 5 23
1
2
The solutions are 23 and }.
or
x 5 3.4
x54
5x 2 2 5 8
5x 2 2 5 28
or
5x 5 10
5x 5 26
x52
x 5 2}5
6
6
The solutions are 2}5 and 2.
7x 1 4 5 11
7x 1 4 5 211
or
7x 5 7
7x 5 215
x51
x 5 2}
7
15
15
The solutions are 2}
and 1.
7
27. {3x 2 11{ 5 4
19. {x 2 1.3{ 5 2.1
x 2 1.3 5 2.1
x 5 12
26. {7x 1 4{ 5 11
18. {4x 1 5{ 5 7
4x 1 5 5 7
0.5x 5 2
25. {5x 2 2{ 5 8
17. {3x 2 1{ 5 8
or
0.5x 2 4 5 22
or
0.5x 5 6
The solutions are 4 and 12.
The solutions are 1.5 and 2.5.
3x 2 1 5 8
x 5 14
The solutions are 25 and 14.
85x
The solutions are 0 and 8.
2 2 x 5 0.5
22x 5 228
x 5 25
4 2 x 5 24
or
9 2 2x 5 219
or
22x 5 10
15. {4 2 x{ 5 4
42x54
8x 5 218
9
14. {x 1 6{ 5 2
x1652
8x 1 1 5 217
or
8x 5 16
x 2 1.3 5 22.1
x 5 20.8
The solutions are 20.8 and 3.4.
3x 2 11 5 4
or
3x 2 11 5 24
3x 5 15
3x 5 7
x55
x 5 }3
7
7
The solutions are }3 and 5.
116
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
210 28
Chapter 4,
continued
TAKS Practice (pp. 290–291)
9. D;
P 5 2* 1 2w
1. D; Each time the value of x increases by 1, the value of
f (x) decreases by 4. So, f(x) is a linear function whose
rate of change is 24. Because the function is linear, it
can be written in the form f(x) 5 mx 1 b, where m is the
rate of change. Substitute values from the table to find b.
42 5 2(* 1 w)
21 5 * 1 w
21 2 * 5 w
A 5 *w
f(x) 5 mx 1 b
108 5 *(21 2 *)
6 5 24(22) 1 b
108 5 21* 2 *2
22 5 b
So, the expression 24x 2 2 can be used to find the
values of f(x) in the table.
*2 2 21* 1 108 5 0
(* 2 9)(* 2 12) 5 0
2. F; When x 5 3.2:
*2950
or
*59
or
* 5 12
21 2 9 5 w
or
21 2 12 5 w
216x 2 1 250 5 216(3.2)2 1 250
5 216(10.24) 1 250
12 5 w
5 2163.84 1 250
5 86.16
The approximate height of the ball 3.2 seconds after it is
released is 86 feet.
* 2 12 5 0
95w
The dimensions of the rectangle are 12 meters by
9 meters.
10. G; Because the roots of the function correspond to the
x-intercepts of the graph of the function, the roots are
(21, 0) and (3, 0).
3. A;
4(x 1 2)(x 2 5) 2 2(x 2 7x 1 8)
2
5 4(x 2 2 3x 2 10) 2 2(x 2 2 7x 1 8)
5 4x 2 2 12x 2 40 2 2x 2 1 14x 2 16
5 2x 2 1 2x 2 56
11. A;
Number
Cost per
+ of bags 1
bag of
of bagels
bagels
4. F; Each time the value of t increases by 1, the value of
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
r(t) increases by 0.5. So, r(t) is a linear function whose
rate of change is 0.5. Because the function is linear, it can
be written in the form r(t) 5 mt 1 b, where m is the rate
of change. Substitute values from the table to find b.
12. J; Multiply the x-coordinate of a point by 21 to find
0.5 5 0.5(1) 1 b
05b
So, the expression 0.5t can be used to find the values of
r(t) in the table.
5. B; 5(x 1 4) 2 3(6x 1 7) 5 5x 1 20 2 18x 2 21
5 21 2 13x
the corresponding point in the image after a reflection
across the y-axis. Because point Z in the image, nXYZ,
corresponds to point C in the original triangle, nABC,
the coordinates of Z are (25, 2).
13. C; The boundary line passes through the points (21, 0)
and (0, 1), the x- and y-intercepts. Using these points, the
120
0 2 (21)
1
slope of the boundary line is m 5 } 5 }1 5 1.
6. H; When x 5 26:
4x 2 1 3x 2 19 5 4(26)2 1 3(26) 2 19
5 4(36) 1 3(26) 2 19
5 144 2 18 2 19
The y-intercept is 1, so the equation of the boundary line
is y 5 x 1 1. The half-plane below the line is shaded. So,
the graph of y a x 1 1 is shown.
14. H; The number of students surveyed was about
5 107
180 1 150 1 100 1 70 1 80 5 580. Because about
180 students chose comedy, the percentage of students
7. D;
23(2x 1 1) 1 5x(2x 2 7) 5 26x 2 3 2 5x 2 2 35x
5 25x 2 2 41x 2 3
}
Total
cost
a
3.5 + b 1 2.75 + d a 12
r(t) 5 mt 1 b
8. G;
Cost per
Number of
box of
+ boxes of
doughnuts
doughnuts
}
}
180
who chose comedy is about }
ø 0.31 5 31%. So,
580
about 30% of the students chose comedy.
15. Let a, b, and c be the sides of the original triangle.
* 5 Ï122 1 92 5 Ï 144 1 81 5 Ï225 5 15
P 5 a 1 b 1 c 5 45
S 5 :r* 5 :(12)(15) ø 565.49
Pnew 5 3a 1 3b 13c
The approximate lateral surface area of the cone is
565 square feet.
5 3(a 1 b 1 c)
5 3(45)
5 135
The perimeter of the new triangle is 135 feet.
Geometry
Worked-Out Solution Key
117