...

Document 1807233

by user

on
Category: Documents
2

views

Report

Comments

Transcript

Document 1807233
Chapter 7
1. P 5 36 units2
Prerequisite Skills (p. 430)
1. The triangle is an equilateral triangle.
Q 5 18 units2
2. The triangle is a right triangle.
R 5 9 units2
3. The triangle is an acute triangle.
S 5 18 units2
4. The triangle is an obtuse isosceles triangle.
T 5 18 units2
}
}
}
Sum of areas 5 2P 1 Q 1 2R 1 S 1 T
5. Ï 45 5 Ï 9 + 5 5 3Ï 5
}
}
6. (3Ï 7 ) 5 3 (Ï 7 ) 5 9(7) 5 63
5 2(36) 1 18 1 2(9) 1 18 1 18
7. Ï 3 + Ï 5 5 Ï 15
5 144 units2
}
2
2
2
}
}
}
}
7 + Ï2
7Ï 2
7
8. }
}
} 5 }
} 5 }
2
Ï2 + Ï2
Ï2
12
3
}
9.
x5}
16
The side lengths of the squares are equal to the lengths of
the two legs of Triangle P.
x
2
10. } 5 }
18
3
12x 5 48
3x 5 36
x54
x 5 12
1
x15
}5}
2
4
11.
12.
6
x14
}5}
5
x24
2(x 1 5) 5 4
5(x 1 4) 5 6(x 2 4)
2x 1 10 5 4
5x 1 20 5 6x 2 24
2x 5 26
44 5 x
x 5 23
2. Area of square in Step 4 5 2P 1 Q 1 2R 1 S 1 T 5
144 units2. The side length of this square is equal to the
length of the hypotenuse of Triangle P.
3. The sum of the areas of the two squares in Step 3 is
equal to the area of the square in Step 4. So, the sum of
the squares of the legs of a right triangle is equal to the
square of the hypotenuse.
4. The legs of Triangle P are congruent, and they meet to
form a right angle. The conjecture in Exercise 3 is not
true for all isosceles triangles.
For example:
Lesson 7.1
Investigating Geometry Activity 7.1 (p. 432)
Step 3:
R
S
Q
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
T
The conjecture is true for all right triangles. In the figures
below, you can see that A 1 B 1 4T 5 C 1 4T. So,
A 1 B 5 C.
R
P
P
C
A
Step 4:
T
A
T
T
T
T
C
T
T
B
T
T
B
T
P
R
Q
S
P
R
7.1 Guided Practice (pp. 433–436)
1. The unknown side is a leg.
52 5 x 2 1 32
5 2 32 5 x2
2
25 2 9 5 x 2
16 5 x 2
45x
2. The unknown side is a hypotenuse.
x 2 5 62 1 42
x 2 5 36 1 16
x 2 5 52
}
x 5 2Ï 13
Geometry
Worked-Out Solution Key
189
Chapter 7,
continued
3. (Length of ladder)2 5
7.1 Exercises (pp. 436–439)
(Distance from house)2 1 (Height of ladder)2
Skill Practice
x 2 5 62 1 232
1. A set of three positive integers a, b, and c that satisfies
x 2 5 36 1 529 5 565
the equation c 2 5 a 2 1 b 2 is called a Pythagorean triple.
}
The length of the ladder is about 23.8 feet.
4. The Pythagorean Theorem is only true for right triangles.
1 2
2. In order to use the Pythagorean Theorem, you must have
the lengths of two of the sides of a right triangle.
3. x 2 5 502 1 1202
1
5. The base of each right triangle is } (30) 5 15.
2
x 2 5 2500 1 14,400
x 2 5 16,900
182 5 152 1 h2
324 5 225 1 h2
99 5 h2
x 5 130
4. x 2 5 332 1 562
}
3Ï 11 5 h
x 2 5 1089 1 3136
x 2 5 1600 1 1764
x 5 4225
x2 5 3364
2
1
Area 5 }2 (base)(height)
}
1
5 }2(30)(3Ï11 )
x 5 65
x 5 58
6. In the Pythagorean Theorem, b and c were
substituted incorrectly.
}
5 45Ï11
a 2 1 b 2 5 c2
ø 149.25
102 1 242 5 262
The area of the triangle is about 149.25 square feet.
7. It is not algebraically correct to simplify 72 1 242
as (7 1 24)2. The solution should be:
6. The base of each right triangle is 10 m.
262 5 102 1 h2
x 2 5 72 1 242
676 5 100 1 h2
x 2 5 49 1 576
576 5 h2
x 2 5 625
24 5 h
x 5 25
1
1
Area 5 }2 (base)(height) 5 }2 (20)(24) 5 240
8.
199.68 5 h2
7. Using the Pythagorean Theorem:
14.13 ø h
x 2 5 92 1 122
x 2 5 81 1 144
x 2 5 225
The height of the fire escape landing is about 14.13 feet.
9.
5 + 3 5 15
So, the length of the hypotenuse is 15 inches.
8. Using the Pythagorean Theorem:
x 2 5 142 1 482
x 2 5 196 1 2304
x 5 2500
2
x 5 50
Using a Pythagorean triple: a common triple is 7, 24, 25.
Multiply each number by 2.
7 + 2 5 14
24 + 2 5 48
25 + 2 5 50
So, the length of the hypotenuse is 50 centimeters.
Geometry
Worked-Out Solution Key
13.42 5 9.82 1 h2
179.56 5 96.04 1 h2
x 5 15
Using a Pythagorean triple: A common triple is 3, 4, 5.
Multiply each number by 3.
3+359
4 + 3 5 12
16.72 5 8.92 1 h 2
278.89 5 79.21 1 h 2
The area of the triangle is 240 square meters.
190
5. x 2 5 402 1 422
83.52 5 h2
9.14 ø h
The height of the backboard frame is about 9.14 inches.
10.
5.72 5 4.92 1 b 2
32.49 5 24.01 1 b 2
8.48 5 b 2
2.9 ø b
The base of the frame is about 2.9 feet.
1
11. The base of each right triangle is } (16) 5 8 meters.
2
172 5 82 1 h 2
289 5 64 1 h2
225 5 h2
15 5 h
1
1
Area 5 }2 (base)(height) 5 }2 (16)(15) 5 120
The area of the triangle is 120 square meters.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 5 Ï 565 ø 23.8
Chapter 7,
continued
1
12. The base of each right triangle is } (32) 5 16 feet.
2
Multiply the lengths by 5.
202 5 162 1 h2
3 + 5 5 15
400 5 256 1 h2
4 + 5 5 20
144 5 h2
5 + 5 5 25
12 5 h
1
1
Area 5 }2 (base)(height) 5 }2 (32)(12) 5 192
The area of the triangle is 192 square feet.
13. The base of each right triangle is 6 centimeters.
10 5 6 1 h
2
2
2
100 5 36 1 h2
64 5 h
2
The missing side is a leg with length of 15.
20. 28 and 96; A common Pythagorean triple is 7, 24, 25.
Multiply the lengths by 4.
7 + 4 5 28
24 + 4 5 96
25 + 4 5 100
The missing side is a hypotenuse with length of 100.
85h
21. 20 and 48; A common Pythagorean triple is 5, 12, 13.
1
1
Area 5 }2 (base)(height) 5 }2 (12)(8) 5 48
Multiply the lengths by 4.
The area of the triangle is 48 square centimeters.
12 + 4 5 48
14. A common triple is 7, 24, 25. Multiply each length by 3.
7 + 3 5 21
Multiply the lengths by 5.
So, x 5 75.
8 + 5 5 40
15. A common triple is 3, 4, 5. Multiply each length by 10.
3 + 10 5 30
15 + 5 5 75
17 + 5 5 85
4 + 10 5 40
The missing side is a leg with length of 40.
5 + 10 5 50
23. 72 and 75; A common Pythagorean triple is 7, 24, 25.
So, x 5 40.
Multiply the lengths by 3.
16. A common triple is 8, 15, 17. Multiply each length by 4.
8 + 4 5 32
7 + 3 5 21
24 + 3 5 72
15 + 4 5 60
25 + 3 5 75
17 + 4 5 68
The missing side is a leg with length of 21.
So, x 5 32.
24. x 2 5 6 2 1 32
17. B;
2
2
c 5 8 1 15
2
2
c 5 64 1 225
2
c 2 5 289
c 5 17
The length of the hypotenuse is 17 inches.
18. 24 and 51; A common Pythagorean triple is 8, 15, 17.
Multiply the lengths by 3.
8 + 3 5 24
15 + 3 5 45
17 + 3 5 51
25. x 2 5 112 1 112
x 5 36 1 9
x 2 5 121 1 121
x 2 5 45
x 2 5 242
2
c 5a 1b
2
13 + 4 5 52
22. 75 and 85; A common Pythagorean triple is 8, 15, 17.
25 + 3 5 75
2
5 + 4 5 20
The missing side is a hypotenuse with length of 52.
24 + 3 5 72
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
19. 20 and 25; A common Pythagorean triple is 3, 4, 5.
}
x 5 3Ï 5
}
x 5 11Ï2
26. Let h represent the height.
52 5 32 1 h 2
25 5 9 1 h 2
16 5 h 2
45h
x 2 5 42 1 72
x 2 5 16 1 49
x 2 5 65
}
x 5 Ï 65
The missing side is a leg with length of 45.
Geometry
Worked-Out Solution Key
191
Chapter 7,
continued
So, b1 5 14 2 9 5 5.
27. A;
392 5 152 1 b 2
132 5 52 1 x 2
1521 5 225 1 b 2
169 5 25 1 x 2
1296 5 b
144 5 x 2
2
36 5 b
12 5 x
1
1
Area 5 }2 (base)(height) 5 }2 (15)(36) 5 270
Problem Solving
The area of the triangle is 270 square feet.
31.
x 2 5 902 1 902
2nd base
x 2 5 8100 1 8100
(4x 2 4)2 5 (2x)2 1 (2x 1 4)2
x 2 5 16,200
16x 2 2 32x 1 16 5 4x 2 1 4x 2 1 16x 1 16
x ø 127.3
x
8x2 2 48x 5 0
8x(x 2 6) 5 0
x50
90 ft
or x 5 6
Home plate
Because x cannot be zero, the value of x is 6.
29.
x
u6
s
36
10
t
A ball thrown from home plate to second base must go
about 127.3 feet.
102 5 62 1 x 2
32.
100 5 36 1 x 2
r
10 ft
64 5 x 2
x
9
85x
39
392 5 362 1 r 2
152 5 92 1 s 2
1521 5 1296 1 r
225 5 r
2
225 5 81 1 s
2
144 5 s 2
2
15 5 r
12 5 s
102 5 62 1 t 2
6 ft
The balloon is 8 feet above the ground.
33. The hypotenuse is 65, because it must be the longest of
the three sides.
34. a.
6400 5 1225 1 x 2
100 5 36 1 t 2
5175 5 x 2
64 5 t 2
71.9 ø x
85t
P ø 35 1 80 1 71.9 ø 186.9
u 5 s 2 t 5 12 2 8 5 4
The perimeter of the field is about 186.9 feet.
x 2 5 62 1 42
186.9 4 10 5 18.7
You will need about 19 dogwood seedlings.
}
x 5 2Ï13
b1 1 b2 5 14
30.
(14 2 b2)2 1 x 2 5 132
13
and b22 1 x 2 5 152
15
x
b1
(14 2 b2)2 1 x 2 5 132
196 2 28b2 1 b22 1 x 2 5 169
228b2 1 b22 1 x 2 5 227
28b2 2 b22 2 x 2 5 27
Solve the system of equations.
28b2 2 b22 2 x 2 5 27
b22 1 x 2 5 225
5 252
b2 5 9
192
c.
Number of dogwoods 3 cost of dogwoods 5 total cost
19
3
228
The trees will cost $228.
b2
14
28b2
P 4 10 5 Number of dogwoods to plant
b.
x 2 5 36 1 16
x 2 5 52
802 5 352 1 x 2
Geometry
Worked-Out Solution Key
12
5 total cost
5 total cost
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
28.
Chapter 7,
continued
35. a–b.
BC
AC
CE
}
}
Ï102 1 602 ø 60.8
10
Ï1102 1 302 ø 114
}
20
Ï20 1 60 ø 63.2
30
Ï302 1 602 ø 67.1
2
2
Ï802 1 302 ø 85.4
Ï702 1 302 ø 76.2
}
60
Ï60 1 60 ø 84.9
70
Ï702 1 602 ø 92.2
Ï60 1 30 ø 67.1
2
2
2
4. f 2 5 a 2 1 e 2
4. Substitution Property of
Equality
5. a 2 1 e 2 5 d 2 1 e 2
5. Substitution Property of
Equality
6. a 2 5 d 2
6. Subtraction Property of
Equality
7. a 5 d
} }
8. BC > EF
7. Definition of square roots
154.3
152
}
Ï502 1 302 ø 58.3
}
150.5
}
Ï802 1 602 5 100
Ï402 1 302 5 50
}
150
}
Ï902 1 602 ø 108.2
90
157.5
}
}
80
3. Given
}
Ï502 1 602 ø 78.1
2
3. b 5 e, c 5 f
}
}
Ï302 1 302 ø 42.4
}
150.6
}
100 Ï1002 1 602 ø 116.6
Ï202 1 302 ø 36.1
}
Ï102 1 302 ø 31.6
}
156.9
Ï02 1 302 5 30
164.2
b. The shortest distance that you must travel is 150 feet.
9. Right Angles Congruence
Theorem
10. n ABC > nDEF
10. SAS Congruence Postulate
38. Row 1
A
5 ft
100 ft
60 ft
E
50 ft
C
B
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
9. Ž ACB > Ž DFE
}
120 Ï1202 1 602 ø 134.2
36.
8. Definition of congruent
segments
152.7
}
110 Ï1102 1 602 ø 125.3
c.
1. n ABC and nDEF 1. Given
are right triangles.
162
}
Ï402 1 602 ø 72.1
120 ft
y
D
(x1, y1)
y2 2
5 ft
If the trees are staggered so that trees in row 2 are
halfway between the trees in row 1, at the minimum
distance of 5 feet, an equilateral triangle is formed with
each leg having a length of 5 feet.
Bisect the base to form two right triangles.
x2 2 x1
52 5 2.52 1 h2
O
5 ft
h
5 ft
By the Pythagorean Theorem,
2.5 ft
(x2 2 x1)2 1 (y2 2 y1)2 5 d 2.
}}
So, d 5 Ï (x2 2 x1)2 1 (y2 2 y1)2 .
37. Given: n ABC and nDEF
are right triangles; b 5 e, c 5 f
Prove: n ABC > nDEF
A
c
f
e
2.5 ft
25 5 6.25 1 h2
18.75 5 h2
4.3 ø h
The minimum distance between the rows is
about 4.3 feet.
Mixed Review for TAKS
39. D;
If Rob can unload a box in 5 minutes and Jen can unload
a box in 3 minutes, then to find whether they can unload
all of the boxes in 30 minutes if they work together, the
total number of boxes is also needed.
D
b
5 ft
Row 2
30 ft
(x2,
d
Reasons
2. Pythagorean Theorem
2
Ï902 1 302 ø 94.9
Statements
2. c 2 5 a 2 1 b 2,
f 2 5 d 2 1 e2
Ï100 1 30 ø 104.4 167.6
2
}
50
174.8
}
}
40
AC 1
CE
40. G;
C
a
B
F
d
E
Determine which of the 4 shapes gives a perimeter of
about 90 feet and encloses the largest area.
Choice F, rectangle with a length of 30 feet and a width
of 15 feet:
P 5 2* 1 2w 5 2(30) 1 2(15) 5 60 1 30 5 90 feet
A 5 *w 5 30(15) 5 450 square feet
Choice G, a square with a length of 22.5 feet:
P 5 4s 5 4(22.5) 5 90 feet
Geometry
Worked-Out Solution Key
193
Chapter 7,
continued
A 5 s 2 5 22.52 5 506.25 square feet
Choice H, an equilateral triangle with a side length of
30 feet:
7.2 Guided Practice (pp. 441–443)
}
1. 82 0 42 1 (4Ï 3 )2
64 0 16 1 48
P 5 3s 5 3(30) 5 90 feet
s 5 30 ft
2. 142 0 102 1 112
196 0 100 1 121
15 ft
196 Þ 221
302 5 152 1 h2
900 5 225 1 h2
The triangle is not a right triangle.
}
3. (Ï 61 )2 0 52 1 62
675 5 h2
61 0 25 1 36
26 ø h
61 5 61 1
1
A 5 }2 sh ø }2 (30)(26) ø 390 square feet
Choice J, an isosceles right triangle with a leg length of
26 feet:
The triangle is a right triangle.
4. 3 1 4 5 7
31659
7>6
9>4
c 2 ? a2 1 b2
62 ? 32 1 42
36 ? 9 1 16
b 5 26 ft
36 25
c 2 5 262 1 262 5 676 1 676 5 1352
The triangle is obtuse.
}
c 5 Ï1352 ø 37
P 5 a 1 b 1 c ø 26 1 26 1 37 ø 89 feet
5. No, triangles with side lengths 2, 3, and 4 could not be
used to verify that you have perpendicular lines, because
the side lengths do not form a right triangle.
1
A 5 }2(26)(26) 5 338 square feet
The shape given in Choice G, a square with a length
of 22.5 feet uses 90 feet of fencing and encloses the
largest area.
42 Þ 22 1 32
7.2 Exercises (pp. 444–447)
Skill Practice
Lesson 7.2
1. The longest side of a right triangle is called a hypotenuse.
Investigating Geometry Activity 7.2 (p. 440)
2. The side lengths of a triangle can be used to classify
a triangle as acute, obtuse, or right by comparing the
square of the length of the longest side to the sum
of the squares of the lengths of the two other sides.
If c 2 5 a 2 1 b 2, the triangle is a right triangle. If
c 2 > a 2 1 b 2, the triangle is an obtuse triangle.
If c 2 < a 2 1 b 2, the triangle is an acute triangle.
1. If a triangle is a right triangle, then the square of the
length of the hypotenuse is equal to the sum of the
squares of the lengths of the legs.
If the square of the length of the longest side of a triangle
is equal to the sum of the squares of the lengths of the
other two sides, then the triangle is a right triangle.
2. The converse of the Pythagorean Theorem is true.
Because the theorem can be stated as an equation, it will
be true in either direction.
3. Let C be the largest angle in n ABC. If mŽ C < 908,
then AB 2 < AC 2 1 CB 2. If mŽ C 5 908, then AB 2 5
AC 2 1 CB 2. If mŽC > 908, then AB 2 > AC 2 1 CB 2.
4. If AB 2 > AC 2 1 CB 2, then the triangle is an
obtuse triangle.
5. If AB 2 < AC 2 1 CB 2, then the triangle is an
acute triangle.
6. If AB 2 5 AC 2 1 CB 2, then the triangle is a
right triangle.
3.
972 0 652 1 722
9409 0 4225 1 5184
9409 5 9409 The triangle is a right triangle.
4. 232 0 11.42 1 21.22
529 0 129.96 1 449.44
529 Þ 579.4
The triangle is not a right triangle.
}
5. (3Ï 5 ) 2 0 22 1 62
45 0 4 1 36
45 Þ 40
The triangle is not a right triangle.
194
10 > 3
The side lengths 3, 4, and 6 can form a triangle.
c
a 5 26 ft
4 1 6 5 10
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
h
64 5 64 The triangle is a right triangle.
Fly UP