Document 1807248

Chapter 7,
continued
AC
CB AB
AB
37. Prove: } 5 } , } 5 }
DB AC
AD
CB
2. By the Base Angles Theorem and the Corollary to the
Statements
1. n ABC is a right triangle.
}
}
hypotenuse 5 leg + Ï 2
1. Given
}
2. CD > AB
2. Given
3. n ABC , nCBD
3. Theorem 7.5
CB
AB
5}
4. }
DB
CB
4. Definition of similar
triangles
5. n ABC , n ACD
5. Theorem 7.5
}
y 5 Ï4
y52
3. By the Base Angles Theorem and the Corollary to the
Triangle Sum Theorem, the two triangles are 458-458-908
triangles.
}
hypotenuse 5 leg + Ï 2
6. Definition of similar
triangles
AC
AB
6. }
5}
AD
AC
}
d 5 8Ï 2 ø 11.31
}
4. hypotenuse 5 leg + Ï 2
}
38. a. a 5 10, b 5 15
6 5 x + Ï2
2(10)(15)
(10 1 15)
2ab
a1b
300
25
6
} 5 } 5 } 5 12
}
} 5 x
Ï2
}
The harmonic mean of 10 and 15 is 12.
Ï2
6
}
} + }
} 5 x
Ï2 Ï2
b. a 5 6, b 5 14
2(6)(14)
6 1 14
2ab
a1b
}
3Ï 2 5 x
168
20
} 5 } 5 } 5 8.4
}
The length of the hypotenuse is 3Ï2 .
}
The harmonic mean of 6 and 14 is 8.4.
5. longer leg 5 shorter leg + Ï 3
}
c. Strings whose lengths have the ratio 4 : 6 :12 will have
x53
2(4k)(12k)
will sound harmonious if 6k 5 }
.
4k 1 12k
}
x 5 Ï3 + Ï3
lengths 4k, 6k, and 12k, for any constant k. The strings
6. The equilateral triangle has an altitude that forms the
longer leg of two 308-608-908 triangles.
2(4k)(12k)
96k 2
} 5 } 5 6k
16k
4k 1 12k
}
longer leg 5 shorter leg + Ï 3
So, 6k is the harmonic mean of 4k and 12k, and the
strings will sound harmonious.
Mixed Review for TAKS
}
h 5 2Ï 3 ø 3.46
7.
h
14 ft
308
39. A;
Cost
Number
of color + of one 1
color
copies
copy
6
+
x
1
10
}
y 5 Ï2 + Ï2
+
x
1
Cost of
Number
Total
of black
+ one black 5
cost
and white
and white
copy
copies
16
+
y
5 1.22
8
+
y
5 1.66
Lesson 7.4
7.4 Guided Practice (pp. 458–460)
1. By the Base Angles Theorem and the Corollary to the
Triangle Sum Theorem, the triangle is a 458-458-908
triangle.
}
hypotenuse 5 leg + Ï 2
}
}
2Ï 2 5 x + Ï 2
25x
hypotenuse 5 2 + shorter leg
14 5 2 + h
75h
The height of the body of the dump truck is 7 feet.
8. In a 308-608-908 triangle, the shorter side of the triangle
is opposite the 308 angle. The longer side of the triangle
is opposite the 608 angle.
7.4 Exercises (pp. 461–464)
Skill Practice
1. A triangle with two congruent sides and a right angle is
called an isosceles right triangle.
2. The Corollary to the Triangle Sum Theorem requires that
the acute angles of a right triangle are complementary.
Because the triangle is isosceles, its base angles are
congruent. Half of 908 is 458, so each of the acute angles
measures 458.
}
3. hypotenuse 5 leg + Ï 2
}
x 5 7Ï 2
204
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
Triangle Sum Theorem, the triangle is 458-458-908
triangle.
Reasons
Chapter 7,
continued
}
4. hypotenuse 5 leg + Ï 2
}
}
5. hypotenuse 5 leg + Ï 2
}
}
x 5 5Ï2 + Ï2
}
12. hypotenuse 5 2 + shorter leg
3Ï 2 5 x + Ï 2
x 5 5 + 2 5 10
f
f 5 2 + d l d 5 }2
35x
}
longer leg 5 shorter leg + Ï 3
6. C;
}
}
hypotenuse 5 leg + Ï 2
}
}
eÏ 3
Ï3
Ï3
e
Ï3
e 5 d + Ï3 l d 5 }
} + }
} 5 }
3
}
7 5 AC + Ï 2
7
}
} 5 AC
Ï2
}
7Ï 2
} 5 AC
}
}
8Ï 3 + Ï3
3
14
2
}57
}58
d
5
e
5Ï 3
7Ï 3
8Ï 3
f
2 + 5 5 10
14
2 + 8 5 16
}
}
}
2
}
7. hypotenuse 5 leg + Ï 2
}
}
2Ï 2 5 xÏ 2
}
25x
The corner triangles have leg lengths of 2 inches. The
overall side length of the tile is 2 + 2 5 4 inches.
} 5 9Ï 3
e
9Ï3 + Ï3 5 27
f
18Ï3
8. hypotenuse 5 2 + shorter leg
y52+9
y 5 18
longer leg 5 shorter leg + Ï 3
}
}
15
x5}
+ Ï3
2
}
3Ï 3 5 x Ï 3
}
15Ï3
35x
x5}
2
y 5 2(3) 5 6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
15
2
longer leg 5 shorter leg + Ï3
}
14. hypotenuse 5 leg + Ï 2
10. hypotenuse 5 2 + shorter leg
}
}
Ï6 5 mÏ2
}
12Ï 3 5 2y
}
Ï6
}
6Ï 3 5 y
}
} 5 m
Ï2
}
}
longer leg 5 shorter leg + Ï3
}
Ï3 5 m
}
}
x 5 6Ï 3 + Ï 3
n 5 m 5 Ï3
x 5 6(3)
15. hypotenuse 5 2 + shorter leg
x 5 18
24 5 2p
11. a 5 b
12 5 p
}
hypotenuse 5 leg + Ï 2
}
longer leg 5 shorter leg + Ï 3
}
}
c 5 aÏ2
q 5 p + Ï3
}
}
a
7
11
}
10 Ï2
}
} + }
} 5 5Ï 2
Ï2 Ï2
b
7
11
5Ï 2
}
}
}
2 + 4Ï3 5 8Ï 3
}
y 5 2x
}
}
12
}5y
9. hypotenuse 5 2 + shorter leg
7Ï 2 11Ï 2
}
15 5 2y
}
x 5 9Ï 3
c
}
}
} 5 4Ï 3
13. hypotenuse 5 2 + shorter leg
}
longer leg 5 shorter leg + Ï3
}
}
12Ï3
3
}
18Ï 3
2
d
10
q 5 12Ï 3
}
6
Ï5
6
Ï5
16. The altitude s forms the longer leg of two
308-608-908 triangles.
}
}
}
}
}
6Ï2 Ï5 + Ï2 5 Ï10
hypotenuse 5 2 + shorter leg
r 5 21 }
22
18
r 5 18
}
longer leg 5 shorter leg + Ï 3
}
s 5 9Ï 3
Geometry
Worked-Out Solution Key
205
Chapter 7,
17.
continued
22. Abigail’s method does work.
3
}
t
longer leg 5 shorter leg + Ï 3
458
}
9 5 xÏ 3
4
}
}
}
9Ï 3 5 x + 3
4
}
u
3Ï 3 5 x
The triangle formed is a 458-458-908 triangle, so each leg
has a length of 4.
u541357
Her method is algebraically correct, so the equation
simplifies to the same answer found in Example 5.
23.
}
hypotenuse 5 leg + Ï 2
308
608
}
f
g
t 5 4Ï2
10
}
longer leg 5 shorter leg + Ï 3
}
}
longer leg 5 shorter leg + Ï 3
}
}
9Ï 3 5 eÏ 3
10 5 gÏ 3
95e
}
10 Ï3
}
} + }
} 5 g
Ï3 Ï3
}
10Ï3
}5g
3
hypotenuse 5 2 + shorter leg
f 5 2(9)
f 5 18
hypotenuse 5 2 + shorter leg
The lower triangle is a 458-458-908 triangle.
}
1 10Ï3 2
}
f52 }
3
hypotenuse 5 leg + Ï 2
}
f 5 gÏ2
}
20Ï3
}
f5}
3
18 5 gÏ2
18
24.
}
} 5 g
Ï2
}
18 Ï 2
}
} + }
} 5 g
Ï2 Ï2
y
x
608
308
1508
4 2
}
9Ï 2 5 g
}
4Ï2 5 2x
5 5Ï3
19. C; }, }, 10
2 2
}
2Ï 2 5 x
hypotenuse 0 2 + shorter leg
}
longer leg 5 shorter leg + Ï 3
}
5
y 5 x + Ï3
10 0 2 + }2
}
}
y 5 2Ï 6
20. The formula for the longer leg was used instead of the
25.
308
hypotenuse formula. The correct solution is:
608
x
8
hypotenuse 5 2 + shorter leg 5 2 + 7 5 14
308
14
608
y
First triangle:
308
hypotenuse 5 2 + shorter leg
8 5 2x
21. The length of the hypotenuse was incorrectly calculated
}
}
}
}
}
as leg + leg + Ï 2 5 Ï 5 + Ï5 + Ï 2 5 5Ï2 . The correct
solution is:
}
}
}
}
hypotenuse 5 leg + Ï 2 5 Ï 5 + Ï 2 5 Ï 10
45x
Second triangle:
}
longer leg 5 shorter leg + Ï 3
}
10
458
5
x 5 yÏ 3
}
4 5 yÏ 3
4
Ï3
}
} 5 y
}
4Ï3
3
}5y
206
}
y 5 2Ï 2 + Ï 3
10 Þ 5
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
hypotenuse 5 2 + shorter leg
}
5
608
308
18. The upper triangle is a 308-608-908 triangle.
7
}
9Ï3 5 xÏ3 + Ï3
458
Chapter 7,
continued
}}}
26. CB 5 Ï (3 2 (23))2 1 (21 2 (21))2
}
hypotenuse 5 2 + shorter leg
28.
}
5 Ï 62 1 02 5 Ï 36 5 6
h ft
608
284 ft
hypotenuse 5 2 + shorter leg
284 5 2 + h
142 5 h
308
6 5 2AB
}
458
284 5 h + Ï 2
284 ft
h ft
3 5 AB
284
}
}
} 5 h
Ï2
458
longer leg 5 shorter leg + Ï 3
}
142Ï2 5 h
}
AC 5 AB + Ï 3
hypotenuse 5 2 + shorter leg
}
AC 5 3Ï3
284 5 2 + s
Let (x, y) represent the coordinates of A.
h ft 308 284 ft
}}
AB 5 Ï(x 2 3)2 1 ( y 2 (21))2
142 5 s
3 5 Ï(x 2 3) 1 ( y 1 1)
2
2
608
}
h 5 142Ï3
9 5 (x 2 3)2 1 ( y 1 1)2
h ø 246
9 2 (x 2 3) 5 ( y 1 1)
2
2
When the angle is 308, the seagull rises 142 feet;
when the angle is 458, the seagull rises about 200 feet
10 inches; when the angle is 608, the seagull rises about
246 feet.
}}}
AC 5 Ï(x 2 (23))2 1 ( y 2 (21))2
}}
3Ï3 5 Ï(x 1 3)2 1 ( y 1 1)2
27 5 (x 1 3)2 1 ( y 1 1)2
29. You could show that all isosceles right triangles are
27 2 (x 1 3)2 5 (y 1 1)2
similar to each other by showing that the corresponding
angles are congruent. They are all 458-458-908 triangles.
9 2 (x 2 3)2 5 27 2 (x 1 3)2
9 2 x 2 1 6x 2 9 5 27 2 x 2 2 6x 2 9
You could also show that the corresponding side lengths
are always proportional, because in an isosceles }right
triangle, the side lengths are always x, x, and xÏ2 .
}
}
For example, let 1, 1, Ï2 , and 2, 2, 2Ï 2 , be the side
lengths of two isosceles right triangles. The ratios of
12x 5 18
3
x 5 }2
9 2 1 }2 2 3 2 5 (y 1 1)2
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3
2
}
1 2
9
2Ï2
}
} 5 2.
Ï2
30. Because nDEF is a 458-458-908 triangle, DF 5 FE. By
the Pythagorean Theorem, DF 2 1 FE 2 5 DE 2. This
2
equation can be written as 2DF
5 DE 2 or 2FE 2 5 DE 2.
}
Ï
By
the Property of Squares, 2 DF 5 DE and }
}
Ï 2 FE 5 DE. So, the hypotenuse is equal to Ï 2 times the
length of one leg of the 458-458-908 triangle.
9 2 }4 5 y 2 1 2y 1 1
23
0 5 y 2 1 2y 2 }
4
0 5 4y 2 1 8y 2 23
}}
28 6 Ï82 2 4(4)(223)
y 5 }}
2(4)
}
28 6 Ï 432
5}
8
}
3Ï 3
5 21 6 }
2
Point A lies in the first quadrant, so its y-coordinate
is positive.
}
1
2
corresponding side lengths are }1 5 2, }1 5 2, and
3 2
9 2 2}2 5 y 2 1 2y 1 1
3Ï 3
3
A }2 , 21 1 }
2
}
longer leg 5 shorter leg + Ï3
}}
}
}
hypotenuse 5 leg + Ï2
CB 5 2AB
2
Problem Solving
31.
20 in.
h
608
608
20 in.
h
308
608
20 in.
608
10 in.
20 in.
The height h divides the equilateral triangle into two
308-608-908 triangles.
}
longer leg 5 shorter leg + Ï3
}
h 5 10Ï3 ø 17.3
The height of the equilateral triangle is about
17.3 inches.
27. hypotenuse 5 2 + shorter leg
11 5 2h
11
2
}5h
5.5 5 h
The height of the ramp is 5 feet 6 inches.
Geometry
Worked-Out Solution Key
207
Chapter 7,
This process can be duplicated to find the unknown
lengths of the remaining right triangles.
K
608 x
308
L
t 2 5 12 1 s 2
x
t 5 1 1 (Ï3 )
M
t2 5 1 1 3
Because n JML > n JKL, mŽ M 5 mŽ K 5 608. So, all
three angles of n JKM measure 608, and the triangle is
equilateral. So, all of its side lengths are x 1 x 5 2x. It
is given that the shorter leg of n JKL is x. So, n JKL’s
hypotenuse is two times the length of its shorter side.
By the Pythagorean Theorem, JL2 1 x 2 5 (2x)2, }
and
JL2 5 4x 2 2 x 2 5 3x 2. This simplifies to}JL 5 Ï3 x,
showing that the longer leg is equal to Ï3 times the length
of the shorter leg.
33. a. The large orange triangles are 458-458-908 triangles,
because they are isosceles right triangles. The smaller
blue triangles are also 458-458-908 triangles, because
the 458 angles of the orange triangle are complementary
with the acute angles of the blue triangle.
v2 5 1 1 5
w2 5 1 1 6
v2 5 6
w2 5 7
308-608-908 triangle. It is the only one whose side
lengths satisfy the 308-608-908 Triangle Theorem.
35. a.
Q
8
b. QR > SR, QT > ST, and TR is shared by the two
triangles so nRQT > nRST by the Side-Side-Side
Congruence Postulate.
c. RT is longer than QS. ZR is the longer leg of a
308-608-908 triangle, so ZR > ZS. From part (b), nRQT
> nRST, so Ž QTR > Ž STR. Because
mŽ QTS 5 908, mŽ QTR 5 mŽ STR 5 458. Because
nQTS is an isosceles right triangle, its base angles
measure 458. So, nQZT and nTZS are congruent right
isosceles triangles, and QZ 5 TZ 5 ZS.
So, ZR 1 TZ > ZS 1 QZ, and RT > QS.
}
hypotenuse 5 leg + Ï 2
}
}
3Ï 2
} 5 y + Ï2
2
3
2
}5y
458
1.5 5 y
The square of fabric for the small blue triangles should
be 1.5 inches by 1.5 inches.
34. a. The first triangle is a 458-458-908 triangle, because the
legs are both 1.
}
hypotenuse 5 leg + Ï 2
}
r 5 1 + Ï2
}
r 5 Ï2
The }
second triangle is a right triangle with legs 1 and r
5 Ï 2 . Using the Pythagorean Theorem,
s 2 5 12 1 r 2
}
s 2 5 1 1 (Ï 2 )2
s2 5 1 1 2
R
S
The square of fabric for the large orange triangles
should be about 2.12 inches by 2.12 inches.
3 2
2
Z
T
8
}
458
Mixed Review for TAKS
36. B;
Choice A: 142 0 82 1 122
196 0 64 1 144
196 Þ 208
The triangle is not a right triangle.
Choice B: 292 0 202 1 212
841 0 400 1 441
841 5 841 The triangle is a right triangle.
Choice C: 162 0 92 1 122
256 0 81 1 144
256 Þ 225
The triangle is not a right triangle.
2
}
s 5 Ï3 .
Geometry
Worked-Out Solution Key
}
w 5 Ï7
}
3
y
}
}
c. The third triangle, with sides 1, Ï 3 , 2, is the only
2.12 ø x
208
w 2 5 12 1 (Ï6 )2
first one, because it is the only one whose legs are equal
in length.
}
s 53
v 2 5 1 1 (Ï 5 )2
b. The only triangle that is a 458-458-908 triangle is the
}
} 5 x
Ï2
3Ï 2
}5x
2
y
w 2 5 12 1 v 2
608
Ï2
3
}
} + }
} 5 x
Ï2 Ï2
c.
v 2 5 12 1 u 2
v 5 Ï6
}
458
}
u 5 Ï5
}
3 5 x + Ï2
3
x
u2 5 5
t52
}
458
u 2 5 1 1 22
t 54
hypotenuse 5 leg + Ï 2
x
u 2 5 12 1 t 2
2
2
Construct n JML congruent to n JKL. Because they are
congruent, mŽ KJL 5 mŽ LJM 5 308, so mŽ KJM 5
mŽ KJL 1 mŽ LJM 5 308 1 308 5 608.
b.
}
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
32.
continued
Chapter 7,
continued
}
Choice D: 252 0 102 1 242
5. longer leg 5 shorter leg + Ï 3
}
3Ï2 5 aÏ3
625 Þ 676
3Ï2
}
} 5 a
}
The triangle is not a right triangle.
}
The triangle given in Choice B, a triangle with lengths
20 inches, 21 inches, and 29 inches, is a right triangle.
Quiz 7.3–7.4 (p. 464)
1.
Ï3
}
3Ï2 Ï3
}
} + }
} 5 a
Ï3 Ï3
}
Ï6 5 a
hypotenuse 5 2 + shorter leg
b 5 2a
C
B
10
6
B
A
D
A
D
BC
CD
}5}
BC 2 5 CD 2 1 BD 2
102 5 6 2 1 BD 2
64 5 BD
40 in.
402 5 152 1 h2
15 in.
30 in.
1600 5 225 1 h2
h
40 in.
1375 5 h2
1
Area 5 }2(base)(height) ø }2(30)(37.08) 5 556.2 in.2
8 5 BD
Larger sign:
64 in.
642 5 242 1 h2
24 in.
4096 5 576 1 h2
8
6
}5}
h
48 in.
3520 5 h2
6 + AD 5 8 + 8
64 in.
59.33 ø h
64
AD 5 }
6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1. D;
1
2
BD
AD
}5}
CD
BD
1
1
Area 5 }2(base)(height) ø }2(48)(59.33) 5 1423.92 in.2
32
Ratio of areas: 556.2 : 1423.92 ø 0.3906 : 1
AD 5 }
ø 10.67
3
AB
BD
Mixed Review for TEKS (p. 465)
B
37.08 ø h
100 5 36 1 BD 2
AD
8
10
Smaller sign:
BC is the geometric mean of AC and CD.
2.
}
b 5 2Ï 6
C
AC
BC
}
625 0 100 1 576
BC
CD
}5}
10
AB
}5}
6
8
6 + AB 5 8 + 10
3:4
5
Choice B:
5:8
5 0.625 : 1
Choice C:
9 : 16
5 0.5625 : 1
Choice D:
25 : 64
ø 0.3906 : 1
0.75 : 1
The ratio of the area of the smaller sign to the area of the
larger sign is about 25 : 64.
6 + AB 5 80
80
Choice A:
40
5}
ø 13.3
AB 5 }
6
3
}
3. hypotenuse 5 leg + Ï 2
}
x 5 8Ï2
}
10 5 y + Ï 2
10
}
} 5 y
Ï2
}
10Ï 2
}5y
2
Using the formula d 5 r + t (Distance 5 Rate + Time),
t 5 1.5 h.
Sandra:
}
4. hypotenuse 5 Ï 2
2. H;
Tina:
r 5 5 mi/h
r 5 4 mi/h
d 5 (5)(1.5) 5 7.5 mi
d 5 (4)(1.5) 5 6 mi
Because Sandra runs west and Tina runs south, their
paths are legs of a right triangle.
N
7.5 mi
}
W
5Ï 2 5 y
E
S
c
6 mi
c2 5 62 1 7.52 5 36 1 56.25 5 92.25
}
c 5 Ï 92.25 ø 9.6
At 10:30 A.M., Sandra and Tina are about 9.6 miles apart.
Geometry
Worked-Out Solution Key
209
Chapter 7,
continued
3. C;
Step 3
Because 10 < x < 14, x is the length of the longest side of
the triangle. The sum of the squares of the lengths of the
other two sides is 82 1 62 5 64 1 36 5 100. Because
10 < x < 14, 100 < x 2 < 196, so the square of the length
of the longest side is greater than the sum of the squares
of the lengths of the other two sides. So, by Theorem 7.4,
the triangle is obtuse. Therefore, Ž1 is obtuse.
4. F;
The triangle formed by the pushpins is not equilateral
because the 3 sides formed are not congruent.
? 1192 1 2082
? 14,161 1 43,264
50,625
50,625 < 57,425
Because the square of the length of the longest side of the
triangle is less than the sum of the squares of the lengths
of the other 2 sides, by Theorem 7.3 the triangle formed
is acute.
5. C;
n ABC , n ADE. The two triangles are similar because their
corresponding angles are congruent.
Step 4
Sample answer: The ratio of the lengths of the legs in a
right triangle is constant for a given angle measure; answers
will vary.
7.5 Guided Practice (pp. 467–468)
opp. ŽJ
3
24
1. tan J 5 } 5 } 5 } 5 0.75
32
4
adj. to ŽJ
opp. ŽK
32
4
5}
5 }3 ø 1.3333
tan K 5 }
24
adj. to ŽK
opp. ŽJ
8
2. tan J 5 } 5 } ø 0.5333
15
adj. to ŽJ
opp. ŽK
15
tan K 5 }
5}
5 1.875
8
adj. to ŽK
opp.
3.
tan 618 5 }
adj.
x 2 5 32 1 22
22
tan 618 5 }
x
x 5914
2
x + tan 618 5 22
x 2 5 13
}
x 5 Ï13 ø 3.6
22
tan 618
x5}
The length of the plywood is about 3.6 feet.
6. 51;
22
xø}
ø 12.2
1.8040
N
24
Treasure
(24, 45)
Stump
opp.
tan 568 5 }
adj.
4.
x
(30, 20)
45
W
DE
BC
tan 568 5 }
13
(24, 20)
(0, 0)
(30, 0)
13 + tan 568 5 x
E
13(1.4826) ø x
S
If the stump is at point (0, 0) then the hidden treasure is
at point (24, 45). The shortest distance from the treasure
}
}
to the stump is Ï 452 1 242 5 Ï2601 5 51 paces.
19.3 ø x
5.
608
5
5 3
Lesson 7.5
}
longer leg 5 shorter leg + Ï 3
}
Activity (p. 466)
x 5 5Ï 3
Step 1
opp.
}
5Ï3
}
tan 608 5 }
5}
5 Ï3
5
adj.
Check student’s work.
Step 2
7.5 Exercises (pp. 469–472)
Sample answer:
210
Triangle
Adjacent
Leg
Opposite
Leg
}}
n ABC
5 cm
2.9 cm
0.58
n ADE
10 cm
5.8 cm
0.58
n AFG
15 cm
8.7 cm
0.58
Opposite Leg
Adjacent Leg
Skill Practice
1. The tangent ratio compares the length of the leg opposite
the angle to the length of the leg adjacent to the angle.
2. All right triangles with an acute angle measuring n8 will
Geometry
Worked-Out Solution Key
have the same ratio of the leg opposite the angle to the
leg adjacent to the angle because tangent n8 is a constant.
So, the triangles must be similar.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2252
AC
BC
5}
and }
5}
are true because
The proportions }
AE
AE
DE
AC