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Practice BFor use with the lesson “Use Similar Right Triangles”
Complete and solve the proportion.
1. x } 12
5 ? }
8 2.
15 }
x 5
x }
? 3.
9 }
x 5
x }
?
12
8x
x
2015
9
x
11
Find the value(s) of the variable(s).
4. a
612
5.
x
6 8
6. b
10
24
7.
w 1 9
8 18
8. y
z x
14
16
9. bc
a
32
24
Tell whether the triangle is a right triangle. If so, find the length of the altitude to the hypotenuse. Round decimal answers to the nearest tenth.
10.
14
122 85
11.
1519
6 19 12.
3018
6 34
Use the Geometric Mean Theorems to find AC and BD.
13.
30 40
A
D
B C
14. A
B D
C
610
15.
C D A
B
5
35
Name ——————————————————————— Date ————————————
GeometryChapter Resource Book 7-37
Lesson
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16. Complete the proof.
GIVEN: nXYZ is a right triangle with m∠XYZ 5 908.
ZWY
X
VU
} VW i } XY , } YU is an altitude of nXYZ.
PROVE: nYUZ , nVWZ
Statements Reasons
1. nXYZ is a right n with altitude } YU . 1. ?
2. nXYZ , nYUZ 2. ?
3. } VW i } XY 3. ?
4. ∠VWZ > ∠XYZ 4. ?
5. ∠Z > ∠Z 5. ?
6. ? 6. AA Similarity Postulate
7. nYUZ , nVWZ 7. ?
In Exercises 17–19, use the diagram.
17. Sketch the three similar triangles in the diagram.
J LM
K Label the vertices.
18. Write similarity statements for the three triangles.
19. Which segment’s length is the geometric mean of LM and JM?
20. Kite Design You are designing a diamond-shaped B
A
D
C
kite. You know that AB 5 38.4 centimeters, BC 5 72 centimeters, and AC 5 81.6 centimeters. You want to use a straight crossbar } BD . About how long should it be?
Practice B continuedFor use with the lesson “Use Similar Right Triangles”
Name ——————————————————————— Date ————————————
GeometryChapter Resource Book7-38
Lesson
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.3
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Lesson Use the Converse of the Pythagorean Theorem, continued 7. Given: AX 5 BX 5 AY 5 BY 5 7.5 in. DY 5 19.5 2 7.5 5 12 in.
Then, DX 5 Ï}}
(19.5)2 2 (7.5)2 5 18 in. because n ABD is isosceles. Because the pot is tangent to the planter at X, Y, and Z, CX 5 CY 5 CZ 5 5 in., the radius of the pot.
CD 5 DX 2 CX 5 18 2 5 5 13 in.
(CD)2 5 (CY)2 1 (DY)2
(CD)2 5 (CZ)2 1 (DZ)2
132 5 52 1 122
169 5 25 1 144
So, the screw locations hit the pot exactly where it touches the wall of the planter.
8. 0 < x < 3 1 Ï}
17 9. x > 15 1 2 Ï
}
30 } 5
10. x > 0 11. 22 Ï}
2 < x < 0 or 0 < x < 2 Ï}
2
12. x > 5 1 Ï
} 5 }
20 13. 5 < x <
40 1 10 Ï}
10 }
3
14. Quadrant II or III; Sample answer:
XY 5 Ï}
26 , XZ 5 Î}
106 1 }
4 ; YZ 5
Î}
83 1 }
4
So, n XYZ is not a right triangle by the Converse of the Pythagorean Theorem. The point Q needs to be above or below } PX . By Theorem 3, n XPQ is acute. So, Q has to be in either Quadrant II or III.
Lesson Use Similar Right Triangles
Teaching Guide
1. nBCD ~ nACB, nABD ~ nACB, nBCD ~ nABD 2. about 7.2 ft
3. about 12.1 ft, about 4.3 ft
Practice Level A
1. n ABC , n ADB , n BDC
2. n TSR , n TWS , n SWR 3.
DB
D
CBBACA
4.
W
S W
TSSRTR
5. 2.4 6. 7.1 7. C 8. x }
8 5
8 } 4 ; x 5 16
9. 12
} x 5
x } 3 ; x 5 6 10.
5 }
x 5
x } 4 ; x 5 2 Ï
} 5
11. x 5 8 12. y 5 4 13. z 5 2 14. no
15. yes; 14.4 16. yes; 5.4 17. 5 in.
Practice Level B
1. 12; 18 2. 5; 5 Ï}
3 3. 20; 6 Ï}
5 4. a 5 3
5. x 5 4 Ï}
3 6. b 5 25
} 6 7. w 5 3
8. x 5 12.25, y 5 3.75, z 5 7 Ï
}
15 }
4
9. a 5 42 2 }
3 , b 5 40, c 5 53
1 }
3 10. yes; 9.1
11. no 12. yes; 15.4 13. AC 5 50, BD 5 24
14. AC 5 2, BD 5 Ï}
15 15. AC 5 7, BD 5 Ï}
10
16. 1. Given 2. Theorem 5 3. Given 4. Corresponding Angles Postulate 5. Reflexive Property of Congruence 6. nXYZ , nVWZ 7. Transitive Property
17.
L
J
L
M
MJ
K
K
K
18. nLKJ , nKMJ, nLKJ , nLMK, nKMJ , nLMK 19. } KM 20. about 67.8 cm
Practice Level C
1. n FED , n FGE , n EGD; EG
2. n MQP , n MNQ , n QNP; NQ
3. n RST , n RUS , n SUT; RS
4. m 5 4 5. n 5 3 Ï}
6 6. k 5 4 7. w 5 6
8. y 5 3 9. a 5 14 10. yes; 11.5
11. yes; 17.7 12. no 13. AC 5 34, BD 5 240
} 17
14. AC 5 Ï}
30 , BD 5 15 Ï
}
2 }
2
15. AC 5 62 Ï
}
17 }
17 , BD 5 2 Ï
}
14
16. 1. Given; 2. Geometric Mean (Altitude) Theorem 6; 3. Given; 4. Substitution; 5. Solve for UZ; 6. Pythagorean Theorem; 7. Solve for YZ; 8. Substitution; 9. Simplify
17. Sample Answer:
Because AB
} CB
5 CB
} DB , (BC)2 5 (CB)2 5 AB p BD.
an
sw
er
s
A4GeometryChapter Resource Book
7.2
7.3