Lesson Warm Up 39 1. exterior 2. 78° 3. C Lesson Practice 39 a ...
Lesson
39
Warm Up 39 1. exterior 2. 78° 3. C Lesson Practice 39 _ _ _
a. EF , DE , DF b. ∠Q, ∠P, ∠R c. a = 120°. Using the Exterior Angle Theorem and the given values of 50° for vertex Y and 70° for vertex X, it can clearly be seen that a is larger than the angle at either of the remote interior angles. d. 50 ft < x < 380 ft
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 39–1
Saxon Geometry
Lesson Practice 39
39
8. ST and UV each equal 9 units, TU and SV each equal 3 units.
1. Since ∠G and ∠K are right angles, GHJ and KLM are_ right triangles; legs GH and _ KL are congruent; and acute angles ∠H and ∠L are congruent; so by Leg-Angle Congruence Theorem, GHJ KLM.
9. To show that AB = CD, the Transitive Property must be used, not the Reflexive Property. 10.
N
O C
_ _ _
M
2. BC , AB , AC 3. 7.1 in.
11. For any two sections, by definition of congruent segments, both pairs of legs are congruent; by the Leg-Leg Congruence Theorem, the sections are congruent.
4. 5 cameras 5. The orthocenter is located on the vertex of the right angle; yes.
12. 143 ft2 orthocenter
13. (0.67, -4)
6. D
14. ∠RTS
7. Since 34 + 14 < 51, these three side lengths cannot form a triangle. © 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
x+7 15. _ 2x
16. B
LSN 39–2
Saxon Geometry
Lesson
39
17. 3796 ft2; See student work; Sample: The buildings will be arranged to form a park that is in the shape of a right triangle. Based on proof model of the Pythagorean Theorem, the sum of the squares sharing the sides with the legs of a right triangle are equal to the sum of the square sharing the hypotenuse side, which is the third building. 18. approximately 35.71 19. The central angles will always sum to 360°, so as the number of sides increases, the number of central angles will increase so that there are fewer degrees for each angle. 20.
Statements 1. 16 = 4(3x - 8) 4(3x - 8) 16 =_ 2. _ 4
4
Reasons 1. Given 2. Division Property of Equality
3. 4 = 3x- 8 4. 4 + 8 = 3x - 8 + 8 5. 12 = 3x 3x 12 = _ 6. _
6. Division Property of Equality
7. 4 = x 8. x = 4
7. Simplify. 8. Symmetric Property of Equality
3
3
3. Simplify. 4. Addition Property of Equality 5. Simplify.
21. y = 2x + 1 113 22. x > √ 23. 21.38 cm © 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 39–3
Saxon Geometry
Lesson 24.
39
2x + 1 = 15 Given 2x + 1 - 1 = 15 - 1 Subtraction Property of Equality 2x = 14 Simplify. 2x 14 _ _ = Division Property of Equality 2 2 x=7 Simplify. Therefore, Preetha is 7 years old. _
_
25. It is given that AB DE . The Exterior Angle Theorem says that the measure of an exterior angle is equal to the sum of the two remote interior angles, so in ABC, _ we_ know that m∠CAB + m∠ACB = m∠ABE. Since AB DE , the corresponding angles formed by a transversal are congruent. ∠ABE and ∠DEF are corresponding angles, so m∠ABE = m∠DEF. Since they are equal to one another, ∠DEF can be substituted for ∠ABE. Doing this with the earlier equation yields m∠CAB + m∠ACB = m∠DEF. 26. 34.0 m 27. a. obtuse b. obtuse c. right _
28. TU is the smallest piece 29. They are parallel. 5 in. 30. 12 √
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 39–4
Saxon Geometry
39
Warm Up 39 1. exterior 2. 78° 3. C Lesson Practice 39 _ _ _
a. EF , DE , DF b. ∠Q, ∠P, ∠R c. a = 120°. Using the Exterior Angle Theorem and the given values of 50° for vertex Y and 70° for vertex X, it can clearly be seen that a is larger than the angle at either of the remote interior angles. d. 50 ft < x < 380 ft
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 39–1
Saxon Geometry
Lesson Practice 39
39
8. ST and UV each equal 9 units, TU and SV each equal 3 units.
1. Since ∠G and ∠K are right angles, GHJ and KLM are_ right triangles; legs GH and _ KL are congruent; and acute angles ∠H and ∠L are congruent; so by Leg-Angle Congruence Theorem, GHJ KLM.
9. To show that AB = CD, the Transitive Property must be used, not the Reflexive Property. 10.
N
O C
_ _ _
M
2. BC , AB , AC 3. 7.1 in.
11. For any two sections, by definition of congruent segments, both pairs of legs are congruent; by the Leg-Leg Congruence Theorem, the sections are congruent.
4. 5 cameras 5. The orthocenter is located on the vertex of the right angle; yes.
12. 143 ft2 orthocenter
13. (0.67, -4)
6. D
14. ∠RTS
7. Since 34 + 14 < 51, these three side lengths cannot form a triangle. © 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
x+7 15. _ 2x
16. B
LSN 39–2
Saxon Geometry
Lesson
39
17. 3796 ft2; See student work; Sample: The buildings will be arranged to form a park that is in the shape of a right triangle. Based on proof model of the Pythagorean Theorem, the sum of the squares sharing the sides with the legs of a right triangle are equal to the sum of the square sharing the hypotenuse side, which is the third building. 18. approximately 35.71 19. The central angles will always sum to 360°, so as the number of sides increases, the number of central angles will increase so that there are fewer degrees for each angle. 20.
Statements 1. 16 = 4(3x - 8) 4(3x - 8) 16 =_ 2. _ 4
4
Reasons 1. Given 2. Division Property of Equality
3. 4 = 3x- 8 4. 4 + 8 = 3x - 8 + 8 5. 12 = 3x 3x 12 = _ 6. _
6. Division Property of Equality
7. 4 = x 8. x = 4
7. Simplify. 8. Symmetric Property of Equality
3
3
3. Simplify. 4. Addition Property of Equality 5. Simplify.
21. y = 2x + 1 113 22. x > √ 23. 21.38 cm © 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 39–3
Saxon Geometry
Lesson 24.
39
2x + 1 = 15 Given 2x + 1 - 1 = 15 - 1 Subtraction Property of Equality 2x = 14 Simplify. 2x 14 _ _ = Division Property of Equality 2 2 x=7 Simplify. Therefore, Preetha is 7 years old. _
_
25. It is given that AB DE . The Exterior Angle Theorem says that the measure of an exterior angle is equal to the sum of the two remote interior angles, so in ABC, _ we_ know that m∠CAB + m∠ACB = m∠ABE. Since AB DE , the corresponding angles formed by a transversal are congruent. ∠ABE and ∠DEF are corresponding angles, so m∠ABE = m∠DEF. Since they are equal to one another, ∠DEF can be substituted for ∠ABE. Doing this with the earlier equation yields m∠CAB + m∠ACB = m∠DEF. 26. 34.0 m 27. a. obtuse b. obtuse c. right _
28. TU is the smallest piece 29. They are parallel. 5 in. 30. 12 √
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 39–4
Saxon Geometry
