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Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle5-5
Indirect Proof and Inequalities
in One Triangle
Holt Geometry
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Warm Up
1. Write a conditional from the sentence “An
isosceles triangle has two congruent sides.”
2. Write the contrapositive of the conditional “If it
is Tuesday, then John has a piano lesson.”
3. Show that the conjecture “If x > 6, then 2x >
14” is false by finding a counterexample.
If a ∆ is isosc., then it has 2 ≅ sides.
If John does not have a piano lesson, then it is
not Tuesday.
x = 7
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Write indirect proofs.
Apply inequalities in one triangle.
Objectives
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
indirect proof
Vocabulary
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
So far you have written proofs using direct reasoning.
You began with a true hypothesis and built a logical
argument to show that a conclusion was true. In an
indirect proof, you begin by assuming that the
conclusion is false. Then you show that this
assumption leads to a contradiction. This type of
proof is also called a proof by contradiction.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
When writing an indirect proof, look for a
contradiction of one of the following: the given
information, a definition, a postulate, or a
theorem.
Helpful Hint
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 1: Writing an Indirect Proof
Step 1 Identify the conjecture to be proven.
Given: a > 0
Step 2 Assume the opposite of the conclusion.
Write an indirect proof that if a > 0, then
Prove:
Assume
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 1 Continued
Step 3 Use direct reasoning to lead to a contradiction.
However, 1 > 0.
1 ≤ 0
Given, opposite of conclusion
Zero Prop. of Mult. Prop. of Inequality
Simplify.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Step 4 Conclude that the original conjecture is true.
Example 1 Continued
The assumption that is false.
Therefore
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 1
Write an indirect proof that a triangle cannot
have two right angles.
Step 1 Identify the conjecture to be proven.
Given: A triangle’s interior angles add up to 180°.
Prove: A triangle cannot have two right angles.
Step 2 Assume the opposite of the conclusion.
An angle has two right angles.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 1 Continued
Step 3 Use direct reasoning to lead to a contradiction.
However, by the Protractor Postulate, a triangle
cannot have an angle with a measure of 0°.
m∠1 + m∠2 + m∠3 = 180°
90° + 90° + m∠3 = 180°
180° + m∠3 = 180°
m∠3 = 0°
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Step 4 Conclude that the original conjecture is true.
The assumption that a triangle can have
two right angles is false.
Therefore a triangle cannot have two right
angles.
Check It Out! Example 1 Continued
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
The positions of the longest and shortest sides of
a triangle are related to the positions of the
largest and smallest angles.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 2A: Ordering Triangle Side Lengths and
Angle Measures
Write the angles in order from
smallest to largest.
The angles from smallest to largest are ∠F, ∠H and ∠G.
The shortest side is , so the
smallest angle is ∠F.
The longest side is , so the largest angle is ∠G.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 2B: Ordering Triangle Side Lengths and
Angle Measures
Write the sides in order from
shortest to longest.
m∠R = 180° – (60° + 72°) = 48°
The smallest angle is ∠R, so the
shortest side is .
The largest angle is ∠Q, so the longest side is .
The sides from shortest to longest are
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 2a
Write the angles in order from
smallest to largest.
The angles from smallest to largest are ∠B, ∠A, and ∠C.
The shortest side is , so the
smallest angle is ∠B.
The longest side is , so the largest angle is ∠C.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 2b
Write the sides in order from
shortest to longest.
m∠E = 180° – (90° + 22°) = 68°
The smallest angle is ∠D, so the shortest side is .
The largest angle is ∠F, so the longest side is .
The sides from shortest to longest are
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
A triangle is formed by three segments, but not
every set of three segments can form a triangle.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
A certain relationship must exist among the lengths
of three segments in order for them to form a
triangle.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 3A: Applying the Triangle Inequality
Theorem
Tell whether a triangle can have sides with the
given lengths. Explain.
7, 10, 19
No—by the Triangle Inequality Theorem, a triangle
cannot have these side lengths.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 3B: Applying the Triangle Inequality
Theorem
Tell whether a triangle can have sides with the
given lengths. Explain.
2.3, 3.1, 4.6
Yes—the sum of each pair of lengths is greater
than the third length.
 
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 3C: Applying the Triangle Inequality
Theorem
Tell whether a triangle can have sides with the
given lengths. Explain.
n + 6, n2
– 1, 3n, when n = 4.
Step 1 Evaluate each expression when n = 4.
n + 6
4 + 6
10
n2
– 1
(4)2
– 1
15
3n
3(4)
12
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 3C Continued
Step 2 Compare the lengths.
Yes—the sum of each pair of lengths is greater
than the third length.
  
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 3a
Tell whether a triangle can have sides with the
given lengths. Explain.
8, 13, 21
No—by the Triangle Inequality Theorem, a triangle
cannot have these side lengths.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 3b
Tell whether a triangle can have sides with the
given lengths. Explain.
6.2, 7, 9
Yes—the sum of each pair of lengths is greater
than the third side.
  
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 3c
Tell whether a triangle can have sides with the
given lengths. Explain.
t – 2, 4t, t2
+ 1, when t = 4
Step 1 Evaluate each expression when t = 4.
t – 2
4 – 2
2
t2
+ 1
(4)2
+ 1
17
4t
4(4)
16
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 3c Continued
Step 2 Compare the lengths.
Yes—the sum of each pair of lengths is greater
than the third length.
  
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 4: Finding Side Lengths
The lengths of two sides of a triangle are 8
inches and 13 inches. Find the range of
possible lengths for the third side.
Let x represent the length of the third side. Then
apply the Triangle Inequality Theorem.
Combine the inequalities. So 5 < x < 21. The length
of the third side is greater than 5 inches and less
than 21 inches.
x + 8 > 13
x > 5
x + 13 > 8
x > –5
8 + 13 > x
21 > x
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 4
The lengths of two sides of a triangle are 22
inches and 17 inches. Find the range of possible
lengths for the third side.
Let x represent the length of the third side. Then
apply the Triangle Inequality Theorem.
Combine the inequalities. So 5 < x < 39. The length
of the third side is greater than 5 inches and less
than 39 inches.
x + 22 > 17
x > –5
x + 17 > 22
x > 5
22 + 17 > x
39 > x
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Example 5: Travel Application
The figure shows the
approximate distances
between cities in California.
What is the range of distances
from San Francisco to Oakland?
Let x be the distance from San Francisco to Oakland.
x + 46 > 51
x > 5
x + 51 > 46
x > –5
46 + 51 > x
97 > x
5 < x < 97 Combine the inequalities.
Δ Inequal. Thm.
Subtr. Prop. of
Inequal.
The distance from San Francisco to Oakland is
greater than 5 miles and less than 97 miles.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Check It Out! Example 5
The distance from San Marcos to Johnson City is
50 miles, and the distance from Seguin to San
Marcos is 22 miles. What is the range of
distances from Seguin to Johnson City?
Let x be the distance from Seguin to Johnson City.
x + 22 > 50
x > 28
x + 50 > 22
x > –28
22 + 50 > x
72 > x
28 < x < 72 Combine the inequalities.
Δ Inequal. Thm.
Subtr. Prop. of
Inequal.
The distance from Seguin to Johnson City is greater
than 28 miles and less than 72 miles.
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Lesson Quiz: Part I
1. Write the angles in order from smallest to
largest.
2. Write the sides in order from shortest to
longest.
∠C, ∠B, ∠A
Holt Geometry
5-5
Indirect Proof and Inequalities
in One Triangle
Lesson Quiz: Part II
3. The lengths of two sides of a triangle are 17 cm
and 12 cm. Find the range of possible lengths for
the third side.
4. Tell whether a triangle can have sides with
lengths 2.7, 3.5, and 9.8. Explain.
No; 2.7 + 3.5 is not greater than 9.8.
5 cm < x < 29 cm
5. Ray wants to place a chair so it is
10 ft from his television set. Can
the other two distances
shown be 8 ft and 6 ft? Explain.
Yes; the sum of any two lengths is
greater than the third length.

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Gch5 l5

  • 1. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle5-5 Indirect Proof and Inequalities in One Triangle Holt Geometry Warm UpWarm Up Lesson PresentationLesson Presentation Lesson QuizLesson Quiz
  • 2. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Warm Up 1. Write a conditional from the sentence “An isosceles triangle has two congruent sides.” 2. Write the contrapositive of the conditional “If it is Tuesday, then John has a piano lesson.” 3. Show that the conjecture “If x > 6, then 2x > 14” is false by finding a counterexample. If a ∆ is isosc., then it has 2 ≅ sides. If John does not have a piano lesson, then it is not Tuesday. x = 7
  • 3. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Write indirect proofs. Apply inequalities in one triangle. Objectives
  • 4. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle indirect proof Vocabulary
  • 5. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle So far you have written proofs using direct reasoning. You began with a true hypothesis and built a logical argument to show that a conclusion was true. In an indirect proof, you begin by assuming that the conclusion is false. Then you show that this assumption leads to a contradiction. This type of proof is also called a proof by contradiction.
  • 6. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle
  • 7. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle When writing an indirect proof, look for a contradiction of one of the following: the given information, a definition, a postulate, or a theorem. Helpful Hint
  • 8. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 1: Writing an Indirect Proof Step 1 Identify the conjecture to be proven. Given: a > 0 Step 2 Assume the opposite of the conclusion. Write an indirect proof that if a > 0, then Prove: Assume
  • 9. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 1 Continued Step 3 Use direct reasoning to lead to a contradiction. However, 1 > 0. 1 ≤ 0 Given, opposite of conclusion Zero Prop. of Mult. Prop. of Inequality Simplify.
  • 10. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Step 4 Conclude that the original conjecture is true. Example 1 Continued The assumption that is false. Therefore
  • 11. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 1 Write an indirect proof that a triangle cannot have two right angles. Step 1 Identify the conjecture to be proven. Given: A triangle’s interior angles add up to 180°. Prove: A triangle cannot have two right angles. Step 2 Assume the opposite of the conclusion. An angle has two right angles.
  • 12. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 1 Continued Step 3 Use direct reasoning to lead to a contradiction. However, by the Protractor Postulate, a triangle cannot have an angle with a measure of 0°. m∠1 + m∠2 + m∠3 = 180° 90° + 90° + m∠3 = 180° 180° + m∠3 = 180° m∠3 = 0°
  • 13. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Step 4 Conclude that the original conjecture is true. The assumption that a triangle can have two right angles is false. Therefore a triangle cannot have two right angles. Check It Out! Example 1 Continued
  • 14. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle The positions of the longest and shortest sides of a triangle are related to the positions of the largest and smallest angles.
  • 15. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 2A: Ordering Triangle Side Lengths and Angle Measures Write the angles in order from smallest to largest. The angles from smallest to largest are ∠F, ∠H and ∠G. The shortest side is , so the smallest angle is ∠F. The longest side is , so the largest angle is ∠G.
  • 16. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 2B: Ordering Triangle Side Lengths and Angle Measures Write the sides in order from shortest to longest. m∠R = 180° – (60° + 72°) = 48° The smallest angle is ∠R, so the shortest side is . The largest angle is ∠Q, so the longest side is . The sides from shortest to longest are
  • 17. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 2a Write the angles in order from smallest to largest. The angles from smallest to largest are ∠B, ∠A, and ∠C. The shortest side is , so the smallest angle is ∠B. The longest side is , so the largest angle is ∠C.
  • 18. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 2b Write the sides in order from shortest to longest. m∠E = 180° – (90° + 22°) = 68° The smallest angle is ∠D, so the shortest side is . The largest angle is ∠F, so the longest side is . The sides from shortest to longest are
  • 19. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle A triangle is formed by three segments, but not every set of three segments can form a triangle.
  • 20. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle A certain relationship must exist among the lengths of three segments in order for them to form a triangle.
  • 21. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 3A: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain. 7, 10, 19 No—by the Triangle Inequality Theorem, a triangle cannot have these side lengths.
  • 22. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 3B: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain. 2.3, 3.1, 4.6 Yes—the sum of each pair of lengths is greater than the third length.  
  • 23. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 3C: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain. n + 6, n2 – 1, 3n, when n = 4. Step 1 Evaluate each expression when n = 4. n + 6 4 + 6 10 n2 – 1 (4)2 – 1 15 3n 3(4) 12
  • 24. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 3C Continued Step 2 Compare the lengths. Yes—the sum of each pair of lengths is greater than the third length.   
  • 25. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3a Tell whether a triangle can have sides with the given lengths. Explain. 8, 13, 21 No—by the Triangle Inequality Theorem, a triangle cannot have these side lengths.
  • 26. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3b Tell whether a triangle can have sides with the given lengths. Explain. 6.2, 7, 9 Yes—the sum of each pair of lengths is greater than the third side.   
  • 27. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3c Tell whether a triangle can have sides with the given lengths. Explain. t – 2, 4t, t2 + 1, when t = 4 Step 1 Evaluate each expression when t = 4. t – 2 4 – 2 2 t2 + 1 (4)2 + 1 17 4t 4(4) 16
  • 28. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3c Continued Step 2 Compare the lengths. Yes—the sum of each pair of lengths is greater than the third length.   
  • 29. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 4: Finding Side Lengths The lengths of two sides of a triangle are 8 inches and 13 inches. Find the range of possible lengths for the third side. Let x represent the length of the third side. Then apply the Triangle Inequality Theorem. Combine the inequalities. So 5 < x < 21. The length of the third side is greater than 5 inches and less than 21 inches. x + 8 > 13 x > 5 x + 13 > 8 x > –5 8 + 13 > x 21 > x
  • 30. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 4 The lengths of two sides of a triangle are 22 inches and 17 inches. Find the range of possible lengths for the third side. Let x represent the length of the third side. Then apply the Triangle Inequality Theorem. Combine the inequalities. So 5 < x < 39. The length of the third side is greater than 5 inches and less than 39 inches. x + 22 > 17 x > –5 x + 17 > 22 x > 5 22 + 17 > x 39 > x
  • 31. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Example 5: Travel Application The figure shows the approximate distances between cities in California. What is the range of distances from San Francisco to Oakland? Let x be the distance from San Francisco to Oakland. x + 46 > 51 x > 5 x + 51 > 46 x > –5 46 + 51 > x 97 > x 5 < x < 97 Combine the inequalities. Δ Inequal. Thm. Subtr. Prop. of Inequal. The distance from San Francisco to Oakland is greater than 5 miles and less than 97 miles.
  • 32. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Check It Out! Example 5 The distance from San Marcos to Johnson City is 50 miles, and the distance from Seguin to San Marcos is 22 miles. What is the range of distances from Seguin to Johnson City? Let x be the distance from Seguin to Johnson City. x + 22 > 50 x > 28 x + 50 > 22 x > –28 22 + 50 > x 72 > x 28 < x < 72 Combine the inequalities. Δ Inequal. Thm. Subtr. Prop. of Inequal. The distance from Seguin to Johnson City is greater than 28 miles and less than 72 miles.
  • 33. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Lesson Quiz: Part I 1. Write the angles in order from smallest to largest. 2. Write the sides in order from shortest to longest. ∠C, ∠B, ∠A
  • 34. Holt Geometry 5-5 Indirect Proof and Inequalities in One Triangle Lesson Quiz: Part II 3. The lengths of two sides of a triangle are 17 cm and 12 cm. Find the range of possible lengths for the third side. 4. Tell whether a triangle can have sides with lengths 2.7, 3.5, and 9.8. Explain. No; 2.7 + 3.5 is not greater than 9.8. 5 cm < x < 29 cm 5. Ray wants to place a chair so it is 10 ft from his television set. Can the other two distances shown be 8 ft and 6 ft? Explain. Yes; the sum of any two lengths is greater than the third length.