Geometry of Triangles Ex 7.4 (NCERT Class 9 Maths)

Understanding Triangle Inequality Theorems

In triangles, when sides or angles are not equal, specific rules govern their relationships. These are known as triangle inequality theorems. They are fundamental in geometry and have practical applications in various fields.

Theorem 1: Angle Opposite Longer Side is Larger

If two sides of a triangle are unequal, the angle opposite to the longer side is larger.

• Explanation: Consider a triangle ABC. If side AC is longer than side AB (AC > AB), then the angle opposite to AC (which is ∠B) will be greater than the angle opposite to AB (which is ∠C). So, ∠B > ∠C. This makes intuitive sense: to accommodate a longer side, the angle opening up to it must be wider.

Theorem 2: Side Opposite Larger Angle is Longer (Converse of Theorem 1)

If two angles of a triangle are unequal, the side opposite to the larger angle is longer.

• Explanation: This is the converse of the first theorem. If, in triangle ABC, angle B is greater than angle C (∠B > ∠C), then the side opposite to ∠B (which is AC) will be longer than the side opposite to ∠C (which is AB). So, AC > AB. This theorem is incredibly useful for proving relationships between side lengths when angles are known.

Theorem 3: The Triangle Inequality Theorem (Sum of Two Sides)

The sum of any two sides of a triangle is greater than the third side.

• Explanation: For any triangle ABC, the following three conditions must always hold true:

1. AB + BC > AC
2. BC + CA > AB
3. CA + AB > BC

• This theorem essentially states that the shortest distance between two points is a straight line. If you were to walk from A to C, it would always be shorter to go directly (length AC) than to take a detour via B (length AB + BC). This theorem helps us determine if a given set of three lengths can even form a triangle.

Mastering these three theorems is key to solving problems in Exercise 7.4 and building a strong foundation in geometry.

Key Definitions

Triangle Inequality Theorem

A fundamental theorem stating that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Converse Theorem

A theorem formed by interchanging the hypothesis and conclusion of another theorem. For example, the converse of 'If A, then B' is 'If B, then A'.

Opposite Side/Angle

In a triangle, the side directly across from a given angle is its opposite side. Similarly, the angle directly across from a given side is its opposite angle.

Worked Examples on Triangle Inequalities

• Example 1: Ordering Angles from Side Lengths In ΔPQR, PQ = 7 cm, QR = 5 cm, and PR = 9 cm. Arrange the angles in ascending order. A: Step 1: Identify side lengths. Given: PQ = 7 cm, QR = 5 cm, PR = 9 cm. Step 2: Relate sides to opposite angles. Angle opposite PQ (7 cm) is ∠R. Angle opposite QR (5 cm) is ∠P. Angle opposite PR (9 cm) is ∠Q. Step 3: Apply the theorem. We know that the angle opposite the longer side is larger. So, we order the sides first: QR < PQ < PR (5 cm < 7 cm < 9 cm). Step 4: Arrange the angles. Therefore, the angles in ascending order are ∠P < ∠R < ∠Q. Final Answer: ∠P < ∠R < ∠Q Example 2: Checking Triangle Feasibility Can a triangle be formed with side lengths 3 cm, 4 cm, and 8 cm? A: Step 1: Apply the Triangle Inequality Theorem. The sum of any two sides must be greater than the third side. Step 2: Check all three conditions. Condition 1: 3 cm + 4 cm > 8 cm? 7 cm > 8 cm (False) Condition 2: 4 cm + 8 cm > 3 cm? 12 cm > 3 cm (True) Condition 3: 3 cm + 8 cm > 4 cm? 11 cm > 4 cm (True) Step 3: Conclude based on results. Since the first condition (3 + 4 > 8) is false, these lengths cannot form a triangle. Final Answer: No, a triangle cannot be formed with side lengths 3 cm, 4 cm, and 8 cm because the sum of two sides (3+4=7) is not greater than the third side (8).
• Example 3: Ordering Sides from Angle Measures In ΔXYZ, if ∠X = 70°, ∠Y = 50°, find the longest and shortest sides. A: Step 1: Find the third angle. The sum of angles in a triangle is 180°. ∠Z = 180° - (∠X + ∠Y) = 180° - (70° + 50°) = 180° - 120° = 60°. So, ∠X = 70°, ∠Y = 50°, ∠Z = 60°. Step 2: Identify the largest and smallest angles. Largest angle is ∠X = 70°. Smallest angle is ∠Y = 50°. Step 3: Apply the theorem. We know that the side opposite the larger angle is longer. Step 4: Determine the longest and shortest sides. Side opposite ∠X (70°) is YZ. So, YZ is the longest side. Side opposite ∠Y (50°) is XZ. So, XZ is the shortest side. Final Answer: Longest side is YZ, shortest side is XZ.

Exam Tip: Avoiding Common Mistakes

When working with triangle inequalities, students often make a few common errors:

1. Confusing Inequalities with Congruence: Remember, congruence is about triangles being identical (same shape and size), while inequalities deal with relationships when sides or angles are different. Don't try to apply congruence criteria when inequality theorems are required.
2. Incorrectly Identifying Opposite Sides/Angles: Always double-check which side is opposite which angle. For angle A, the opposite side is BC. For side AB, the opposite angle is C. A simple sketch can help prevent this mistake.
3. Forgetting to Check All Conditions for Triangle Formation: When using the theorem "sum of two sides is greater than the third side," make sure to check all three possible pairs of sides against the remaining third side. Just checking one pair is insufficient.
4. Not Calculating the Third Angle: If only two angles are given, always calculate the third angle first. This ensures you have all necessary information to correctly order sides based on angle measures.

Practice Questions with Solutions

• Q: In ΔABC, if AB = 6 cm, BC = 8 cm, and CA = 5 cm, arrange the angles ∠A, ∠B, ∠C in descending order. A: Step 1: Identify the side lengths and their opposite angles. Side AB (6 cm) is opposite ∠C. Side BC (8 cm) is opposite ∠A. Side CA (5 cm) is opposite ∠B. Step 2: Order the side lengths from largest to smallest. BC > AB > CA (8 cm > 6 cm > 5 cm) Step 3: Apply the theorem: the angle opposite the longer side is larger. Since BC is the longest side, ∠A is the largest angle. Since CA is the shortest side, ∠B is the smallest angle. Step 4: Arrange the angles in descending order. ∠A > ∠C > ∠B. Final answer: ∠A > ∠C > ∠B
• Q: Is it possible to construct a triangle with side lengths 10 cm, 7 cm, and 2 cm? Justify your answer. A: Step 1: Apply the Triangle Inequality Theorem: The sum of any two sides must be greater than the third side. Step 2: Check all three possible combinations. Condition 1: 10 cm + 7 cm > 2 cm? (17 cm > 2 cm) - True Condition 2: 7 cm + 2 cm > 10 cm? (9 cm > 10 cm) - False Condition 3: 10 cm + 2 cm > 7 cm? (12 cm > 7 cm) - True Step 3: Conclude based on the conditions. Since Condition 2 (7 cm + 2 cm > 10 cm) is false, it is not possible to construct a triangle with these side lengths. Final answer: No, it is not possible because 7 + 2 = 9, which is not greater than 10.
• Q: In ΔPQR, if ∠P = 65° and ∠Q = 45°, determine the shortest side of the triangle. A: Step 1: Find the measure of the third angle, ∠R. The sum of angles in a triangle is 180°. ∠R = 180° - (∠P + ∠Q) = 180° - (65° + 45°) = 180° - 110° = 70°. Step 2: Identify all angle measures. ∠P = 65°, ∠Q = 45°, ∠R = 70°. Step 3: Identify the smallest angle. The smallest angle is ∠Q = 45°. Step 4: Apply the theorem: the side opposite the smaller angle is shorter. The side opposite ∠Q is PR. Final answer: The shortest side of ΔPQR is PR.
• Q: In a quadrilateral ABCD, prove that AB + BC + CD + DA > AC + BD. A: Step 1: Consider triangle ABC. According to the Triangle Inequality Theorem, AB + BC > AC (Equation 1). Step 2: Consider triangle ADC. According to the Triangle Inequality Theorem, CD + DA > AC (Equation 2). Step 3: Consider triangle ABD. According to the Triangle Inequality Theorem, AB + DA > BD (Equation 3). Step 4: Consider triangle BCD. According to the Triangle Inequality Theorem, BC + CD > BD (Equation 4). Step 5: Add Equation 1 and Equation 2. (AB + BC) + (CD + DA) > AC + AC AB + BC + CD + DA > 2AC (Equation 5) Step 6: Add Equation 3 and Equation 4. (AB + DA) + (BC + CD) > BD + BD AB + BC + CD + DA > 2BD (Equation 6) Step 7: Add Equation 5 and Equation 6. 2(AB + BC + CD + DA) > 2AC + 2BD Step 8: Divide by 2 to get the final proof. AB + BC + CD + DA > AC + BD. Final answer: Proven by applying the Triangle Inequality Theorem to the four constituent triangles within the quadrilateral.

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