Exercise 7.4 — Triangle Inequalities

Inequalities in a triangle.

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Inequalities in a Triangle — What This Exercise Covers

Exercise 7.4 of Class 9 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) explores the inequality relationships between the sides and angles of a triangle. Unlike congruence, which deals with equal triangles, this topic deals with unequal triangles — where one side is longer than another, or one angle is greater than another.

Understanding these inequalities is essential for proving geometric results and for solving problems involving the relative sizes of sides and angles. This exercise contains 7 problems ranging from classic proofs (the hypotenuse is the longest side in a right triangle) to finding how many distinct triangles can be formed with given side lengths.

Triangle Inequalities Side vs Angle Comparison Triangle Inequality Theorem External Angle Property Valid Triangle Conditions

💡 Core Principle: In any triangle, the larger angle is opposite the longer side — and conversely, the longer side is opposite the larger angle. These two facts drive every proof in this exercise.

Key Theorems on Inequalities in a Triangle

Before attempting the problems, make sure you understand the four fundamental theorems stated in the chapter. These are tested directly in board exams by Telangana and Andhra Pradesh state boards as well as CBSE.

#	Theorem	What It Means
1	Larger angle → longer opposite side	If ∠A > ∠B, then the side opposite ∠A (i.e., BC) > side opposite ∠B (i.e., AC)
2	Longer side → larger opposite angle	If BC > AC, then ∠A > ∠B (the angle across from the longer side is bigger)
3	Triangle Inequality (Sum)	The sum of any two sides of a triangle is always greater than the third side
4	Triangle Inequality (Difference)	The difference of any two sides is always less than the third side

Sum rule: AB + BC > CA, BC + CA > AB, CA + AB > BC

Difference rule: |AB − BC| < CA

📌 For board exams: When asked to "show" or "prove" a result about sides or angles, always state which theorem you are using as a reason in each step. Telangana and AP board marking schemes award marks for correct reasoning, not just the final answer.

Question 1 — Hypotenuse is the Longest Side in a Right Triangle

Question 1

Show that in a right-angled triangle, the hypotenuse is the longest side.

△ABC with ∠C = 90° — AB is the hypotenuse

Given: In △ABC, ∠C = 90° and AB is the hypotenuse.

To prove: AB is the longest side, i.e., AB > BC and AB > AC.

Step 1: Since ∠C = 90°, and the angle sum of a triangle = 180°, ∠A + ∠B = 180° − 90° = 90° ← This means both ∠A and ∠B together equal 90° Step 2: Therefore, ∠A < 90° and ∠B < 90° ← Neither ∠A nor ∠B can be 90° or more on their own Step 3: ∠C > ∠A ⟹ AB > BC ← side opposite larger angle is longer Step 4: ∠C > ∠B ⟹ AB > AC ← same reasoning ∴ AB (hypotenuse) is longer than both remaining sides. ✓

✅ Key idea: The right angle (90°) is always the largest angle in a right triangle. Since the side opposite the largest angle is always the longest, the hypotenuse (opposite 90°) must be the longest side.

Question 2 — Comparing Sides Using Exterior Angles

Question 2

Sides AB and AC of △ABC are extended to P and Q respectively. Given ∠PBC < ∠QCB, show that AC > AB.

AB extended to P; AC extended to Q

Given: ∠PBC < ∠QCB. To Prove: AC > AB.

Step 1: ∠PBC is the exterior angle at B, so ∠PBC = ∠A + ∠ACB ∠QCB is the exterior angle at C, so ∠QCB = ∠A + ∠ABC ← Exterior angle = sum of two non-adjacent interior angles Step 2: Given: ∠PBC < ∠QCB ⟹ ∠A + ∠ACB < ∠A + ∠ABC ⟹ ∠ACB < ∠ABC ← cancel ∠A from both sides Step 3: ∠ACB < ∠ABC ⟹ AB < AC ← side opposite larger angle is longer ∴ AC > AB ✓

💡 Trick used: Both ∠PBC and ∠QCB are exterior angles, and they both contain ∠A. Subtracting ∠A from both sides isolates the interior angles ∠ACB and ∠ABC — then the side-angle inequality does the rest.

Question 3 — Comparing Diagonals Using Two Triangles

Question 3

In the figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Lines AB and CD intersect at O

Given: ∠B < ∠A and ∠C < ∠D. To Prove: AD < BC.

In △OAB: ∠B < ∠A (given) ⟹ OA < OB …(1) ← side opposite larger angle is longer In △OCD: ∠C < ∠D (given) ⟹ OD < OC …(2) ← same reasoning Adding (1) and (2): OA + OD < OB + OC ∴ AD < BC ✓

📌 Strategy: The key insight is that AD = OA + OD and BC = OB + OC (because O lies between A and D, and between B and C). Adding the two inequalities gives the required result directly.

Question 4 — Angles at Ends of Smallest and Longest Sides

Question 4

AB is the smallest side and CD is the longest side of quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Quadrilateral ABCD with diagonals AC and BD drawn

Construction: Join A to C and B to D (draw both diagonals).

This splits the quadrilateral into four triangles. We use the inequality theorem in each triangle separately.

Proving ∠B > ∠D

In △BCD: CD > BC (CD is longest, BC is not the smallest) ⟹ ∠1 > ∠2 …(1) ← ∠1 is at B in △BCD; ∠2 is at D In △ABD: AD > AB (AB is the smallest side) ⟹ ∠4 > ∠3 …(2) ← ∠4 is at B in △ABD; ∠3 is at D Adding (1) and (2): ∠1 + ∠4 > ∠2 + ∠3 ∴ ∠B > ∠D ✓

Proving ∠A > ∠C

In △ABC: BC > AB (AB is the smallest) ⟹ ∠7 > ∠6 …(3) ← ∠7 is at A in △ABC; ∠6 is at C In △ACD: CD > AD (CD is the longest) ⟹ ∠8 > ∠5 …(4) ← ∠8 is at A in △ACD; ∠5 is at C Adding (3) and (4): ∠7 + ∠8 > ∠6 + ∠5 ∴ ∠A > ∠C ✓

✅ Summary: Vertices at the ends of the longest side (CD) have larger angles. Vertices at the ends of the smallest side (AB) also have larger angles than their diagonal counterparts. Drawing the diagonals is the essential construction trick for this proof.

Question 5 — Angle Bisector and Inequality

Question 5

In △PQR, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

PS is the angle bisector of ∠QPR

Given: In △PQR, PR > PQ. PS bisects ∠QPR, so ∠QPS = ∠RPS.
To Prove: ∠PSR > ∠PSQ.

Step 1: In △PQR, PR > PQ (given) ⟹ ∠Q > ∠R …(1) ← angle opp. longer side is larger Step 2: PS bisects ∠QPR ⟹ ∠QPS = ∠RPS …(2) ← definition of angle bisector Step 3: Adding (1) and (2): ∠Q + ∠QPS > ∠R + ∠RPS Step 4: ∠PSR is an exterior angle of △PSQ ∠PSR = ∠Q + ∠QPS (exterior angle theorem) ∠PSQ = 180° − ∠PSR, but more directly: ∠PSR is the exterior angle of △PQS ⟹ ∠PSR = ∠Q + ∠QPS ∠PSQ is the exterior angle of △PRS ⟹ ∠PSQ = ∠R + ∠RPS ∴ ∠PSR > ∠PSQ ✓

💡 Key property used: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles. Here, ∠PSR (exterior to △PQS) = ∠Q + ∠QPS, and ∠PSQ (exterior to △PRS) = ∠R + ∠RPS. Since ∠Q + ∠QPS > ∠R + ∠RPS, the result follows.

Question 6 — Finding All Possible Third Side Lengths

Question 6

Two sides of a triangle measure 4 cm and 6 cm. Find all possible positive integer values for the third side. How many distinct triangles can be formed?

This is a direct application of both the sum and difference forms of the Triangle Inequality Theorem.

Let the third side = x cm. From the difference rule: x > 6 − 4 ⟹ x > 2 From the sum rule: x < 6 + 4 ⟹ x < 10 ∴ 2 < x < 10

Since x must be a positive integer, the possible values are:

x ∈ { 3, 4, 5, 6, 7, 8, 9 } — a total of 7 values

Third Side (cm)	Difference Check (> 2)	Sum Check (< 10)	Valid?
1	1 > 2? ✗	1 < 10? ✓	❌ No
2	2 > 2? ✗	2 < 10? ✓	❌ No
3	3 > 2? ✓	3 < 10? ✓	✅ Yes
4	4 > 2? ✓	4 < 10? ✓	✅ Yes
5	5 > 2? ✓	5 < 10? ✓	✅ Yes
6	6 > 2? ✓	6 < 10? ✓	✅ Yes
7	7 > 2? ✓	7 < 10? ✓	✅ Yes
8	8 > 2? ✓	8 < 10? ✓	✅ Yes
9	9 > 2? ✓	9 < 10? ✓	✅ Yes
10	10 > 2? ✓	10 < 10? ✗	❌ No

✅ Answer: 7 distinct triangles are possible, with the third side being 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, or 9 cm.

Question 7 — Can a Triangle with 5 cm, 8 cm, 1 cm be Constructed?

Question 7

Try to construct a triangle with sides 5 cm, 8 cm, and 1 cm. Is it possible? Give justification.

Given sides: 5 cm, 8 cm, 1 cm Check: difference of the two given sides: 8 − 5 = 3 Third side must be > 3 cm But the third side given is 1 cm. 1 is NOT greater than 3. ∴ A triangle CANNOT be constructed with these measurements. ✗

⛔ Why it fails: The Triangle Inequality Theorem requires that every side must be greater than the absolute difference of the other two. Here, 1 < |8 − 5| = 3. The three points would be collinear (falling on a straight line) and cannot form a closed triangle.

📌 Quick test for any three sides (a, b, c): Check all three conditions — a + b > c, b + c > a, a + c > b. All three must hold. If even one fails, no triangle is possible.

Common Mistakes to Avoid

Confusing the direction of the inequality: The larger angle is opposite the longer side. Do not reverse this — the longer side is NOT opposite the smaller angle.
Forgetting to construct diagonals in Q4: Without drawing AC and BD, you cannot split the quadrilateral into triangles needed for the proof.
Using ≥ instead of > in the Triangle Inequality: The third side must be strictly greater than the difference and strictly less than the sum. Equality would make a degenerate (flat) triangle.
Missing the exterior angle step in Q2 and Q5: Both proofs rely on the Exterior Angle Theorem. Always state this as your reason when using it.
Counting endpoints in Q6: The values 2 cm and 10 cm are excluded (strict inequalities). Students sometimes include them — this is wrong because 2 and 10 do not satisfy the strict inequalities.

⛔ Board Exam Pitfall: When writing proofs, always write the reason (∵) next to each step. Telangana and AP board examiners deduct marks if conclusions are stated without citing the theorem used.

Quick Summary — All Questions at a Glance

Q	What to Prove / Find	Key Theorem Used	Result
1	Hypotenuse is longest in right △	Larger angle → longer opposite side	AB > BC and AB > AC
2	AC > AB when ∠PBC < ∠QCB	Exterior angle theorem + inequality	AC > AB
3	AD < BC when ∠B < ∠A, ∠C < ∠D	Adding two inequalities	AD < BC
4	∠A > ∠C and ∠B > ∠D in quad ABCD	Triangle inequality in sub-triangles	∠A > ∠C; ∠B > ∠D
5	∠PSR > ∠PSQ when PR > PQ	Exterior angle theorem	∠PSR > ∠PSQ
6	Third side range for sides 4 cm, 6 cm	Triangle inequality (sum & difference)	3 to 9 cm → 7 triangles
7	Can 5, 8, 1 form a triangle?	Triangle inequality (difference rule)	Not possible

What This Exercise Prepares You For

Exercise 7.4 wraps up the chapter on Triangles by moving from equality (congruence) to inequality. The reasoning skills you build here — especially using the Exterior Angle Theorem and adding inequalities — are directly used in Class 10 in the chapter on Similar Triangles, where proportional sides and angle relationships are central.

The Triangle Inequality Theorem also reappears in Coordinate Geometry when you verify whether three given points form a triangle. For students preparing for CBSE board exams, Telangana SSC exams, or AP Board exams, this theorem is a guaranteed 2–4 mark question every year — either as a direct proof or a fill-in-the-blank numerical (like Q6 and Q7 in this exercise).

📐 Board Exam Tip (Telangana & AP): For proof questions like Q1–Q5, write each step on a new line with the reason in brackets. For Q6-type problems, always show both the difference bound and the sum bound — never just one. Full marks require both conditions stated explicitly.