Inequalities Class 9 ICSE — Mathematics Chapter 11
Inequalities class 9 ICSE, Chapter 11 of the Mathematics book, teaches you how to compare sides and angles of a triangle using the signs \gt and \lt. The chapter proves three theorems: the greater side has the greater angle opposite to it, its converse, and the rule that the perpendicular is the shortest line from a point to a line. Two corollaries give you the triangle inequality tests.
Download Inequalities Class 9 ICSE Mathematics Chapter 11 PDF
Download the Inequalities Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 8 pages, 4 sections, 25 questions.
What is inside Chapter 11
8 pages · 4 sections · 6 worked examples · 25 exercise questions
| Section | Title |
|---|---|
| 1 | INTRODUCTION |
| 2 | Theorem 3 |
| 3 | Theorem 4 |
| 4 | Theorem 5 |
- EXERCISE 11: 25 questions
1 INTRODUCTION
The chapter opens by fixing the two comparison signs. The sign \gt means “is greater than”. If a is greater than b, we write a \gt b. The sign \lt means “is less than”. If a is less than b, we write a \lt b.
These two signs carry the whole chapter. The signs compare two lengths such as AB \gt AC, or two angles such as \angle B \lt \angle C. For instance, 9 \gt 5 states that 9 is greater than 5, and 3 \lt 7 states that 3 is less than 7. The theorems that follow turn comparisons of sides into comparisons of angles, and the other way round.
2 Theorem 3
Theorem 3 states the first comparison rule: if two sides of a triangle are unequal, the greater side has the greater angle opposite to it. Consider triangle ABC with AB \gt AC. The goal is to prove \angle ACB \gt \angle B.
The construction cuts AD = AC from AB and joins C with D. This creates the smaller triangle ACD. In triangle ACD, sides AC and AD are equal, so the angles opposite them are equal: \angle ACD = \angle ADC. This uses the isosceles triangles chapter result that angles opposite equal sides are equal.
In triangle BDC, \angle ADC is an exterior angle, so it is greater than the interior opposite angle \angle B. Hence \angle ACD \gt \angle B. Because \angle ACD is only part of \angle ACB, the full angle is larger still:
\angle ACB \gt \angle ACD \gt \angle B.
The chain explains the logic: the constructed angle beats the interior angle, and the whole angle beats its own part. The construction exists because the full angles cannot be compared directly. The exterior-angle property turns the comparison into a known one.
Worked pattern: take a triangle with sides AB = 12\ \text{cm}, BC = 9\ \text{cm} and CA = 7\ \text{cm}. The greatest side is AB, so the greatest angle is the one opposite AB, namely \angle C. No angle has to be measured; the theorem decides the order directly.
Two mistakes are common when applying Theorem 3.
- Mixing opposite with adjacent: the side AB is opposite \angle C, not \angle A.
- Wrong pairing: in triangle ABC, side BC is opposite \angle A, side CA is opposite \angle B, and side AB is opposite \angle C.
3 Theorem 4
Theorem 4 is the converse of Theorem 3: if two angles of a triangle are unequal, the greater angle has the greater side opposite to it. Where Theorem 3 moved from sides to angles, Theorem 4 moves from angles back to sides.
Given triangle ABC with \angle CAB \gt \angle B, the claim is BC \gt AC. The construction draws \angle BAD = \angle B, with D on BC. In triangle ABD, the two equal angles give AD = BD, because sides opposite equal angles are equal.
Now apply the triangle inequality to triangle ADC: AD + DC \gt AC. Since AD = BD, this becomes BD + DC \gt AC. But BD + DC is exactly BC, so
BC \gt AC.
The construction earns its place by manufacturing an isosceles triangle inside ABC. Once AD is known to equal BD, the triangle inequality finishes the proof.
Worked pattern: in triangle ABC, the angles are \angle A = 80^\circ, \angle B = 60^\circ, \angle C = 40^\circ. The greatest angle is \angle A, so the greatest side is BC, the side opposite it. The full order is BC \gt AC \gt AB, matching the angle order \angle A \gt \angle B \gt \angle C. This side–angle pairing is the tool used throughout the chapter.
4 Theorem 5
Theorem 5 answers a different question: which line from a point outside a line to the line itself is the shortest? The answer is the perpendicular.
Given a point O outside line AB, with OP perpendicular to AB, join O to any other point Q on AB. In the right triangle OPQ, the angle at P is 90^\circ. The right angle is the greatest angle of the triangle, because the other two angles add to 90^\circ and each is smaller than a right angle. So \angle OPQ \gt \angle OQP.
By Theorem 4, the side opposite the greater angle is greater, giving
OQ \gt OP.
Q was chosen anywhere on AB, so every other segment from O to AB is longer than OP. Hence OP is the shortest line from O to AB. The rule applies whenever the point lies outside the line; the perpendicular is the single shortest path, and any oblique segment is longer.
Two corollaries follow. Corollary 1: the sum of the lengths of any two sides of a triangle is always greater than the third side. Corollary 2: the difference between the lengths of any two sides is always less than the third side.
The corollaries work as quick filters. Try lengths 5 cm, 7 cm and 13 cm. Since 5 + 7 = 12 \lt 13, the two shorter sides cannot span the longest one. No triangle exists with these lengths.
The same idea gives the allowed range for a missing side. If two sides are 6 cm and 10 cm, the third side x must satisfy 10 - 6 \lt x \lt 10 + 6, so 4 \lt x \lt 16. This ICSE chapter then applies the same reasoning to medians and interior points.
The section extends the idea to medians. For a median AD, the proof extends AD to E, shows triangles ADB and EDC are congruent, and concludes AB + AC \gt 2AD. Adding the three median results gives: the perimeter of a triangle is greater than the sum of its medians.
For a point P inside triangle ABC, the proof produces BP to meet AC at M and applies the triangle inequality in two small triangles. It ends with PA + PB \lt AC + BC.
What to practise from Inequalities Class 9 ICSE
The chapter has one exercise block.
- Exercise 11 — 25 questions
The set moves from short questions to full proofs. Questions 1 to 3 check basic ordering and the allowed range of a third side. Questions 4 to 10 arrange sides and angles and apply Theorems 3, 4 and 5 directly. Questions 11 to 25 are longer proofs on quadrilaterals, medians, angle bisectors and interior points.
The textbook’s own solutions use a statement-reason format. In ICSE inequality proofs, state the theorem you are applying and the reason at each step, such as “side opposite to greater angle is greater”. A proof that names the rule earns the credit.
For the full chapter list of this book, see the Class 9 Maths hub, or browse the Class 9 books page. For official syllabus details, visit the CISCE website.
Inequalities Class 9 ICSE frequently asked questions
What is the main idea of inequalities class 9 ICSE?
The chapter compares the sides and angles of one triangle. Theorem 3 says the greater side has the greater angle opposite it. Theorem 4 is the converse. Theorem 5 proves the perpendicular is the shortest line from a point to a line.
Why is the sum of any two sides of a triangle greater than the third side in ICSE Class 9 Maths?
It is Corollary 1 of Theorem 5. If the sum of two sides were not greater than the third, the two shorter segments could not meet to close the triangle. With 5, 7 and 13, the sum 12 is too short, so no triangle forms.
How do I find the largest angle in a triangle in Class 9 ICSE Maths?
Find the longest side first; the largest angle is opposite it. In a triangle with sides 7 cm, 9 cm and 12 cm, the angle opposite 12 cm is largest. When an angle bisector is drawn, compute the angles of the small triangles before applying Theorem 3.
What does Theorem 5 say about the shortest distance from a point to a line in Class 9 ICSE?
The perpendicular is the shortest segment from the point to the line. The proof joins the point to any other point on the line and shows that oblique segment is longer, because it lies opposite the right angle in the right triangle.
Which exercise should I practise for inequalities class 9 ICSE?
Exercise 11 is the one exercise block for this chapter; it has 25 questions. Questions 1 to 3 are entry-level, Questions 4 to 10 apply the theorems, and Questions 11 to 25 are proofs. Use the statement-reason format shown in the chapter.
Reference: ICSE Class 9 Mathematics textbook material, chapter 11: Inequalities — CISCE-aligned.