MathsICSE Class 9 · Triangles: Congruency and Inequalities

Inequalities in a Triangle: Greater Side, Triangle Inequality and the Shortest Distance (ICSE Class 9)

Q: Is the shortest distance result proved?

Yes, the perpendicular result is proved using the greater-angle-greater-side theorem.

Q: Can the inequality ever be equal?

If a + b = c the points are collinear and no triangle forms.

Q: Do I need all three inequalities?

Checking that the two shorter sides add to more than the longest is enough.

Updated 17 October 2026 · Fact-checked

In a triangle the greater angle is opposite the greater side, and the greater side is opposite the greater angle. The sum of any two sides exceeds the third, so the third side lies between the difference and the sum of the other two. Of all segments from a point to a line, the perpendicular is the shortest.

Each side faces its angle: the bigger the angle, the longer the side opposite it, so 75° faces the longest side.

Illustration drawn for QuestionBuddy.

Understand Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality

Greater side, greater angle.

If $A B > A C$ then $∠ C > ∠ B$ (the greater side has the greater angle opposite it).
If $∠ B > ∠ C$ then $A C > A B$ .

So in a triangle with angles $4 5^{\circ}, 6 0^{\circ}, 7 5^{\circ}$ the sides opposite them are in increasing order. The hypotenuse (opposite $9 0^{\circ}$ ) is the longest side of a right triangle.

*Proof of the first (outline).* On $A B$ mark $D$ with $A D = A C$ . In isosceles $△ A C D$ , $∠ A C D = ∠ A D C$ . The exterior angle $∠ A D C > ∠ B$ , and $∠ A C B > ∠ A C D$ , so $∠ A C B > ∠ B$ .

Triangle inequality (prove). In $△ A B C$ , $A B + A C > B C$ . *Proof.* Extend $B A$ to $D$ with $A D = A C$ . In isosceles $△ A C D$ , $∠ A C D = ∠ A D C$ . Since $∠ B C D > ∠ A C D = ∠ B D C$ , the side opposite the greater angle is greater: $B D > B C$ , i.e. $B A + A C > B C$ .

Consequences. (i) Sides $a, b, c$ form a triangle exactly when each is less than the sum of the other two; it suffices to test the largest. (ii) The third side $c$ satisfies $∣ a - b ∣ < c < a + b$ : for sides $7$ and $12$ , the third lies between $5$ and $19$ . (iii) For $D$ on $B C$ : $A B + B D > A D$ and $A C + C D > A D$ , so $A B + A C + B C > 2 A D$ .

Perpendicular is shortest (prove). From point $P$ to line $l$ let $P M ⊥ l$ and $N$ any other point of $l$ . In $△ P M N$ , $∠ M = 9 0^{\circ}$ is the greatest angle, so the opposite side $P N$ is the greatest: $P N > P M$ . Numerically, if $P$ is $5$ cm above $l$ and $N$ is $12$ cm along $l$ from $M$ , $P N = 13 > 5$ .

Using angle sums. Find the angles first (from an angle sum or ratio) and then order the sides.

Level. Class IX is a school-exam class. Order-of-sides questions are routine; the inequality proofs and range-of-integers questions are at the hardest school-exam level.

Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality: formulas to remember

Greater angle: $∠ B > ∠ C \Rightarrow A C > A B$; Sides opposite angles.
Triangle inequality: $a + b > c$; For every pair.
Third side range: $∣ a - b ∣ < c < a + b$; Open interval.
Shortest distance: $P M ⊥ l \Rightarrow P M < P N$; N any other point.

How to solve Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality questions

Find the angles or lengths, order them, and cite the theorem.

1List the angles (find the third by the angle sum if needed).
2Order the angles; the sides opposite them have the same order.
3To test a triangle, check that the sum of the two shorter sides exceeds the longest.
4For a missing side, write $∣ a - b ∣ < c < a + b$ and list integer values strictly between.
5For shortest distance, name the right-angled triangle and use the greatest angle.
6Cite the theorem by name.

Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Choosing a road length

Suppose two villages $A$ and $C$ are joined to $B$ by roads of length 8 km ( $A B$ ) and 6 km ( $B C$ ), and $A C$ is to be a straight road of whole kilometres.

What lengths are possible for $A C$ ?

Try it first, then show the solutionSolution

$8 - 6 < A C < 8 + 6$ , so $2 < A C < 14$ .
Whole numbers from $3$ to $13$ : $11$ possible lengths.

Answer: AC between 2 km and 14 km; 11 integer values (3 to 13).

Memory hook: Third side between the difference and the sum.

Shortest connection to a highway

Imagine a farm 5 km from a straight highway (measured perpendicular). A service road joins it to a point on the highway 12 km along from the foot of that perpendicular.

Compare the two road lengths.

Try it first, then show the solutionSolution

Perpendicular $= 5$ km.
Slant road $= 5^{2} + 1 2^{2} = 13$ km $> 5$ km.

Answer: 5 km (shortest) against 13 km.

Memory hook: Perpendicular is the shortest distance.

Tricks to remember Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality

Short memory aids. Say each one aloud once, then use it on a question the same day.

Big angle, big side
The largest angle faces the longest side.
Use it when: Ordering sides.
Two shorter must beat the longest
a + b > c must hold for the largest c.
Use it when: Existence of a triangle.
Right angle is greatest
In a right triangle the hypotenuse is the longest side, because the right angle is the greatest angle. So the perpendicular from a point to a line is shorter than any slant segment from that point to the line.
Use it when: Shortest distance proofs.

Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Order of sides · 1 mark · MCQBoard level

In $△ P QR$ , $∠ P = 6 5^{\circ}$ and $∠ Q = 5 5^{\circ}$ . The longest side is:
(a) $P Q$
(b) $QR$
(c) $R P$
(d) cannot be determined

Try it first, then show the solutionSolution

$∠ R = 180 - 120 = 6 0^{\circ}$ ; the greatest angle is $∠ P = 6 5^{\circ}$ , opposite side $QR$ .

Answer: (b) QR

Example 2: Can these be sides? · 3 marks · Short answerBoard level

Can a triangle have sides (i) $5, 7, 13$ cm (ii) $6, 8, 13$ cm? Give reasons.

Try it first, then show the solutionSolution

(i) $5 + 7 = 12 < 13$ : the two shorter sides do not exceed the longest, so no.
(ii) $6 + 8 = 14 > 13$ (and $6 + 13 > 8$ , $8 + 13 > 6$ ): yes.

Answer: (i) No (ii) Yes

Example 3: Range of the third side · 3 marks · Short answerHarder board level

Two sides of a triangle are $7$ cm and $12$ cm. Find the range of the third side and the range of possible perimeters.

Try it first, then show the solutionSolution

$12 - 7 < c < 12 + 7$ , so $5 < c < 19$ .
Perimeter $= 19 + c$ , so $24 < perimeter < 38$ .

Answer: $5 < c < 19$ ; perimeter between $24$ cm and $38$ cm

Example 4: Ordering the sides from algebra · 4 marks · Short answerHarder board level

In $△ A B C$ , $∠ A = (2 x + 10)^{\circ}$ , $∠ B = (3 x)^{\circ}$ and $∠ C = (x + 20)^{\circ}$ . Find the angles and arrange the sides in increasing order.

Try it first, then show the solutionSolution

$2 x + 10 + 3 x + x + 20 = 180$ , so $6 x = 150$ and $x = 25$ .
$∠ A = 6 0^{\circ}$ , $∠ B = 7 5^{\circ}$ , $∠ C = 4 5^{\circ}$ .
Increasing angles: $C < A < B$ , so the sides opposite are $A B < B C < A C$ .

Answer: $6 0^{\circ}, 7 5^{\circ}, 4 5^{\circ}$ ; $A B < B C < A C$

Example 5: The triangle inequality · 5 marks · Long answerHOTS / competency

Prove that the sum of any two sides of a triangle is greater than the third. Test it for the sides $5, 6, 10$ .

Try it first, then show the solutionSolution

In $△ A B C$ extend $B A$ to $D$ so that $A D = A C$ . Join $C D$ .
In $△ A C D$ , $A D = A C$ , so $∠ A C D = ∠ A D C$ .
$∠ B C D = ∠ B C A + ∠ A C D > ∠ A C D = ∠ B D C$ .
In $△ B C D$ , the side opposite the greater angle is greater: $B D > B C$ , i.e. $B A + A C > B C$ . Hence proved.
Test: $5 + 6 = 11 > 10$ , $5 + 10 > 6$ , $6 + 10 > 5$ : a triangle exists.

Answer: Proved; 5, 6, 10 form a triangle.

Where students lose marks in Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality

The slips examiners see most often, and what to write instead.

Mistake: Testing the wrong sum
Add the two shorter sides and compare with the longest.
Wrong working
$7 + 13 > 5$ , so a triangle exists.
Right working
$5 + 7 = 12 < 13$ : no triangle.
Do this: Sort and test the longest.
Usually costs: Whole answer.
Mistake: Including the endpoints
The inequality is strict.
Wrong working
$5 \leq c \leq 19$ .
Right working
$5 < c < 19$ .
Do this: Use strict inequalities.
Usually costs: Final answer.
Mistake: Ordering sides in the wrong direction
The greater angle faces the greater side.
Wrong working
Smallest angle opposite the longest side.
Right working
Smallest angle opposite the shortest side.
Do this: Match angle to side.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; the triangle inequality proof is a standard 4 to 5 mark theorem.
Name the triangle in which you compare angles.
Use strict inequalities, never ≤.
Find missing angles before comparing sides.

Practice questions on Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which set can be the sides of a triangle?
(a) $3, 4, 8$
(b) $5, 6, 10$
(c) $2, 3, 5$
(d) $1, 2, 4$
Hint
Largest less than the sum of the others.
Answer
(b)
Q2 · 4 marks · Short answerHarder board level
$D$ is a point on side $B C$ of $△ A B C$ . Prove that $A B + B C + C A > 2 A D$ . Check it for $A B = 7$ , $A C = 9$ , $B C = 10$ , $B D = 5$ .
Hint
Use two triangle inequalities.
Answer
$A B + B D > A D$ and $A C + C D > A D$ ; adding, $A B + A C + B C > 2 A D$ . Numbers: $A D < 7 + 5 = 12$ and $A D < 9 + 5 = 14$ , so $2 A D < 26 = 7 + 9 + 10$ .
Q3 · 3 marks · Short answerHarder board level
The angles of a triangle are in the ratio $2 : 3 : 4$ . Find them and name the longest side as the side opposite the largest angle.
Hint
Parts: 180 ÷ 9.
Answer
Each part $= 2 0^{\circ}$ : $4 0^{\circ}, 6 0^{\circ}, 8 0^{\circ}$ ; the longest side is opposite $8 0^{\circ}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a right-angled triangle the hypotenuse is the longest side.
Reason (R): The right angle is the greatest angle, and the greatest angle is opposite the longest side.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
The other two angles are acute.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Villages $A$ , $B$ , $C$ have roads $A B = 8$ km and $B C = 6$ km. A straight road $A C$ of whole kilometres is to be built. Separately, a farm is $5$ km from a straight highway (perpendicular distance) and a service road meets the highway $12$ km along from the foot of the perpendicular.
(a) Find the range of $A C$ .
(b) How many whole-kilometre lengths are possible?
(c) Find the length of the service road and say which is shorter.
Hint
|8 − 6| < AC < 8 + 6.
Answer
(a) 2 km < AC < 14 km (b) 11 (3 to 13) (c) 13 km; the 5 km perpendicular is shorter.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which side is opposite the greatest angle?Show answer
The longest side.
2.State the triangle inequality.Show answer
The sum of any two sides is greater than the third.
3.Range of the third side with sides a and b?Show answer
|a − b| < c < a + b.
4.Shortest segment from a point to a line?Show answer
The perpendicular.

Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality: frequently asked questions

Is the shortest distance result proved?

Yes, the perpendicular result is proved using the greater-angle-greater-side theorem.

Can the inequality ever be equal?

If a + b = c the points are collinear and no triangle forms.

Do I need all three inequalities?

Checking that the two shorter sides add to more than the longest is enough.