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Inequalities Class 9 ICSE Important Questions with Answers

The 13 questions below are Inequalities Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 5 topics of ICSE Class 9 Maths in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
Questions
5
Topics
3
Key concepts
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With answers
Quick answer

ICSE Class 9 Inequalities questions apply three triangle results (proofs not required): the greater side is opposite the greater angle, any two sides together exceed the third, and the perpendicular is the shortest distance from a point to a line. Expect ordering sides from angles, testing whether lengths form a triangle and finding the range of a third side.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Inequalities

Inside the ICSE Class 9 Maths chapter Inequalities (in Triangles) you examine how the sides and angles of a triangle are related: the greater angle is always opposite the greater side and vice versa, the sum of any two sides is greater than the third side, the difference of any two sides is less than the third, and the perpendicular is the shortest line segment from a point to a line. The chapter is proof-based and uses the exterior-angle theorem.

Greater side opposite greater angleGreater angle opposite greater side (converse)Sum of two sides greater than the thirdDifference of two sides less than the thirdPerpendicular is the shortest segment to a line

Key concepts & formulas

Side-angle relation

In a triangle, if two sides are unequal, the greater side has the greater angle opposite it; conversely the greater angle has the greater side opposite it. So the longest side faces the largest angle.

Triangle inequality

The sum of any two sides of a triangle is greater than the third side: AB+BC>CAAB+BC>CAAB+BC>CA. Equivalently, the difference of any two sides is less than the third side: ∣AB−BC∣<CA|AB-BC|<CA|AB-BC|<CA.

Shortest segment

Of all line segments drawn from a point to a given line, the perpendicular is the shortest. If PM⊥ℓPM\perp\ellPM⊥ ℓ and PNPNPN is any other segment to ℓ\ellℓ, then PM<PNPM<PNPM<PN.

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Inequalities Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

In △ABC\triangle ABC△ ABC, if AB=7AB=7AB=7 cm, BC=5BC=5BC=5 cm and CA=6CA=6CA=6 cm, the greatest angle is:

  1. (a)

    ∠A\angle A∠ A

  2. (b)

    ∠B\angle B∠ B

  3. (c)

    ∠C\angle C∠ C

  4. (d)

    ∠A=∠C\angle A=\angle C∠ A=∠ C

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Answer: (c) ∠C\angle C∠ C.

The greatest angle lies opposite the greatest side. Comparing the sides, AB=7>CA=6>BC=5AB=7>CA=6>BC=5AB=7>CA=6>BC=5, so the greatest side is ABABAB. Side ABABAB is opposite vertex CCC, hence the greatest angle is ∠C\angle C∠ C.

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Q2MCQEasy1 mark

Which of the following can be the sides of a triangle?

  1. (a)

    333 cm, 444 cm, 888 cm

  2. (b)

    555 cm, 666 cm, 101010 cm

  3. (c)

    222 cm, 333 cm, 555 cm

  4. (d)

    444 cm, 444 cm, 999 cm

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Answer: (b) 555 cm, 666 cm, 101010 cm.

A triangle exists only if the sum of any two sides exceeds the third. Here 5+6=11>105+6=11>105+6=11>10, 5+10>65+10>65+10>6, 6+10>56+10>56+10>5, so all conditions hold. In (a) 3+4=7<83+4=7<83+4=7<8, in (c) 2+3=52+3=52+3=5 (not >5>5>5), in (d) 4+4=8<94+4=8<94+4=8<9 fail.

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Q3MCQModerate1 mark

In △PQR\triangle PQR△ PQR, ∠P=50∘\angle P=50^\circ∠ P=50° and ∠Q=60∘\angle Q=60^\circ∠ Q=60°. The sides in increasing order of length are:

  1. (a)

    PQ<QR<RPPQ<QR<RPPQ<QR<RP

  2. (b)

    QR<RP<PQQR<RP<PQQR<RP<PQ

  3. (c)

    RP<QR<PQRP<QR<PQRP<QR<PQ

  4. (d)

    PQ<RP<QRPQ<RP<QRPQ<RP<QR

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Answer: (b) QR<RP<PQQR<RP<PQQR<RP<PQ.

∠R=180∘−50∘−60∘=70∘\angle R=180^\circ-50^\circ-60^\circ=70^\circ∠ R=180°-50°-60°=70°. Ordering angles: ∠P(50∘)<∠Q(60∘)<∠R(70∘)\angle P(50^\circ)<\angle Q(60^\circ)<\angle R(70^\circ)∠ P(50°)<∠ Q(60°)<∠ R(70°). The side opposite the smaller angle is smaller. Side opposite ∠P\angle P∠ P is QRQRQR, opposite ∠Q\angle Q∠ Q is RPRPRP, opposite ∠R\angle R∠ R is PQPQPQ. Hence QR<RP<PQQR<RP<PQQR<RP<PQ.

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Q4MCQHOTS1 mark

Two sides of a triangle are 888 cm and 555 cm. The length of the third side xxx (in cm) must satisfy:

  1. (a)

    x<13x<13x<13

  2. (b)

    3<x<133<x<133<x<13

  3. (c)

    x>3x>3x>3

  4. (d)

    5<x<85<x<85<x<8

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Answer: (b) 3<x<133<x<133<x<13.

By the triangle inequality the third side is less than the sum and greater than the difference of the other two: 8−5<x<8+58-5<x<8+58-5<x<8+5, i.e. 3<x<133<x<133<x<13.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): In △ABC\triangle ABC△ ABC, if ∠B>∠C\angle B>\angle C∠ B>∠ C then AC>ABAC>ABAC>AB.

Reason (R): In a triangle, the side opposite the greater angle is longer.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) ∠B\angle B∠ B is opposite side ACACAC and ∠C\angle C∠ C is opposite side ABABAB. Since ∠B>∠C\angle B>\angle C∠ B>∠ C, the side opposite the greater angle is longer, so AC>ABAC>ABAC>AB. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

In △ABC\triangle ABC△ ABC, AB=ACAB=ACAB=AC and ∠B=70∘\angle B=70^\circ∠ B=70°. Arrange the sides ABABAB, BCBCBC, CACACA in ascending order of length.

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Since AB=ACAB=ACAB=AC, the triangle is isosceles, so ∠B=∠C=70∘\angle B=\angle C=70^\circ∠ B=∠ C=70°.

Then ∠A=180∘−70∘−70∘=40∘\angle A=180^\circ-70^\circ-70^\circ=40^\circ∠ A=180°-70°-70°=40°.

The smallest angle is ∠A=40∘\angle A=40^\circ∠ A=40°, opposite side BCBCBC, so BCBCBC is the smallest. As ∠B=∠C\angle B=\angle C∠ B=∠ C, sides AC=ABAC=ABAC=AB.

Ascending order: BC<AB=CABC<AB=CABC<AB=CA.

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Q7Very ShortModerate2 marks

Is it possible to draw a triangle with sides 666 cm, 777 cm and 141414 cm? Justify your answer.

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For a triangle, the sum of any two sides must be greater than the third side.

Check the two smaller sides: 6+7=136+7=136+7=13 cm.

But the third side is 141414 cm, and 13<1413<1413<14.

Since the sum of two sides is not greater than the third side, no such triangle can be drawn.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

In △ABC\triangle ABC△ ABC, ∠A=2∠B\angle A=2\angle B∠ A=2∠ B and ∠C=3∠B\angle C=3\angle B∠ C=3∠ B. Find the three angles and arrange the sides in ascending order of length.

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∠A+∠B+∠C=180∘⇒2x+x+3x=180∘\angle A+\angle B+\angle C=180^\circ\Rightarrow2x+x+3x=180^\circ∠ A+∠ B+∠ C=180° ⇒2x+x+3x=180° with ∠B=x\angle B=x∠ B=x, so x=30∘x=30^\circx=30°: ∠B=30∘\angle B=30^\circ∠ B=30°, ∠A=60∘\angle A=60^\circ∠ A=60°, ∠C=90∘\angle C=90^\circ∠ C=90°.

The smaller angle has the smaller side opposite it: side opposite ∠B\angle B∠ B is ACACAC, opposite ∠A\angle A∠ A is BCBCBC, opposite ∠C\angle C∠ C is ABABAB.

Ascending order: AC<BC<ABAC<BC<ABAC<BC<AB.

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Q9Short AnswerModerate3 marks

Two sides of a triangle are 6 cm6\text{ cm}6 cm and 9 cm9\text{ cm}9 cm. Between what limits must the third side lie? If the third side is a whole number of centimetres, how many triangles are possible?

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The third side xxx must be less than the sum and greater than the difference of the other two: 9−6<x<9+69-6<x<9+69-6<x<9+6, i.e. 3<x<153<x<153<x<15.

Whole-number values: 4,5,6,…,144,5,6,\dots,144,5,6,…,14, which is 111111 possible lengths, so 111111 triangles.

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Q10Short AnswerHOTS3 marks

PPP is a point 5 cm5\text{ cm}5 cm from a straight road ℓ\ellℓ, and PM⊥ℓPM\perp\ellPM⊥ ℓ with MMM on ℓ\ellℓ. NNN is another point on the road with PN=13 cmPN=13\text{ cm}PN=13 cm. Find MNMNMN, and explain why PMPMPM is the shortest path from PPP to the road.

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In right △PMN\triangle PMN△ PMN (right angle at MMM): MN=PN2−PM2=169−25=144=12 cmMN=\sqrt{PN^2-PM^2}=\sqrt{169-25}=\sqrt{144}=12\text{ cm}MN=√(PN²-PM²)=√(169-25)=√144=12 cm.

In △PMN\triangle PMN△ PMN, ∠PMN=90∘\angle PMN=90^\circ∠ PMN=90° is the greatest angle, so the side opposite it, PNPNPN, is the longest side; hence PM<PNPM<PNPM<PN. This holds for every point NNN on the road, so the perpendicular PMPMPM is the shortest distance from PPP to the road.

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Long answer questions (5 marks)

Q11Long AnswerHOTS5 marks

In △ABC\triangle ABC△ ABC, ∠B=65∘\angle B=65^\circ∠ B=65° and ∠C=35∘\angle C=35^\circ∠ C=35°. The bisector of ∠A\angle A∠ A meets BCBCBC at DDD.

(a) Arrange the sides of △ABC\triangle ABC△ ABC in descending order.

(b) Find ∠ADB\angle ADB∠ ADB and state which is longer, ABABAB or ADADAD.

(c) Find ∠ADC\angle ADC∠ ADC and state which is longer, ADADAD or DCDCDC.

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(a) ∠A=180∘−65∘−35∘=80∘\angle A=180^\circ-65^\circ-35^\circ=80^\circ∠ A=180°-65°-35°=80°. Greatest angle AAA faces BCBCBC, then ∠B\angle B∠ B faces ACACAC, then ∠C\angle C∠ C faces ABABAB: BC>AC>ABBC>AC>ABBC>AC>AB.

(b) ∠BAD=12×80∘=40∘\angle BAD=\tfrac12\times80^\circ=40^\circ∠ BAD=1/2×80°=40°, so ∠ADB=180∘−65∘−40∘=75∘\angle ADB=180^\circ-65^\circ-40^\circ=75^\circ∠ ADB=180°-65°-40°=75°. In △ABD\triangle ABD△ ABD, ABABAB faces ∠ADB=75∘\angle ADB=75^\circ∠ ADB=75° and ADADAD faces ∠B=65∘\angle B=65^\circ∠ B=65°, so AB>ADAB>ADAB>AD.

(c) ∠ADC=180∘−75∘=105∘\angle ADC=180^\circ-75^\circ=105^\circ∠ ADC=180°-75°=105°. In △ADC\triangle ADC△ ADC, DCDCDC faces ∠DAC=40∘\angle DAC=40^\circ∠ DAC=40° and ADADAD faces ∠C=35∘\angle C=35^\circ∠ C=35°, so DC>ADDC>ADDC>AD.

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Q12Long AnswerHOTS5 marks

In quadrilateral ABCDABCDABCD, AB=4 cmAB=4\text{ cm}AB=4 cm, BC=5 cmBC=5\text{ cm}BC=5 cm, CD=6 cmCD=6\text{ cm}CD=6 cm and DA=3 cmDA=3\text{ cm}DA=3 cm. Using the triangle inequality in △ABC\triangle ABC△ ABC and △ACD\triangle ACD△ ACD, find the range of possible lengths of the diagonal ACACAC. If ACACAC is a whole number, list its possible values. If AC=8 cmAC=8\text{ cm}AC=8 cm, which angle of △ACD\triangle ACD△ ACD is the greatest?

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In △ABC\triangle ABC△ ABC: 5−4<AC<5+45-4<AC<5+45-4<AC<5+4, i.e. 1<AC<91<AC<91<AC<9.

In △ACD\triangle ACD△ ACD: 6−3<AC<6+36-3<AC<6+36-3<AC<6+3, i.e. 3<AC<93<AC<93<AC<9.

Both must hold, so 3<AC<93<AC<93<AC<9; whole-number values: 4,5,6,7,84,5,6,7,84,5,6,7,8 cm.

If AC=8AC=8AC=8 cm, it is the longest side of △ACD\triangle ACD△ ACD (8>6>38>6>38>6>3), so the greatest angle is the one opposite it, ∠D\angle D∠ D.

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Case study questions (4 marks)

Q13Case-basedModerate4 marks

A surveyor marks three points AAA, BBB and CCC on a plot. She measures ∠A=80∘\angle A=80^\circ∠ A=80° and ∠B=45∘\angle B=45^\circ∠ B=45° in △ABC\triangle ABC△ ABC. She also plans a straight path from a fixed point PPP to a boundary line ℓ\ellℓ, and finds PM⊥ℓPM\perp\ellPM⊥ ℓ with MMM on ℓ\ellℓ.

(i) Find ∠C\angle C∠ C.

(ii) Arrange the sides a=BCa=BCa=BC, b=CAb=CAb=CA, c=ABc=ABc=AB in descending order of length.

(iii) Which side is the longest, and why?

(iv) If PNPNPN is any other segment from PPP to a point NNN on ℓ\ellℓ, compare PMPMPM and PNPNPN.

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(i) ∠C=180∘−80∘−45∘=55∘\angle C=180^\circ-80^\circ-45^\circ=55^\circ∠ C=180°-80°-45°=55°.

(ii) The larger the angle, the longer the opposite side. Angles: ∠A=80∘>∠C=55∘>∠B=45∘\angle A=80^\circ>\angle C=55^\circ>\angle B=45^\circ∠ A=80°>∠ C=55°>∠ B=45°. Opposite sides: a=BCa=BCa=BC (opp ∠A\angle A∠ A), b=CAb=CAb=CA (opp ∠B\angle B∠ B), c=ABc=ABc=AB (opp ∠C\angle C∠ C).

So descending order of sides: BC>AB>CABC>AB>CABC>AB>CA, i.e. a>c>ba>c>ba>c>b.

(iii) The longest side is BCBCBC (=a=a=a), because it lies opposite the largest angle ∠A=80∘\angle A=80^\circ∠ A=80°.

(iv) The perpendicular from a point to a line is the shortest segment. Since PM⊥ℓPM\perp\ellPM⊥ ℓ and PNPNPN is a slant segment, PM<PNPM<PNPM<PN.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • What are the inequality results for triangles in ICSE Class 9?
    The greater side lies opposite the greater angle (and conversely), the sum of any two sides is greater than the third, and the perpendicular is the shortest segment from a point to a line. Proofs are not required; this page has 13 questions applying them.
  • Between what limits must the third side of a triangle lie?
    It must be greater than the difference and less than the sum of the other two sides. With sides 888 cm and 555 cm, the third side xxx satisfies 3<x<133<x<133<x<13; with 666 cm and 999 cm, 3<x<153<x<153<x<15.
  • How do you arrange the sides of a triangle from its angles?
    Order the angles, then name the side opposite each. If ∠A=2∠B\angle A=2\angle B∠ A=2∠ B and ∠C=3∠B\angle C=3\angle B∠ C=3∠ B, then 6∠B=180∘6\angle B=180^\circ6∠ B=180°, giving 30∘,60∘,90∘30^\circ, 60^\circ, 90^\circ30°, 60°, 90°. The sides opposite are AC<BC<ABAC<BC<ABAC<BC<AB.
  • Can a triangle have sides 6 cm, 7 cm and 14 cm?
    No, because 6+7=136+7=136+7=13 is less than 141414, so the two shorter sides cannot reach each other's ends. A quick test is to check only the two shortest sides against the longest.
  • Which ICSE Class 9 chapters come before and after Inequalities?
    Isosceles Triangles (Chapter 10) comes before Inequalities (Chapter 11) and Mid-point and Its Converse (Chapter 12) after. Class 10's Linear Inequations is a different, algebra chapter with its own page.
  • What is the most common mistake in triangle inequality questions?
    Naming the wrong side as opposite an angle. Mark each angle on a sketch and draw an arrow to the side across from it before ordering anything; the assertion–reason item on this page checks exactly that skill.
  • Can I download ICSE Class 9 Inequalities questions?
    Yes. Enter your email in the free PDF box on this page and all 13 triangle-inequality questions, the surveyor case study included, open with answers as a printable worksheet.

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