# Frank Solutions for Chapter 13 Inequalities in Triangles Class 9 Mathematics ICSE

**Exercise 13.1**

**1. Name the greatest and the smallest sides in the following triangles:**

**(a) ****△****ABC, ****∠****A = 56**°**, ****∠****B = 64****°**** and ****∠****C = 60****°**

**(b) ****△****DEF, ****∠****D = 32****°****, ****∠****E = 56****°**** and ****∠****F = 92****°**

**(c) ****△****XYZ, ****∠****X = 76****°****, ****∠****Y = 84****°**

**Answer**

**(a)** In the given △ABC,

The greatest angle is ∠B and the opposite side to the ∠B is AC

Therefore,

The greatest side is AC

The smallest angle in the △ABC is ∠A and the opposite side to the ∠A is BC

Therefore,

The smallest side is BC

**(b)** In the given △DEF,

The greatest angle is ∠F and the opposite side to the ∠F is DE

Therefore,

The greatest side is DE

The smallest angle in the △DEF is ∠D and the opposite side to the ∠D is EF

Therefore,

The smallest side is EF

**(c) **In △XYZ,

∠X + ∠Y + ∠Z = 180°

⇒ 76° + 84° + ∠Z = 180°

⇒ 160° + ∠Z = 180°

⇒ ∠Z = 180° – 160°

We get,

∠Z = 20°

Therefore,

∠X = 76°, ∠Y = 84°, ∠Z = 20°

In the given △XYZ,

The greatest angle is ∠Y and the opposite side to the ∠Y is XZ

Therefore,

The greatest side is XZ

The smallest angle in the △XYZ is ∠Z and the opposite side to the ∠Z is XY

Therefore,

The smallest side is XY

**2. ****Arrange the sides of the following triangles in an ascending order:**

**(a) ****△****ABC, ****∠****A = 45****°****, ****∠****B = 65****°**

**(b) ****△****DEF, ****∠****D = 38****°****, ****∠****E = 58****°**

**Answer**

**(a)** In △ABC,

∠A + ∠B + ∠C = 180°

⇒ 45° + 65° + ∠C = 180°

⇒ 110° + ∠C = 180°

⇒ ∠C = 180° – 110°

We get,

∠C = 70°

Hence,

∠A = 45°, ∠B = 65°, ∠C = 70°

⇒ 45° < 65° < 70°

Hence,

Ascending order of the angles in the given triangle is,

∠A < ∠B < ∠C

Therefore,

Ascending order of sides in triangle is,

BC, AC, AB

**(b) **In △DEF,

∠D + ∠E + ∠F = 180°

⇒ 38° + 58° + ∠F = 180°

⇒ 96° + ∠F = 180°

⇒ ∠F = 180° – 96°

We get,

∠F = 84°

Hence,

∠D = 38°, ∠E = 58°, ∠F = 84°

⇒ 38° < 58° < 84°

Hence,

Ascending order of the angles in the given triangle is,

∠D < ∠E < ∠F

Therefore,

Ascending order of sides in triangle is,

EF, DF, DE

**3.** **Name the smallest angle in each of these triangles:**

**(i) In ****△****ABC, AB = 6.2 cm, BC = 5.6 cm and AC = 4.2 cm**

**(ii) In ****△****PQR, PQ = 8.3 cm, QR = 5.4 cm and PR = 7.2 cm**

**(iii) In ****△****XYZ, XY = 6.2 cm, XY = 6.8 cm and YZ = 5 cm**

**Answer**

**(i)** We know that,

In a triangle, the angle opposite to the smallest side is the smallest

In △ABC,

AC = 4.2 cm is the smallest side

Hence,

∠B is the smallest angle

**(ii)** We know that,

In a triangle, the angle opposite to the smallest side is the smallest

In △PQR,

QR = 5.4 cm is the smallest side

Hence,

∠P is the smallest angle

**(iii)** We know that,

In a triangle, the angle opposite to the smallest side is the smallest

In △XYZ,

YZ = 5 cm is the smallest side

Therefore,

∠X is the smallest angle

**4.** **In a triangle ABC, BC = AC and ****∠****A = 35****°****. Which is the smallest side of the triangle?**

**Answer**

In △ABC,

BC = AC **(given)**

∠A = ∠B = 35° **[angles opposite to the two equal sides are equal]**

Let ∠C = x°

In △ABC,

∠A + ∠B + ∠C = 180°

⇒ 35° + 35° + x° = 180°

⇒ 70° + x° = 180°

⇒ x° = 180° – 70°

We get,

x° = 110°

Hence,

∠C = x° = 110°

Therefore,

Angles in a triangle are ∠A = ∠B = 35° and ∠C = 110°

In △ABC,

The greatest angle is ∠C

Here, the smallest angles are ∠A and ∠B

Hence,

Smallest sides are BC and AC

**5. ****In ****△****ABC, the exterior ****∠****PBC > exterior ****∠****QCB. Prove that AB > AC.**

**Answer**

In the given triangle,

Given that, ∠PBC > ∠QCB **…(1)**

∠PBC + ∠ABC = 180° **(linear pair angles)**

∠PBC = 180° – ∠ABC **…(2)**

Similarly,

∠QCB = 180° – ∠ACB **…(3)**

Now,

From (2) and (3), we get,

180° – ∠ABC > 180° – ∠ACB

⇒ – ∠ABC > – ∠ACB

Therefore,

∠ABC < ∠ACB or ∠ACB > ∠ABC

We know that,

In a triangle, the greater angle has the longer side opposite to it

Therefore,

AB > AC

Hence, proved

**6.** **△****ABC is isosceles with AB = AC. If BC is extended at D, then prove that AD > AB.**

**Answer**

In △ACD,

We have,

∠ACB = ∠CDA + ∠CAD **[Using exterior angle property]**

∠ACB > ∠CDA **…(1)**

Now,

AB = AC **(given)**

∠ACB = ∠ABC **…(2)**

From equations (1) and (2), we get,

∠ABC > ∠CDA

We know that,

In a triangle, the greater angle has the longer side opposite to it.

Now,

In △ABD,

We have,

∠ABC > ∠CDA

Therefore,

AD > AB

Hence, proved

**7. ****Prove that the perimeter of a triangle is greater than the sum of its three medians.**

**Answer**

Given

In △ABC,

AD, BE and CF are its medians

We know that,

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

Hence,

AD is the median bisecting BC

So,

AB + AC > 2 AD **…(1)**

BE is the median bisecting AC

AB + BC > 2BE **…(2)**

CF is the median bisecting AB

BC + AC > 2CF** …(3)**

Now,

Adding equations (1), (2) and (3), we get,

(AB + AC) + (AB + BC) + (BC + AC) > 2AD + 2BE + 2CF

⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)

We get,

AB + BC + AC > AD + BE + CF

Hence, proved

**8. ****Prove that the hypotenuse is the longest side in a right angled triangle.**

**Answer**

By angle sum property of a triangle,

∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

We get,

∠A + ∠C = 90°

Hence,

The other two angles must be acute i.e. less than 90°

Therefore,

∠B is the largest angle in △ABC

∠B > ∠A and

∠B > ∠C

AC > BC and

AC > AB

∵ In any triangle, the side opposite to the greater angle is longer

Hence,

AC is the largest side in △ABC

But, AC is the hypotenuse of △ABC

Therefore,

Hypotenuse is the longest side in a right angled triangle

Hence, proved

**9.** **D is a point on the side of the BC of ****△****ABC. Prove that the perimeter of ****△****ABC ****is greater than twice of AD.**

**Answer**

Construction: Join AD

In △ACD,

AC + CD > AD **…(1)**

∵ Sum of two sides of a triangle is greater than the third side

Similarly,

In △ADB,

AB + BD > AD **…(2)**

Adding equations (1) and (2), we get,

AC + CD + AB + BD > 2 AD

⇒ AB + BC + AC > 2AD (since, CD + BD = BC)

Hence, proved

**10.** **For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.**

**Answer**

PQRS is a quadrilateral

PR and QS are the diagonals of quadrilateral

To prove: PQ + QR + SR + PS > PR + QS

Proof: In △PQR,

PQ + QR > PR (Sum of two sides of a triangle is greater than the third side)

Similarly,

In △PSR,

PS + SR > PR

In △PQS,

PS + PQ > QS and

In △QRS,

QR + SR > QS

Now,

We have,

PQ + QR > PR

PS + SR > PR

PS + PQ > QS

QR + SR > QS

After adding all the above inequalities, we get,

PQ + QR + PS + SR + PS + PQ + QR + SR > PR + PR + QS + QS

⇒ 2PQ + 2QR + 2PS + 2SR > 2PR + 2QS

⇒ 2 (PQ + QR + PS + SR) > 2 (PR + QS)

We get,

PQ + QR + PS + SR > PR + QS

Hence, proved

**11. ****ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2 (AC + BD)**

**Answer**

We know that,

Sum of two sides of a triangle is greater than the third side

Hence,

In △AOB,

OA + OB > AB **…(1)**

In △BOC,

OB + OC > BC **…(2)**

In △COD,

OC + OD > CD **…(3)**

In △AOD,

OA + OD > AD **…(4)**

Now,

Adding equations (1) (2) and (3) and (4), we get,

2 (OA + OB + OC + OD) > AB + BC + CD + AD

⇒ 2[(OA + OC) + (OB + OD)] > AB + BC + CD + AD

We get,

2 (AC + BD) > AB + BC + CD + AD

∵ OA + OC = AC and OB + OD = BD

Therefore,

AB + BC + CD + AD < 2 (AC + BD)

Hence, proved

**12.** **In ****△****ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.**

**Answer**

We know that,

Sum of two sides of a triangle is greater than the third side

Hence,

In △APR,

AP + AR > PR **…(1)**

In △BPQ,

BQ + PB > PQ **…(2)**

In △QCR,

QC + CR > QR **…(3)**

Now,

Adding equations (1), (2) and (3), we get,

AP + AR + BQ + PB + QC + CR > PR + PQ + QR

⇒ (AP + PB) + (BQ + QC) + (CR + AR) > PR + PQ + QR

We get,

AB + BC + AC > PQ + QR + PR

Hence, proved

**13.** **In ****△****PQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.**

**Answer**

In △PQT,

We have

PT = PQ **…(1) (given)**

In △PQR,

PQ + QR > PR

⇒ PQ + QR > PT + TR **(∵ PR = PT + TR)**

⇒ PQ + QR > PQ + TR **[Using equation (1)]**

We get,

QR > TR

Hence, proved

**14. ****ABCD is a trapezium. Prove that:**

**(i) CD + DA + AB + BC > 2AC**

**(ii) CD + DA + AB > BC**

**Answer**

**(i)** In △ABC,

We have,

AB + BC > AC **...(1)**

In △ACD,

We have,

AD + CD > AC **…(2)**

Adding equations (1) and (2), we get,

AB + BC + AD + CD > 2AC

Hence, proved

**(ii)** In △ACD,

We have,

CD + DA > CA

⇒ CD + DA + AB > CA + AB

⇒ CD + DA + AB > BC **(since, AB + AC > BC)**

Hence, proved

**15.** **In ****△****ABC, BC produced to D, such that, AC = CD; ****∠****BAD = 125****° ****and ****∠****ACD = 105****°****. Show that BC > CD.**

**Answer**

In △ACD,

AC = CD **…(given)**

∠CDA = ∠DAC **…(△ACD is an isosceles triangle)**

Let ∠CDA = ∠DAC = x°

∠CDA + ∠DAC + ∠ACD = 180°

⇒ x° + x° + 105° = 180°

⇒ 2x° + 105° = 180°

⇒ 2x° = 180° – 105°

We get,

2x° = 75°

⇒ x = (75°/2)

We get,

x = 37.5°

∠CDA = ∠DAC = x° = 37.5° **…(1)**

∠DAB = ∠DAC + ∠BAC

⇒ 125° = 37.5° + ∠BAC **[from equation (1)]**

⇒ 125° – 37.5° = ∠BAC

⇒ 87.5° = ∠BAC

Also,

∠BCA + ∠ACD = 180°

⇒ ∠BCA + 105° = 180°

We get,

∠BCA = 75°

Now,

In △BAC,

∠ACB + ∠BAC + ∠ABC = 180°

⇒ 75° + 87.5° + ∠ABC = 180°

⇒ ∠ABC = 180° – 75° – 87.5°

We get,

∠ABC = 17.5°

Since, 87.5° > 17.5°

Hence,

∠BAC > ∠ABC

BC > AC

Therefore,

BC > CD **…(Since AC = CD)**

Hence, proved

**16. In the given figure, ****∠****QPR = 50****°**^{ }and **∠****PQR = 60****°****. Show that:**

**(a) PN < RN **

**(b) SN < SR **

**Answer**

**(a)** In the given △PQR,

PS < PR **…(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)**

PN < PR **…(i) (∵ PN < PS)**

Also,

RT < PR **…(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)**

RN < PR **…(ii) (∵ RN < RT)**

Dividing (i) by (ii),

PN/RN < PR/PR

⇒ PN/RN < 1

⇒ PN < RN

**(a) **In △RTO,

∠RTQ + ∠TQR + ∠TRQ = 180°

⇒ 90°^{ }+ 60° + ∠TRQ = 180°

⇒ 150° + ∠TRQ = 180°

⇒ ∠TRQ = 180°^{ }- 150°

⇒ ∠TRQ = 30°

∠TRQ = ∠SRN = 30° **…(iii)**

In △NSR,

∠RNS + ∠SRN = 90°^{ }**(∵ ∠NSR = 90°)**

∠RNS + 30° = 90° **[from (ii)]**

∠RNS = 90°^{ }- 30°

∠RNS = 60°^{ }**…(iv)**

∠SRN < ∠RNS **…(from (iii) and (iv))**

SN < SR

**17. (a) In ****△****PQR, PS ****⊥**** QR; prove that; PQ > QS and PQ > PS **

**(b)**In

**△PQR, PS**

**⊥ QR; prove that PR > PS**

**(c) In**

**△**

**PQR, PS**

**⊥**

**QR; prove that : PQ + PR > QR and PQ + QR > 2PS**

**Answer**

**(a)** In △PQS,

PS < PQ …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)

i.e., PQ > PS

Also, QS < QP …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)

**(b)** In △PQS,

PS ⊥ QR **…(Given)**

PS < PR …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)

i.e., PR > PS

**(c)** In △PQR,

PQ + PR > QR **(∵ Sum of two sides of a triangle is always greater than third side.)**

In △PQS,

PQ + QS > PS (∵ Sum of two sides of a triangle is always greater than third side.)**…(i)**

In △PRS,

PR + SR > PS (∵ Sum of two sides of a triangle is always greater than third side.) **…(ii)**

Adding (i) and (ii),

PQ + QS + PR + SR > 2PS

⇒ PQ + (QS + SR) + PR > 2PS

⇒ PQ + QR + PR > 2PS

Since PQ + PR > QR

⇒ PQ + QR > 2PS

**18. In the given figure, T is a point on the side PR of triangle PQR. Show that **

**(a) PT < QT **

**(b) RT > QT **

**Note: The question is incomplete.**

**Question should be:**

**In the given figure, T is a point on the side PR of an equilateral triangle PQR,**

**Show that:**

**(a) PT < QT**

**(b) RT < QT**

**Answer**

**(a)** In △PQR,

PQ = QR = PR

⇒ ∠P = ∠Q = ∠R = 60°

In △PQT,

∠PQT < 60°

∴ ∠PQT < ∠P

∴ PT < QT

**(b)** In △TQR,

∠TQR < 60°

∴ ∠TQR < ∠R

∴ RT < QT

**19. In ****△PQR**** is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR. **

**Answer**

In △PQR,

PQ + PR > QR ...(∵ Sum of the two sides of a triangles is always greater than the third side.)**…(i)**

Also, In △SQR,

SQ + SR > QR …(∵ Sum of the two sides of a triangle is always greater than the third side.)**…(ii)**

Dividing (i) and (ii),

(PQ + PR)/(SQ + SR) > QR/QR

⇒ (PQ + PR)/(SQ + SR) > 1

⇒ PQ + PR > SQ + SR

i.e., SQ + SR < PQ + PR

**20. Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex on any point on base of the triangle. **

**Answer**

PS ⊥ QR, the straight line joining vertex P to the line QR.

To prove: PQ > PT and PR > PT

In △PSQ,

PS^{2} + SQ^{2} = PQ^{2 }**...(Pythagoras theorem)**

PS^{2} = PQ^{2 }– SQ^{2 }**...(i)**

In △PST,

PS^{2} + ST^{2} = PT^{2} **…(Pythagoras theorem)**

⇒ PS^{2} = PT^{2} - ST^{2} **…(ii)**

⇒ PQ^{2} – SQ^{2} = PT^{2} – ST^{2} **…[from (i) and (ii)]**

⇒ PQ^{2 }– (ST + TQ)^{2 }= PT^{2} – ST^{2 }

⇒ PQ^{2} - (ST^{2 }+ 2ST × TQ + TQ^{2}) = PT^{2} – ST^{2}

⇒ PQ^{2} – ST^{2} – 2ST × TQ – TQ^{2} = PT^{2} – ST^{2}

⇒ PQ^{2} – PT^{2} = TQ^{2} + 2ST × TQ

⇒ PQ^{2} – PT^{2} = TQ × (2ST + TQ)

As, TQ × (2ST + TQ) > 0 always

⇒ PQ^{2} – PT^{2} > 0

⇒ PQ^{2 }> PT^{2}

⇒ PQ > PT

Also, PQ = PR

PR > PT

**21. ****△ABC, in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE. **

**Answer**

∠AEF > ∠ABC **…(Exterior angle property)**

∠AFE = ∠DFC

∠ACB > DFC **…(Exterior angle property)**

⇒ ∠ACB > ∠AFE

Since AB = AC

⇒ ∠ACB = ∠ABC

So, ∠ABC > ∠AFE

⇒ ∠AFE > ∠ABC > ∠AFE that is, ∠AEF > ∠AFE

⇒ AF > AE

**22. In ****△ABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C. **

**Answer**

**23. In △ABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.**

**Answer**