# 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°

(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°

(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

(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?

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.

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.

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,

Hence, proved

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

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.

Let △ABC be a right angled triangle in which the right angle is at B

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.

In △ACD,

AC + CD > AD …(1)

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

Similarly,

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.

Given,

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)

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.

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.

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

(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.

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

(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

(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)

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

(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.

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.

Let the triangle be △PQR,

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

To prove: PQ > PT and PR > PT

In △PSQ,

PS2 + SQ2 = PQ2 ...(Pythagoras theorem)

PS2 = PQ2 – SQ...(i)

In △PST,

PS2 + ST2 = PT2 …(Pythagoras theorem)

⇒ PS2 = PT2 - ST2 …(ii)

⇒ PQ2 – SQ2 = PT2 – ST2 …[from (i) and (ii)]

⇒ PQ2 – (ST + TQ)2 = PT2 – ST2

⇒ PQ2 - (ST2 + 2ST × TQ + TQ2) = PT2 – ST2

⇒ PQ2 – ST2 – 2ST × TQ – TQ2 = PT2 – ST2

⇒ PQ2 – PT2 = TQ2 + 2ST × TQ

⇒ PQ2 – PT2 = TQ × (2ST + TQ)

As, TQ × (2ST + TQ) > 0 always

⇒ PQ2 – PT2 > 0

⇒ PQ2 > PT2

⇒ 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.

∠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.