**Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles** are provided here, which will improve the problem-solving skills of students and also will help them learn the techniques to solve difficult questions with speed and accuracy. Practising these solutions on a daily basis is the most convenient way to score desired marks in the final exams. Students who find Maths a complex subject are advised to practise Frank Solutions, as the answers are available in an interactive manner.

Chapter 13 deals with the study of Inequality in Triangles. This chapter provides formulas and properties of Inequalities in Triangles which are important from an exam perspective. By practising these solutions, students can identify their weak areas and work on them. Besides, it will boost the problem-solving and time-management skills of students, irrespective of their intelligence levels. For in-depth knowledge of the concepts, access Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles, from the link provided below.

## Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles Download PDF

## Access Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles

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

**(a) â–³ABC, âˆ A = 56 ^{0}, âˆ B = 64^{0} and âˆ C = 60^{0}**

**(b) â–³DEF, âˆ D = 32 ^{0}, âˆ E = 56^{0} and âˆ F = 92^{0}**

**(c) â–³XYZ, âˆ X = 76 ^{0}, âˆ Y = 84^{0}**

**Solution:**

(a) In the given â–³ABC,

The greatest angle is âˆ B, and the opposite side to the âˆ B is AC

Therefore,

The greatest side is AC

The smallest angle in the â–³ABC is âˆ A, and the opposite side to the âˆ A is BC

Therefore,

The smallest side is BC

(b) In the given â–³DEF,

The greatest angle is âˆ F, and the opposite side to the âˆ F is DE

Therefore,

The greatest side is DE

The smallest angle in the â–³DEF is âˆ D, and the opposite side to the âˆ D is EF

Therefore,

The smallest side is EF

(c) In â–³XYZ,

âˆ X + âˆ Y + âˆ Z = 180^{0}

76^{0} + 84^{0} + âˆ Z = 180^{0}

160^{0} + âˆ Z = 180^{0}

âˆ Z = 180^{0} – 160^{0}

We get,

âˆ Z = 20^{0}

Therefore,

âˆ X = 76^{0}, âˆ Y = 84^{0}, âˆ Z = 20^{0}

In the given â–³XYZ,

The greatest angle is âˆ Y, and the opposite side to the âˆ Y is XZ

Therefore,

The greatest side is XZ

The smallest angle in the â–³XYZ is âˆ Z, and the opposite side to the âˆ Z is XY

Therefore,

The smallest side is XY

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

**(a) â–³ABC, âˆ A = 45 ^{0}, âˆ B = 65^{0}**

**(b) â–³DEF, âˆ D = 38 ^{0}, âˆ E = 58^{0}**

**Solution:**

(a) In â–³ABC,

âˆ A + âˆ B + âˆ C = 180^{0}

45^{0} + 65^{0} + âˆ C = 180^{0}

110^{0} + âˆ C = 180^{0}

âˆ C = 180^{0} – 110^{0}

We get,

âˆ C = 70^{0}

Hence,

âˆ A = 45^{0}, âˆ B = 65^{0}, âˆ C = 70^{0}

45^{0} < 65^{0} < 70^{0}

Hence,

Ascending order of the angles in the given triangle is,

âˆ A < âˆ B < âˆ C

Therefore,

Ascending order of sides in the triangle is,

BC, AC, AB

(b) In â–³DEF,

âˆ D + âˆ E + âˆ F = 180^{0}

38^{0} + 58^{0} + âˆ F = 180^{0}

96^{0} + âˆ F = 180^{0}

âˆ F = 180^{0} – 96^{0}

We get,

âˆ F = 84^{0}

Hence,

âˆ D = 38^{0}, âˆ E = 58^{0}, âˆ F = 84^{0}

38^{0} < 58^{0} < 84^{0}

Hence,

Ascending order of the angles in the given triangle is,

âˆ D < âˆ E < âˆ F

Therefore,

Ascending order of sides in the triangle is,

EF, DF, DE

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

**(i) In â–³ABC, AB = 6.2 cm, BC = 5.6 cm and AC = 4.2 cm**

**(ii) In â–³PQR, PQ = 8.3 cm, QR = 5.4 cm and PR = 7.2 cm**

**(iii) In â–³XYZ, XY = 6.2 cm, XY = 6.8 cm and YZ = 5 cm**

**Solution:**

(i) We know that,

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

In â–³ABC,

AC = 4.2 cm is the smallest side

Hence,

âˆ B is the smallest angle

(ii) We know that,

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

In â–³PQR,

QR = 5.4 cm is the smallest side

Hence,

âˆ P is the smallest angle

(iii) We know that,

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

In â–³XYZ,

YZ = 5 cm is the smallest side

Therefore,

âˆ X is the smallest angle

**4. In a triangle ABC, BC = AC and âˆ A = 35 ^{0}. Which is the smallest side of the triangle?**

**Solution:**

In â–³ABC,

BC = AC (given)

âˆ A = âˆ B = 35^{0} [angles opposite to the two equal sides are equal]

Let âˆ C = x^{0}

In â–³ABC,

âˆ A + âˆ B + âˆ C = 180^{0}

35^{0} + 35^{0} + x^{0} = 180^{0}

70^{0} + x^{0} = 180^{0}

x^{0} = 180^{0} – 70^{0}

We get,

x^{0} = 110^{0}

Hence,

âˆ C = x^{0} = 110^{0}

Therefore,

Angles in a triangle are âˆ A = âˆ B = 35^{0} and âˆ C = 110^{0}

In â–³ABC,

The greatest angle is âˆ C

Here, the smallest angles are âˆ A and âˆ B

Hence,

The smallest sides are BC and AC

**5. In â–³ABC, the exterior âˆ PBC > exterior âˆ QCB. Prove that AB> AC.**

**Solution:**

In the given triangle,

Given that, âˆ PBC > âˆ QCB â€¦â€¦. (1)

âˆ PBC + âˆ ABC = 180^{0} (linear pair angles)

âˆ PBC = 180^{0} â€“ âˆ ABC â€¦. (2)

Similarly,

âˆ QCB = 180^{0} â€“ âˆ ACB â€¦â€¦(3)

Now,

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

180^{0} â€“ âˆ ABC > 180^{0} â€“ âˆ ACB

– âˆ ABC > – âˆ ACB

Therefore,

âˆ ABC < âˆ ACB or âˆ ACB > âˆ ABC

We know that,

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

Therefore,

AB > AC

Hence proved

**6. â–³ABC is isosceles with AB = AC. If BC is extended at D, then prove that AD > AB.**

**Solution:**

In â–³ACD,

We have,

âˆ ACB = âˆ CDA + âˆ CAD [Using exterior angle property]

âˆ ACB > âˆ CDA â€¦.. (1)

Now,

AB = AC â€¦. (given)

âˆ ACB = âˆ ABC â€¦. (2)

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

âˆ ABC > âˆ CDA

We know that,

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

Now,

In â–³ABD,

We have,

âˆ ABC > âˆ CDA

Therefore,

AD > AB

Hence, proved

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

**Solution:**

Given

In â–³ABC,

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 â€¦â€¦.. (1)

BE is the median bisecting AC

AB + BC > 2BE â€¦â€¦.. (2)

CF is the median bisecting AB

BC + AC > 2CF â€¦â€¦â€¦ (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.**

**Solution:**

Let â–³ABC be a right-angled triangle in which the right angle is at B

By the angle sum property of a triangle,

âˆ A + âˆ B + âˆ C = 180^{0}

âˆ A + 90^{0} + âˆ C = 180^{0}

We get,

âˆ A + âˆ C = 90^{0}

Hence,

The other two angles must be acute i.e less than 90^{0}

Therefore,

âˆ B is the largest angle in â–³ABC

âˆ B > âˆ A and

âˆ B > âˆ C

AC > BC and

AC > AB

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

Hence,

AC is the largest side of â–³ABC

But AC is the hypotenuse of â–³ABC

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 â–³ABC. Prove that the perimeter of â–³ABC**

**is greater than twice of AD.**

**Solution:**

Construction: Join AD

In â–³ACD,

AC + CD > AD â€¦.. (1)

âˆµ The sum of two sides of a triangle is greater than the third side

Similarly,

In â–³ADB,

AB + BD > AD â€¦â€¦… (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.**

**Solution:**

Given

PQRS is a quadrilateral

PR and QS are the diagonals of the quadrilateral

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

Proof: In â–³PQR,

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

Similarly,

In â–³PSR,

PS + SR > PR

In â–³PQS,

PS + PQ > QS and

In â–³QRS,

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)**

**Solution:**

We know that,

The sum of two sides of a triangle is greater than the third side

Hence,

In â–³AOB,

OA + OB > AB â€¦â€¦â€¦ (1)

In â–³BOC,

OB + OC > BC â€¦â€¦â€¦ (2)

In â–³COD,

OC + OD > CD â€¦â€¦â€¦ (3)

In â–³AOD,

OA + OD > AD â€¦â€¦â€¦. (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

âˆµ OA + OC = AC and OB + OD = BD

Therefore,

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

Hence, proved

**12. In â–³ABC, P, Q and R are points on AB, BC and AC, respectively. Prove that AB + BC + AC > PQ + QR + PR.**

**Solution:**

We know that,

The sum of two sides of a triangle is greater than the third side

Hence,

In â–³APR,

AP + AR > PR â€¦â€¦.. (1)

In â–³BPQ,

BQ + PB > PQ â€¦â€¦.. (2)

In â–³QCR,

QC + CR > QR â€¦â€¦… (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 â–³PQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.**

**Solution:**

In â–³PQT,

We have

PT = PQ â€¦â€¦. (1) (given)

In â–³PQR,

PQ + QR > PR

PQ + QR > PT + TR (âˆµ 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**

**Solution:**

(i) In â–³ABC,

We have,

AB + BC > AC â€¦â€¦ (1)

In â–³ACD,

We have,

AD + CD > AC â€¦â€¦â€¦. (2)

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

AB + BC + AD + CD > 2AC

Hence, proved

(ii) In â–³ACD,

We have,

CD + DA > CA

CD + DA + AB > CA + AB

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

Hence, proved

**15. In â–³ABC, BC produced to D, such that, AC = CD; âˆ BAD = 125 ^{0} and âˆ ACD = 105^{0}. Show that BC > CD.**

**Solution:**

In â–³ACD,

AC = CD â€¦ (given)

âˆ CDA = âˆ DAC â€¦â€¦.. (â–³ACD is an isosceles triangle)

Let âˆ CDA = âˆ DAC = x^{0}

âˆ CDA + âˆ DAC + âˆ ACD = 180^{0}

x^{0} + x^{0} + 105^{0} = 180^{0}

2x^{0} + 105^{0} = 180^{0}

2x^{0} = 180^{0} – 105^{0}

We get,

2x^{0} = 75^{0}

x = (75^{0} / 2)

We get,

x = 37.5^{0}

âˆ CDA = âˆ DAC = x^{0} = 37.5^{0} â€¦â€¦â€¦â€¦ (1)

âˆ DAB = âˆ DAC + âˆ BAC

125^{0} = 37.5^{0} + âˆ BAC {from equation (1)}

125^{0} â€“ 37.5^{0} = âˆ BAC

87.5^{0} = âˆ BAC

Also,

âˆ BCA + âˆ ACD = 180^{0}

âˆ BCA + 105^{0} = 180^{0}

We get,

âˆ BCA = 75^{0}

Now,

In â–³BAC,

âˆ ACB + âˆ BAC + âˆ ABC = 180^{0}

75^{0} + 87.5^{0} + âˆ ABC = 180^{0}

âˆ ABC = 180^{0} â€“ 75^{0} â€“ 87.5^{0}

We get,

âˆ ABC = 17.5^{0}

Since 87.5^{0} > 17.5^{0}

Hence,

âˆ BAC > âˆ ABC

BC > AC

Therefore,

BC > CD â€¦. (Since AC = CD)

Hence, proved

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