ML Aggarwal Triangles Exe-10.4 ICSE Class 9 Maths Solutions Ch-10. Step by Step Answer of Exercise-10.4 of Triangles of ML Aggarwal for ICSE Class 9th Mathematics Questions. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Triangles Exe-10.4 ICSE Class 9 Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-10 | Triangles |

Topics | Solution of Exe-10.4 Questions |

Academic Session | 2024-2025 |

### Solution of Exe-10.4 Questions

ML Aggarwal Triangles Exe-10.4 ICSE Class 9 Maths Solutions Ch-10

**Question 1. ****In ∆PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.**

**Answer : **In ∆PQR, ∠P = 70^{o}, ∠R = 30^{o}

But ∠P + ∠Q + ∠R = 180^{o}

⇒ 100^{o }+ ∠Q = 180^{o}

⇒ ∠Q = 180^{o }– 100^{o }= 180^{o}

⇒ ∠Q = 80^{o} the greatest angle

Its opposite side PR is the longest side

**Question 2. ****Show that in a right angled triangle, the hypotenuse is the longest side.**

**Answer : **in right angled ∆ABC, ∠B = 90^{o}

To prove: AC is the longest side

Proof: in ∆ABC,

∠B = 90^{o}

∠A and ∠C are acute angles

That is less then 90^{o}

∠B is the greatest angle

Or ∠B> ∠C and ∠B> ∠A

AC > AB and AC > BC

Hence AC is the longest side

**Question 3. ****PQR is a right angle triangle at Q and PQ : QR = 3:2. Which is the least angle.**

**Answer :**

Here, PQR is a right angle triangle at Q. Also given that

PQ : QR = 3:2

Let PQ = 3x, then, QR = 2x

It is clear that QR is the least side,

Then, we know that the least angle has least side

Opposite to it.

Therefore, ∠P is the least angle

**Question 4. ****In ∆ ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is **

**(i) the greatest angle ?**

**(ii) the smallest angle ?**

**Answer :**

Given that,

AB = 8 cm, BC = 5.6 cm, CA = 6.5 cm.

Here, AB is the greatest side. **(The greatest side has greatest angle opposite to it)**

Then ∠C is the least angle.

Also, BC is the least side.

Then ∠A is the least angle.

**Question 5. ****In ∆ABC, ∠A = 50°, ∠B= 60°, Arrange the sides of the triangle in ascending order**

**Answer :**

**Question 6. ****In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show**

**(i) BD > AD**

**(ii) DC > AD**

**(iii) AC > DC**

**(iv) AB > BD**

**Answer :**

**Question 7. **** In the figure (1) given below, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.**

**Answer : ****$(a)$△ ABC, AD bisects ∠A,**

In △ABC,

Sum of angles = 180

∠A + ∠B + ∠C = 180

∠A + 60 + 40 = 180

∠A = 80º

∠BAD = ∠DAC = 40º

∠A = 80º, ∠C = 40º

Since, ∠A > ∠C

BC > AB (Sides opposite greater angles is greater) (1)

In △ADC

∠ACD = ∠DAC = 40º

Thus, AD=DC (Isosceles triangle property)

Now, In △ABD

Sum of angles = 180

∠ABD + ∠ADB + ∠BAD = 180

60 + ∠ADB + 40 = 180

∠ADB = 80º

∠ABD = 60º and △ADB = 80º

Since, ∠ABD > ∠ADB

Thus, AD > AB

or DC > AB (Since, AD = DC) (2)

and we know BC = BD + DC

Hence, BC > DC (3)

Hence, from (1), (2) and (3)

BC > DC > AB

**Question 8. ****(a) In the figure (1) given below, prove that (i) CF> AF (ii) DC>DF.**

**(b) In the figure (2) given below, AB = AC.**

**Prove that AB > CD.**

**(c) In the figure (3) given below, AC = CD. Prove that BC < CD.**

**Answer : ****(a) ****(i) In triangle ABC**

ACB=180-60-65=55

In triangle BEC

BCE=180-90-65=25

SO, ACE=55-25=30

Angle opposite CF=90

Angle opposite AF=30

Hence CF>AF as angle opp CF is greater than the angle opposite AF

**(ii) In triangle DFC**

DFC= 180-90-DCF ( DCF=25)

DFC=180-90-25=65

Angle opposite DC=65

Angle opp DF= 25

Hence, DC>DF for the same reason as in the first part

**(b) **Update soon

**Question 9. **(a) In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

#### (b) In the figure (ii) given below, D is any point on the side BC of ∆ABC. If AB > AC, show that AB > AD.

**Answer : ****(a) In figure, ∠B < ∠A and ∠C < ∠D.**

To Prove: AD < BC

Proof: ∠B < ∠A | Given

∴ ∠A > ∠B

∴ OB > OA …(1)

| Side opposite to greater angle is longer

∠C < ∠D | Given

∴ ∠D > ∠C

∴ OC > OD …(2)

| Side opposite to greater angle is longer

From (1) and (2), we get

OB + OC > OA + OD

⇒ BC > AD

⇒ AD < BC.

**(b) ****AB**** ****>**** ****AC ****We know that,**

The angle opposite to greater side is greater than the other……….

**Means****, ****ang****. ****ACB**** ****>**** ****ang****. ****ABC****.**

**NOW****, ****angle** ** ****ADB**** ****=**** ****angle **** ****ACB**** ****+**** ****angle **** ****DAC**

**ADB**** ****>**** ****ang****. ****ABC**

**OR****, ****AB****>**** ****AD****.****.****.****.****.****.**

**Because****, ****AB**** ****is**** ****opp****osite**** ****side of**** ****angle**** ****ADB**** ****which**** ****is**** ****greater**** ****t****han**** ****ang****. ****ABC****.**

**Question 10. **

(i) Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer,

(ii) Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.

(iii) Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.

**Answer: ****(i) Length of sides of a triangle are 4 cm, 3 cm and 7 cm**

We know that sum of any two sides of a triangle is greatar than its third side But 4 + 3 = 7 cm

Which is not possible

Hence to construction of a triangle with sides 4 cm, 3 cm and 7 cm is not possible.

**(ii) Length of sides of a triangle are 9 cm, 7 cm and 17 cm**

We know that sum of any two sides of a triangle is greater than its third side Now 9 + 7 = 16 < 17 ∴ It is not possible to construct a triangle with these sides.

**(iii) Length of sides of a triangle are 8 cm, 7 cm and 4 cm We know that sum of any two sides of a triangle is greater than its third side Now 7 + 4 = 11 > 8**

Yes, It is possible to construct a triangle with these sides.

— : End of ML Aggarwal Triangles Exe-10.4 Class 9 ICSE Maths Solutions :–

Return to :- **ML Aggarawal Maths Solutions for ICSE Class-9**

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In question 7 you have proved the wrong sides