**The Triangle and Its Properties ****Exercise 6.4 ****NCERT Solution for Class 7 Maths Chapter 6**

**The Triangle and Its Properties ****Exercise 6.1**

**The Triangle and Its Properties ****Exercise 6.2**

**The Triangle and Its Properties ****Exercise 6.3**

**The Triangle and Its Properties ****Exercise 6.5**

**Exercise 6.4**

**Q.1:- Is it possible to have a triangle with the following sides? **

**(i) 2cm, 3cm, 5cm**

**(ii) 3cm, 6cm, 7cm**

**(iii) 6cm, 3cm, 2cm**

**Solution 1:- **

**Concept:- A triangle is possible if sum of the lengths of any two sides would be greater
than the length of third side. **

**Solution 1(i):- **Let us check this

2 + 3 > 5 No

2 + 5 > 7 Yes

5 + 7 > 2 Yes

This Triangle is not Possible.

**Solution 1(ii):- **Let us check this

3 + 6 > 7 Yes

3 + 7 > 6 Yes

5 + 7 > 3 Yes

This Triangle is Possible.

**Solution 1(iii):- **Let us check this

2 + 3 > 6 No

2 + 6 > 3 Yes

3 + 6 > 2 Yes

This Triangle is not Possible.

** Q.2:- Take any point O in the interior of a triangle PQR. Is**

**(i) OP + OQ > PQ**

**(ii) OQ + OR > QR**

**(iii) OR + OP > RP**

**Solution 2(i):- Join OR,OP and OQ**

Is OP+OQ > PQ ?

Yes, because POQ form a triangle.

**Solution 2(ii):- **Is OQ+OR > QR ?

Yes, because QOR form a triangle.

**Solution 2(iii):- **Is OR+OP > RP ?

Yes, because POR form a triangle.

**Q.3:- AM is a median of a triangle ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles ∆ABM and ∆AMC.)**

**Solution 3:- **Since, the sum of lengths of any two sides in a triangle should be greater than the length

of third side.

In Triangle ABM, AB + BM > AM ……… equation (i)

In Triangle AMC, AC + MC > AM ………. equation (ii)

By adding equation (i) + (ii)

AB + BM + AC + MC > AM + AM

AB + AC + ( BM + MC ) > 2AM

AB + AC + BC > 2 AM

Hence, Yes it is possible.

**Q.4:- ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD? **

**Solution 4:- **Since, the sum of lengths of any two sides in a triangle should be greater than the length

of third side.

In Triangle ABC, AB + BC > AC ……. equation (i)

In Triangle ADC, AD + DC > AC ……..equation (ii)

In Triangle DCB, DC + CB > BD …….equation (iii)

In Triangle DBA, AB + AD > BD …..equation (iv)

Adding equations (i), (ii), (iii) and (iv)

AB + BC + AD + DC + DC + CB + AB + AD > AC + AC + BD + BD

(AB + AB) + (BC + CB) + (AD + AD) + (DC + DC) > 2AC + 2BD

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

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

AB + BC + CD + DA > AC + BD

Yes, it is True/Possible.

**Q.5:- ABCD is quadrilateral. Is
AB + BC + CD + DA < 2 (AC + BD)?**

**Solution 5:-**

Since, the sum of lengths of any two sides in a triangle should be greater than the length

of third side.

In Triangle AOB, AB < OA + OB ……. equation (i)

In Triangle BOC, BC < OB + OC ……..equation (ii)

In Triangle COD, CD < OC + OD …….equation (iii)

In Triangle DOA, DA < OA + OD …..equation (iv)

Adding equations (i), (ii), (iii) and (iv)

AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OA + OD,

AB + BC + CD + DA < (OA + OA) + (OB + OB) + (OC + OC) + (OD + OD),

AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD,

AB + BC + CD + DA < 2(OA + OB + OC + OD),

AB + BC + CD + DA < 2[(OA + OC) + (OB + OD)]

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

Yes, it is True/Possible.

**Q.6:- The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall ?
**

**Solution 6:-**

We know that the sum of the two sides of a triangle is always greater than the third side.

Therefore, third side has to be less than the sum of two side,

Thus the third side should be less than (12cm + 15cm) = 27cm

The side cannot be less than the difference of the two sides.

Thus the third side should be more than (15cm – 12cm) = 3cm

The length of the third side could be any length greater than 3 and less than 27cm.

**Answer :- Between 3cm and 27cm.**

**NCERT Solutions for Class 7 Math, Chapter 6 – The Triangle and Its Properties, ****Exercise-6.4**