**NCERT** Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties are available here. When students feel stressed about searching for the most comprehensive and detailed NCERT Solutions for Class 7 Maths, we at Swiflearn have prepared step by step solutions with detailed explanations. We recommend students to go through these solutions who want to score good marks in Maths and strengthen their knowledge.

Given below are the concepts covered in Chapter 6 – The Triangle and its Properties of Class 7 Maths NCERT Solutions. This chapter contains five exercises that cover problems related to these concepts.

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## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle: EX 6.1 PDF

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle: EX 6.2 PDF

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle: EX 6.3 PDF

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle: EX 6.4 PDF

**Question 1.**

**Is it possible to have a triangle with the following sides?**

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

**Solution**:-

From the question,

(2 + 3) = 5

5 = 5

As the sum of any two of these numbers is not greater than the third side.

Hence, it is not possible to draw this triangle.

(ii) 3 cm, 6 cm, 7 cm

**Solution**:-

From the question,

(3 + 6) = 9 > 7

(6 + 7) = 13 > 3

(7 + 3) = 10 > 6

As, the sum of any two of these numbers is greater than the third side.

Hence, it is possible to draw this triangle.

(iii) 6 cm, 3 cm, 2 cm

**Solution**:-

From the question,

(3 + 2) = 5 < 6

As the sum of any two of these numbers is not greater than the third side.

Hence, it is not possible to draw this triangle.

**Question 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**:-

By taking any point O in the interior of the triangle PQR and joining OR, OP, OQ.

Then, we will get three triangles ΔOPQ, ΔOQR and ΔORP.

As we know,

The sum of any two sides is always greater than the third in a triangle.

(i) Clearly, ΔOPQ has sides OP, OQ and PQ.

Therefore, OP + OQ > PQ

(ii) Clearly, ΔOQR has sides OR, OQ and QR.

Therefore, OQ + OR > QR

(iii) Clearly, ΔORP has sides OR, OP and PR.

Therefore, OR + OP > RP

**Question 3.**

**AM is a median of a triangle ABC.**

**Is AB + BC + CA > 2 AM?**

**(Consider the sides of triangles ΔABM and ΔAMC.)**

**Solution**:-

As we know,

The sum of any two sides is always greater than the third in a triangle.

Considering the ΔABM,

Since, AB + BM > AM ………………………….. [Equation i]

Then, consider the ΔACM

Since, AC + CM > AM ………………………….. [Equation ii]

Adding equation [i] and [ii] ,

AB + BM + AC + CM is greater (>) than sum of AM + AM

According to the figure in question we have, BC = BM + CM

AB + BC + AC is greater (>) than 2 AM

So, the given expression is true.

**Question 4.**

**ABCD is a quadrilateral.**

**Is AB + BC + CD + DA > AC + BD?**

**Solution**:-

As we know,

The sum of any two sides is always greater than the third in a triangle

Considering the ΔABC,

Since, AB + BC > CA ………………………….. [Equation i]

Now, Considering the ΔBCD

Since, BC + CD > DB ………………………….. [Equation ii]

Now, Considering the ΔCDA

Since, CD + DA > AC ………………………….. [Equation iii]

Now, Considering the ΔDAB

Since, DA + AB > DB ………………………….. [Equation iv]

Adding all the equations [i], [ii], [iii] and [iv],

AB + BC + BC + CD + CD + DA + DA + AB is greater (>) than sum of CA + DB + AC + DB

2AB + 2BC + 2CD + 2DA is greater (>) than sum of 2CA + 2DB

Removing out 2 on both the side,

2(AB + BC + CA + DA) is greater (>) than 2(CA + DB)

AB + BC + CA + DA is greater (>) than sum of CA + DB

So, the given expression is true.

**Question 5.**

**ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)**

**Solution**:-

From the question,

Since, ABCD is quadrilateral and P is the point where the diagonals are intersect.

As we know,

The sum of any two sides is always greater than the third in a triangle

Considering the ΔPAB,

PA + PB < AB ………………………….. [Equation i]

Considering the ΔPBC

Since, PB + PC < BC ………………………….. [Equation ii]

Considering the ΔPCD

Since, PC + PD < CD ………………………….. [Equation iii]

Considering the ΔPDA

Since, PD + PA < DA ………………………….. [Equation iv]

Adding all the equation [i], [ii], [iii] and [iv],

PA + PB + PB + PC + PC + PD + PD + PA is less than AB + BC + CD + DA

2PA + 2PB + 2PC + 2PD is less than AB + BC + CD + DA

2PA + 2PC + 2PB + 2PD is less than AB + BC + CD + DA

2(PA + PC) + 2(PB + PD) is less than AB + BC + CD + DA

From the figure,

AC = (PA + PC) & (BD = PB + PD)

Then,

2AC + 2BD is less than (<) sum of AB + BC + CD + DA

2(AC + BD) is less than (<) sum of AB + BC + CD + DA

So, the given expression is true.

**Question 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:-**

As we know,

The sum of any two sides is always greater than the third in a triangle

From the question, two sides of triangle are 12 cm and 15 cm.

So, the third side should be less than the sum of other two sides,

12 + 15 = 27 cm.

Then, it is given that the third side cannot be less than the difference of the two sides, 15 – 12

= 3 cm

Therefore, the length of the third side lies between 3 cm and 27 cm.