**NCERT
Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4**

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its
Properties Ex 6.4 are the part of NCERT Solutions for Class 7 Maths. Here you
can find the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its
Properties Ex 6.4.

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5**

**Ex 6.4 Class 7 Maths Question 1.**

Is it possible to have a triangle with the following
sides?(i) 2 cm, 3 cm, 5 cm

(ii) 3 cm, 6 cm, 7 cm

(iii) 6 cm, 3 cm, 2 cm

**Solution:
**We know that the sum of any two sides of a triangle
must be greater than the third side of the triangle.

(i) Given sides are 2 cm, 3 cm, 5 cm

Sum of the two sides = 2 cm + 3 cm = 5 cm

Third side = 5 cm

We have, sum of the two sides = the third side, i.e., 5 cm = 5 cm

Hence, the triangle is not possible.

(ii) Given sides are 3 cm, 6 cm, 7 cm

Sum of the two sides = 3 cm + 6 cm = 9 cm

Third side = 7 cm

We have, sum of the two sides > the third side, i.e., 9 cm > 7 cm

Hence, the triangle is possible.

(iii) Given sides are 6 cm, 3 cm, 2 cm

Sum of the two sides = 3 cm + 2 cm = 5 cm

Third side = 6 cm

We have, sum of the two sides < the third side, i.e., 5 cm < 6 cm

Hence, the triangle is not possible.

**Ex 6.4 Class 7 Maths 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:
**(i) Yes. In ∆OPQ, we have OP + OQ > PQ

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

(ii) Yes. In ∆OQR, we have OQ + OR > QR

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

(iii) Yes. In ∆OPR, we have OR + OP > RP

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

**Ex 6.4 Class 7 Maths Question 3.**

AM is a median of a triangle ABC.Is AB + BC + CA > 2AM?

(Consider the sides of triangles ∆ABM and ∆AMC.)

**Solution:**

Yes.**
**In ∆ABM, we have AB + BM > AM …(i)

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

In ∆AMC, we have AC + CM > AM …(ii)

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

Adding eqn (i) and (ii), we get

AB + AC + BM + CM > 2AM

AB + AC + BC + > 2AM

AB + BC + CA > 2AM

Thus, AB + BC + CA > 2AM Hence, proved.

**Ex 6.4 Class 7 Maths Question 4.**

ABCD is a quadrilateral.Is AB + BC + CD + DA > AC + BD?

**Solution:****
**Yes.

In
the given quadrilateral ABCD, AC and BD are diagonals.

In ∆ABC, we have AB + BC > AC …(i)

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

In ∆BDC, we have BC + CD > BD …(ii)

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

In ∆ADC, we have CD + DA > AC …(iii)

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

In ∆DAB, we have DA + AB > BD …(iv)

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

Adding eqn. (i), (ii), (iii) and (iv), we get

2AB + 2BC + 2CD + 2DA > 2AC + 2BD

AB
+ BC + CD + DA > AC + BD [Dividing
both sides by 2]

Thus, AB + BC + CD + DA > AC + BD Hence, proved.

**Ex 6.4 Class 7 Maths Question 5.**

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

**Solution:****
**Yes.

In
the given quadrilateral ABCD, AC and BD are diagonals which intersect each
other at O.

[Any side of a triangle is less than the sum of the other two sides]

In ∆BOC, we have BC < BO + CO …(ii)

[Any side of a triangle is less than the sum of the other two sides]

In ∆COD, we have CD < CO + DO …(iii)

[Any side of a triangle is less than the sum of the other two sides]

In ∆AOD, we have DA < DO + AO …(iv)

[Any side of a triangle is less than the sum of the other two sides]

Adding eqn. (i), (ii), (iii) and (iv), we get

AB + BC + CD + DA < 2AO + 2BO + 2CO + 2DO

AB + BC + CD + DA < 2(AO + BO + CO + DO)

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

Thus, AB + BC + CD + DA < 2(AC + BD) Hence, proved.

**Ex 6.4 Class 7 Maths Question 6.**

The length 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:****
**Sum of the two sides = 12 cm + 15
cm = 27 cm

Difference of the two sides = 15 cm – 12 cm = 3 cm

∴ The measure of the third side should fall between 3 cm and 27 cm.

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