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 part of NCERT Solutions for Class 7 Maths. Here we have given 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

Question 1.
Is it possible to have a triangle with the following sides?

1. 2 cm, 3 cm, 5 cm
2. 3 cm, 6 cm, 7 cm
3. 6 cm, 3 cm, 2 cm.

Solution:
(i) 2 cm, 3 cm, S cm We have 2 + 3 = 5
⇒ Sum of the lengths of two sides = Length of the third side
This is impossible since the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

(ii) 3 cm, 6 cm, 7 cm
We see that 3 + 6 > 7
6 + 7 > 3
7 + 3 > 6
Therefore, it is possible to have a triangle with side lengths 3 cm, 6 cm, 7 cm.

(iii) 6 cm, 3 cm, 2 cm
We see that 6 + 3 = 9 > 2
3 + 2 = 5 \(\ngtr \) 6
2 + 6 = 8 > 3
Therefore, it is not possible to have a triangle with side lengths 6 cm, 3 cm, 2 cm.

Question 2.
Take any point O in the interior of a triangle PQR. Is

1. OP + OQ > PQ ?
2. OQ + OR > QR?
3. OR + OP > RP ?

Solution:

1. Yes ! OP + OQ > PQ …(1)
   Sum of the lengths of any two sides of a triangle is greater than the length of the third side
2. Yes ! OQ + OR > QR …(2)
   Sum of the lengths of any two sides of a triangle is greater than the length of the third side
3. Yes ! OR + OP > RP …(3)
   Sum of the lengths of any two sides of a triangle is greater than the length of the third side

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

Solution:
In ∆ ABM, AB + BM > AM …(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ ACM,
CA + CM > AM …(2)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Sum (1) and (2),
(AB + BM) + (CA + CM) > AM + AM
⇒ AB + (BM + CM) + CA > 2 AM
⇒ AB + BC + CA > 2 AM.

Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD ?

Solution:
In ∆ ABC, AB + BC > AC …(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ ACD, CD + DA > AC …(2)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (1) and (2),
AB + BC + CD + DA > 2AC …(3)
In ∆ ABD, AB + DA > BD …(4)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ BCD, BC + CD > BD …(5)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (4) and (5),
AB + BC + CD + DA > 2BD …(6)
Adding (3) and (6),
2 [AB + BC + CD + DA] > 2 (AC + BD)
⇒ AB + BC + CD + DA > AC + BD.

Question 5.
ABCD is a quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

Solution:
In ∆ OAB, OA + OB > AB ….(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
In ∆ OBC, OB + OC > BC ….(2)
Sum of the lengths of any two sides of a triangle la greater than the length of the third side.
In ∆ OCA,OC + OA > CA ….(3)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ OAD, OA + OD > AD ….(4)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (1), (2), (3) and (4),
2(OA + OB + OC + OD) > AB + BC + CD + DA
⇒ 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).

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:
Let x cm be the length of the third side.
∴ Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
∴ We should have

∴ The length of the third side should be any length between 3 cm and 27 cm.

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