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

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.



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.

In ∆AOB, we have AB < AO + BO      …(i)
[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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