- 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
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 (Rationalised Contents)
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
(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.
Related Links:
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10