The Triangle and Its Properties Exercise 6.4 NCERT Solution for Class 7 Maths Chapter 6
The Triangle and Its Properties Exercise 6.1
The Triangle and Its Properties Exercise 6.2
The Triangle and Its Properties Exercise 6.3
The Triangle and Its Properties Exercise 6.5
Exercise 6.4
Q.1:- Is it possible to have a triangle with the following sides?
(i) 2cm, 3cm, 5cm
(ii) 3cm, 6cm, 7cm
(iii) 6cm, 3cm, 2cm
Solution 1:-
Concept:- A triangle is possible if sum of the lengths of any two sides would be greater
than the length of third side.
Solution 1(i):- Let us check this
2 + 3 > 5 No
2 + 5 > 7 Yes
5 + 7 > 2 Yes
This Triangle is not Possible.
Solution 1(ii):- Let us check this
3 + 6 > 7 Yes
3 + 7 > 6 Yes
5 + 7 > 3 Yes
This Triangle is Possible.
Solution 1(iii):- Let us check this
2 + 3 > 6 No
2 + 6 > 3 Yes
3 + 6 > 2 Yes
This Triangle is not Possible.
Q.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 2(i):- Join OR,OP and OQ
Is OP+OQ > PQ ?
Yes, because POQ form a triangle.
Solution 2(ii):- Is OQ+OR > QR ?
Yes, because QOR form a triangle.
Solution 2(iii):- Is OR+OP > RP ?
Yes, because POR form a triangle.
Q.3:- AM is a median of a triangle ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles ∆ABM and ∆AMC.)
Solution 3:- Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
In Triangle ABM, AB + BM > AM ……… equation (i)
In Triangle AMC, AC + MC > AM ………. equation (ii)
By adding equation (i) + (ii)
AB + BM + AC + MC > AM + AM
AB + AC + ( BM + MC ) > 2AM
AB + AC + BC > 2 AM
Hence, Yes it is possible.
Q.4:- ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?
Solution 4:- Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
In Triangle ABC, AB + BC > AC ……. equation (i)
In Triangle ADC, AD + DC > AC ……..equation (ii)
In Triangle DCB, DC + CB > BD …….equation (iii)
In Triangle DBA, AB + AD > BD …..equation (iv)
Adding equations (i), (ii), (iii) and (iv)
AB + BC + AD + DC + DC + CB + AB + AD > AC + AC + BD + BD
(AB + AB) + (BC + CB) + (AD + AD) + (DC + DC) > 2AC + 2BD
2AB + 2BC + 2AD + 2DC > 2(AC + BD)
2 (AB+BC+AD+DC) > 2(AC + BD)
AB + BC + CD + DA > AC + BD
Yes, it is True/Possible.
Q.5:- ABCD is quadrilateral. Is
AB + BC + CD + DA < 2 (AC + BD)?
Solution 5:-
Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
In Triangle AOB, AB < OA + OB ……. equation (i)
In Triangle BOC, BC < OB + OC ……..equation (ii)
In Triangle COD, CD < OC + OD …….equation (iii)
In Triangle DOA, DA < OA + OD …..equation (iv)
Adding equations (i), (ii), (iii) and (iv)
AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OA + OD,
AB + BC + CD + DA < (OA + OA) + (OB + OB) + (OC + OC) + (OD + OD),
AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD,
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)
Yes, it is True/Possible.
Q.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 6:-
We know that the sum of the two sides of a triangle is always greater than the third side.
Therefore, third side has to be less than the sum of two side,
Thus the third side should be less than (12cm + 15cm) = 27cm
The side cannot be less than the difference of the two sides.
Thus the third side should be more than (15cm – 12cm) = 3cm
The length of the third side could be any length greater than 3 and less than 27cm.
Answer :- Between 3cm and 27cm.
NCERT Solutions for Class 7 Math, Chapter 6 – The Triangle and Its Properties, Exercise-6.4