NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4

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NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4

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

Chapter 6 of the Mathematics subject in Class 7 covers the concepts of Triangles and their Properties and introduces topics like the Medians and Altitudes of a Triangle, the Exterior Angle of a Triangle, the Angle Sum Property of a Triangle, Two Special Triangles which are Equilateral and Isosceles, Sum of the lengths of Two sides of a Triangle, and Right-angled Triangles and Pythagoras Property. All the questions and exercises can be decoded by students with the guidance of the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4.

Furthermore, Exercise 6.4 covers the following topics:-

1) Students have to specify whether the given numbers can be considered to make a triangle.

2) Other questions are related to proving which value is greater or smaller than the other.

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Q.1

Is it possible to have a triangle with the following sides?i 2cm,3cm,5cmii 3cm,6cm,7cmiii 6cm,3cm,2cm

Ans.

For a triangle, the sum of lengths of either two sides is always greater than the third side. (i) Given sides of the triangle are 2 cm, 3 cm and 5 cm. 2 cm + 3 cm = 5 cm So, the sum of two side is not greater than the third side. Therefore, this triangle is not possible. (ii) Given sides of the triangle are 3 cm, 6 cm and 7 cm. 3 cm + 6 cm = 9 cm So, the sum of two side is greater than the third side. Therefore, this triangle is possible. (iii) Given sides of the triangle are 6 cm, 3 cm and 2 cm. 2 cm + 3 cm = 5 cm So, the sum of two side is not greater than the third side. Therefore, 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?

Ans.

If O is a point in the interior of a given triangle, then three triangles ΔOPQ, ΔOQR, and ΔORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side. (i) Yes, as ΔOPQ is a triangle with side OP, OQ and PQ. So, OP + OQ > PQ (ii) Yes, as ΔOQR is a triangle with side OP, OQ and PQ. So, OQ + OR > QR (iii) Yes, as ΔORP is a triangle with side OP, OQ and PQ. So, OP + OR > PR MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGjbGaaeOzaiaabccacaqGpbGaaeii aiaabMgacaqGZbGaaeiiaiaabggacaqGGaGaaeiCaiaab+gacaqGPb GaaeOBaiaabshacaqGGaGaaeyAaiaab6gacaqGGaGaaeiDaiaabIga caqGLbGaaeiiaiaabMgacaqGUbGaaeiDaiaabwgacaqGYbGaaeyAai aab+gacaqGYbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGa ae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeiDaiaabkhaca qGPbGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGSaGaaeiiaiaa bshacaqGObGaaeyzaiaab6gacaqGGaGaaeiDaiaabIgacaqGYbGaae yzaiaabwgaaeaacaqG0bGaaeOCaiaabMgacaqGHbGaaeOBaiaabEga caqGSbGaaeyzaiaabohacaqGGaGaeyiLdqKaae4taiaabcfacaqGrb GaaeilaiaaysW7cqGHuoarcaqGpbGaaeyuaiaabkfacaqGSaGaaGjb VlaabggacaqGUbGaaeizaiaaysW7cqGHuoarcaqGpbGaaeOuaiaabc facaqGGaGaae4yaiaabggacaqGUbGaaeiiaiaabkgacaqGLbGaaeii aiaabogacaqGVbGaaeOBaiaabohacaqG0bGaaeOCaiaabwhacaqGJb GaaeiDaiaabwgacaqGKbGaaeOlaaqaaiaabMeacaqGUbGaaeiiaiaa bggacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaae iBaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG ZbGaaeyDaiaab2gacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabI gacaqGLbGaaeiiaiaabYgacaqGLbGaaeOBaiaabEgacaqG0bGaaeiA aiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyzaiaabMgacaqG0b GaaeiAaiaabwgacaqGYbGaaeiiaiaabshacaqG3bGaae4Baiaabcca caqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccaaeaacaqGPbGaae 4CaiaabccacaqGHbGaaeiBaiaabEhacaqGHbGaaeyEaiaabohacaqG GaGaae4zaiaabkhacaqGLbGaaeyyaiaabshacaqGLbGaaeOCaiaabc cacaqG0bGaaeiAaiaabggacaqGUbGaaeiiaiaabshacaqGObGaaeyz aiaabccacaqG0bGaaeiAaiaabMgacaqGYbGaaeizaiaabccacaqGZb GaaeyAaiaabsgacaqGLbGaaeOlaaqaaiaabIcacaqGPbGaaeykaiaa bccacaqGzbGaaeyzaiaabohacaqGSaGaaeiiaiaabggacaqGZbGaae iiaiabgs5aejaab+eacaqGqbGaaeyuaiaabccacaqGPbGaae4Caiaa bccacaqGHbGaaeiiaiaabshacaqGYbGaaeyAaiaabggacaqGUbGaae 4zaiaabYgacaqGLbGaaeiiaiaabEhacaqGPbGaaeiDaiaabIgacaqG GaGaae4CaiaabMgacaqGKbGaaeyzaiaabccacaqGpbGaaeiuaiaabY cacaqGGaGaae4taiaabgfacaqGGaGaaeyyaiaab6gacaqGKbGaaeii aiaabcfacaqGrbGaaeOlaaqaaiaabofacaqGVbGaaeilaiaabccaca qGpbGaaeiuaiaabccacaqGRaGaaeiiaiaab+eacaqGrbGaaeiiaiaa b6dacaqGGaGaaeiuaiaabgfaaeaacaqGOaGaaeyAaiaabMgacaqGPa GaaeiiaiaabMfacaqGLbGaae4CaiaabYcacaqGGaGaaeyyaiaaboha caqGGaGaeyiLdqKaae4taiaabgfacaqGsbGaaeiiaiaabMgacaqGZb GaaeiiaiaabggacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6ga caqGNbGaaeiBaiaabwgacaqGGaGaae4DaiaabMgacaqG0bGaaeiAai aabccacaqGZbGaaeyAaiaabsgacaqGLbGaaeiiaiaab+eacaqGqbGa aeilaiaabccacaqGpbGaaeyuaiaabccacaqGHbGaaeOBaiaabsgaca qGGaGaaeiuaiaabgfacaqGUaaabaGaae4uaiaab+gacaqGSaGaaeii aiaab+eacaqGrbGaaeiiaiaabUcacaqGGaGaae4taiaabkfacaqGGa GaaeOpaiaabccacaqGrbGaaeOuaaqaaiaabIcacaqGPbGaaeyAaiaa bMgacaqGPaGaaGjbVlaabMfacaqGLbGaae4CaiaabYcacaqGGaGaae yyaiaabohacaqGGaGaeyiLdqKaae4taiaabkfacaqGqbGaaeiiaiaa bMgacaqGZbGaaeiiaiaabggacaqGGaGaaeiDaiaabkhacaqGPbGaae yyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaae4DaiaabMgacaqG 0bGaaeiAaiaabccacaqGZbGaaeyAaiaabsgacaqGLbGaaeiiaiaab+ eacaqGqbGaaeilaiaabccacaqGpbGaaeyuaiaabccacaqGHbGaaeOB aiaabsgacaqGGaGaaeiuaiaabgfacaqGUaaabaGaae4uaiaab+gaca qGSaGaaeiiaiaab+eacaqGqbGaaeiiaiaabUcacaqGGaGaae4taiaa bkfacaqGGaGaaeOpaiaabccacaqGqbGaaeOuaaaaaa@A737@

Q.3

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

(Consider the sides of triangles ΔABM and ΔAMC.)

Ans.

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔABM AB+BM>AM …( i ) Similary in ΔACM, we get AC+CM>AM …( ii ) Adding (i) and (ii) to get AB+BM+AC+CM>2AM AB+CA+(BM+CM)>2AM AB+AC+BC>2AM AB+BC+CA>2AM Therefore, the given expression is true.

Q.4

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

Ans.

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔABC AB+BC>CA …( i ) Similary in ΔBCD, we get BC+CD>DB …( ii ) In ΔCDA, we get CD+DA>AC …( iii ) In ΔDAB, we get DA+AB>DB …( iv ) Adding (i), (ii), (iii) and (iv) to get AB+BC+BC+CD+CD+DA+DA+AB>CA+DB+AC+DB 2( AB+BC+CD+DA )+2(AC+DB) AB+BC+CD+DA>AC+BD Therefore, the given expression is true.

Q.5

ABCD is a quadrilateral. IsAB+BC+CD+DA<2AC+BD?

Ans.

Consider a quadrilateral ABCD.

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔOAB OA+OB>AB …( i )
Similary in ΔOBC, we get OC+OB>BC …( ii )
In ΔOCD, we get OD+OC>CD …( iii )
In ΔODA, we get OA+OD>DA …( iv )
Adding (i), (ii), (iii) and (iv) to get OA+OB+OC+OB+OD+OC+OA+OD>AB+BC+CD+DA
2( OA+OB+OC+OD )>2( AC+BD ) ( OA+OB+OC+OD )>( AC+BD )
Therefore, the given expression is true.

Q.6

The lengths of two sides of a triangle are 12cm and 15cm.Between what two measures should the length of the thirdside fall?

Ans.

In a triangle, the sum of either two sides is always greater than the third side and also, the difference of the lengths of either two sides is always lesser than the third side. Here, the third side will be lesser than the sum of these two (i.e., 12+15=27) and also, it will be greater than the difference of these two (i.e., 15−12=3). Therefore, those two measures are 27 cm and 3 cm.

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