Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4

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:

(i) 2 cm, 3 cm, 5 cm

∵ 2 cm + 3 cm= 5 cm and the third side = 5 cm.

∴ Sum of the measures of lengths of two sides = length of third side which is not possible

i.e. A triangle cannot be possible with these sides.

(ii) 3 cm, 6 cm, 7 cm

∵ 3 cm + 6 cm = 9 cm and 9 cm > 7 cm

3 cm + 7 cm = 10 cm and 10 cm > 6 cm

6 cm + 7 cm = 13 cm and 13 cm > 3 cm

Thus, a triangle can be possible with these sides.

(iii) 6 cm, 3 cm and 2 cm

∵ 6 cm + 3 cm = 9 cm and 9 cm > 2 cm

3 cm + 2 cm = 5 cm and 5 cm > 6 cm

2 cm + 6 cm = 8 cm and 8 cm > 5 cm

∴ A triangle cannot be possible with these sides.

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, OP + OQ > PQ

[ ∵ The sum of lengths of any two sides of a triangle is greater than the length of the third side.]

Similarly,

(ii) Yes, OQ + OR > QR

(iii) Yes, OR + OP > RP

Question 3.

AM is a median of a triangle ABC.

Is AB + BC + CA > 2AM? (Consider the sides of triangles ∆ABM and ∆AMC.)

Solution:

Since, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

∴ In ∆ABM, we have

(AB + BM) >AM … (1)

Similarly, in ∆ACM,

(CA + CM) >AM … (2)

Adding (1) and (2), we have [(AB + BM) + (CA + CM)] > AM + AM

⇒ [AB + (BM + CM) + CA] > 2AM

⇒ [AB + (BC) + CA] > 2AM

Thus, (AB + BC + CA) > 2AM

Question 4.

ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD?

Solution:

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

∵ In AABC, we have

(AB + BC) >AC … (1)

Similarly, in ∆ACD, we have

(CD + DA) >AC … (2)

Adding (1) and (2), we have [(AB + BC) + (CD + DA)] > 2AC … (3)

Again,

In ∆ABD, we have

AB + DA >BD … (4)

In ABCD, we have

BC + CD >BD … (5)

Adding (4) and (5), we have [(AB + DA) + (BC + CD)] > 2BD … (6)

Now, adding (3) and (6), we have

2[(AB + BC) + (CD + DA)] > 2(AC + BD) or (AB + BC + CD + DA) > (AC + BD)

Question 5.

ABCD is quadrilateral.

Is AB + BC + CD + DA < 2(AC + BD)? Solution: Since, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

∴ In ∆OAB, we have (OA + OB) >AB … (1)

Similarly,

In ∆OBC, we have

(OB + OC) >BC … (2)

In AOCD, we have

(OC + OD) >CD … (3)

In ∆OAD, we have

(OA + OD) >AD … (4)

Adding (1), (2), (3) and (4), we have 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:

Since, the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

∴ Sum of 12 cm and 15 cm is greater than the length of the third side.

i.e. (12 cm + 15 cm) > (Third side)

i.e. 27 cm > Third side

or Third side < 27 cm

Also, the difference of the lengths of any two sides is less than the length of the third side.

∴ (15 cm – 12 cm) < Third side

or 3 cm < Third side.

Thus, we have 3 cm < Third side < 27 cm

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