These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 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

Answer:

(i) 2 cm, 3 cm, 5 cm

2 cm + 3 cm = 5 cm

third side = 5 cm

∵ Sum of the measures of length of two sides = length of the third side

∴ It is not possible with these sides to form a triangle.

(ii) 3 cm, 6 cm, 7 cm

3 cm + 6 cm = 9 cm

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

Thus, 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?

Answer:

(i) Yes OP + QQ> PQ

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

(ii) Yes, OQ + OR > QR

(iii) Yes, OR + OP > RP

Question 3.

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

(Consider the sides of triangles ΔABM and ΔAMC.)

Answer:

Since, the sum of the length of any two sides of a triangle is greater than the length

of the third side

∴ In ΔABM, we have

(AB + BM) > AM …………. (i)

Similarly in ΔACM

(CA + CM) > AM ………. (ii)

Adding (1) and (2), we have

[ (AB + BM) + (CA + CM)] > AM + AM

[ AB + (BM + CM) + CA] > 2 AM

[ AB + BC + CA]> 2AM

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

Question 4.

ABCD is a quadrilateral.

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

Answer:

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

In ABC, we have

AB + BC > AC …………(i)

Similarly, in ΔACD, we have

CD + DA > AC …………(ii)

Adding (i) and (ii), we get

[(AB + BC) + (CD + DA)] > 2 AC …………(iii)

Again In ΔABD, we have

AB + DA > BD …………….(iv)

In ΔBCD, we have

BC + CD > BD …(v)

Adding (iv) and (v), we get

[(AB + DA) + (BC + CD)] > 2 BD …………..(vi)

Now Adding (iii) and (vi), we get

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

Question 5.

ABCD is a quadrilateral.

Is AB + BC + CD + DA < 2(AC + BD)?

Answer:

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

∴ In ΔAOB, we have (OA + OB) > AB …………(i)

Similarly,

In ΔOBC, we have

(OB + OC) > BC …………..(ii)

In ΔOCD, we have

(OC + OD) > CD ………….(iii)

In ΔOAD we have

(OA + OD) > AD …………..(iv)

Adding (i), (ii), (iii) and (iv), 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?

Answer:

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 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 3 cm < Third side

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

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