**The Triangle and its Properties**

**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)** Since, 2 + 3 > 5

So the given side lengths cannot form a triangle.**(ii)** We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3

i. e., the sum of any two sides is greater than the third side.

So, these side lengths form a triangle.**(iii)** We have, 6 + 3 > 2, 3 + 2 6

So, the given side lengths cannot form a triangle.

**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 because on joining OP and OQ, we get a ∆OPQ and in a triangle, sum of the lengths of any two sides is always greater than the third side.

**(ii)** Yes, OQ + OR > QR, because on joining OQ and

OR, we get a ∆OQR and in a triangle, sum of the length of any two sides is always greater than the third side.

**(iii)** Yes, OR + OP > RP, because on joining OR and OP, we get a ∆OPR and in a triangle, sum of the lengths of any two sides is always greater than the third side.

**Question 3.****AM is median of a triangle ABC. Is AB + BC + CA > 2AM?**(Consider the sides of triangles ∆ABM and ∆AMC.)

**Solution:**

Using triangle inequality property in triangles ABM and AMC, we have

AB + BM > AM …(1) and, AC + MC > AM …(2)

Adding (1) and (2) on both sides, we get

AB + (BM + MC) + AC > AM + AM

⇒ AB + BC + AC > 2AM

**Question 4.****ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?**

**Solution:**

**Question 5.****ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?**

**Solution:**

Let ABCD be a quadrilateral and its diagonals AC and BD intersect at O. Using triangle inequality property, we have

In ∆OAB

OA + OB > AB ……(1)

**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:**

Let x cm be the length of the third side.

Thus, 12 + 15 > x, x + 12 > 15 and x + 15 > 12

⇒ 27 > x, x > 3 and x > -3

The numbers between 3 and 27 satisfy these.

∴ The length of the third side could be any length between 3 cm and 27 cm