# 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**

Sol :

In a triangle, the sum of the lengths of either two sides is always greater than the third side

(i) Given that , the sides of the triangle are 2 cm , 3 cm , 5 cm

It can be observed that ,

2 + 3 = 5 cm

However, 5 cm = 5 cm

Hence, this triangle is not possible.

(ii) Given that, the sides of the triangle are 3 cm, 6 cm, 7 cm.

Here, 3 + 6 = 9 cm > 7 cm

6 + 7 = 13 cm > 3 cm

3 + 7 = 10 cm > 6 cm

Hence, this triangle is possible.

(iii) Given that, the sides of the triangle are 6 cm, 3 cm, 2 cm.

Here, 6 + 3 = 9 cm > 2 cm

However, 3 + 2 = 5 cm < 6 cm

Hence, this triangle is not possible.

QUESTION 2

**Take any point O in the interior of a triangle PQR . ls**

**(i) OP + OQ > PQ ?**

**(ii) OQ + OR > QR ?**

**(iii) OR + OP > RP ?**

Sol :

If O is a point in the interior of a given triangle, then three triangles △OPQ, △OQR, and △ORP 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 △OPQ is a triangle with sides OR , OQ , and PQ.

OP + OQ > PQ

(ii) yes , as △OQR is a triangle with sides OR , OQ and QR

OQ + OR > QR

(iii) yes , as △ORP is a triangle with sides OR , OP and PR

OR + OP > PR

QUESTION 3

**AM is a median of a triangle ABC.**

**Is AB + BC + CA > 2 AM ?**

**(Consider the sides of triangles △ABM and △AMC)**

Sol :

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

In △ABM,

AB + BM > AM ….(i)

Similarly, in △ACM ,

AC + CM > AM …(ii)

Adding equation (i) and (ii) ,

AB + BM + MC + AC > AM + AM

AB + BC + AC > 2 AM

Yes, the given expression is true.

QUESTION 4

**ABCD is quadrilateral.**

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

Sol :

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

Considering △ABC.

AB + BC > CA …(i)

In △BCD ,

BC + CD > DB …(ii)

In △CDA

CD + DA > AC

In △DAB,

DA + AB > DB ..(iii)

Adding equations (i) , (ii) , (iii) and (iv) , we obtain

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

2AB + 2BC + 2CD + 2DA > 2AC + 2BD

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

(AB + BC + CD + DA ) > (AC + BD)

Yes , the given expression is true

QUESTION 5

**ABCD is quadrilateral.**

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

Sol :

In a triangle , the sum of the lengths of either two sides is always greater than the third side.

Considering △OAB ,

OA + OB > AB ..(1)

In △OBC

OB + OC > BC ..(ii)

In △OCD ,

OC + OD > CD ..(iii)

In △ODA ,

OD + OA > DA ..(iv)

Adding equations (i) , (ii) , (iii) and (iv) , we obtain

OA + OB + OB + OC + OC + OD + OA > AB + BC + CD + DA

2OA + 2OB + 2OC + 2OD > AB + BC + CD + DA

2OA + 2OC + 2OB + 2OD > AB + BC + CD + DA

2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA

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

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

Yes , the given expression is true