**Chapter 6 The Triangle 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

Solution:

We know that for a triangle, the sum of any two sides must be greater than the third side.

(i) Given sides are 2 cm, 3 cm, 5 cm

Sum of the two sides = 2 cm + 3 cm = 5 cm Third side = 5 cm

We have, Sum of any two sides = the third side i.e. 5 cm = 5 cm

Hence, the triangle is not possible.

(ii) Given sides are 3 cm, 6 cm, 7 cm

Sum of the two sides = 3 cm + 6 cm = 9 cm Third side = 7 cm

We have sum of any two sides > the third

side. i.e. 9 cm > 7 cm

Hence, the triangle is possible.

(iii) Given sides are 6 cm, 3 cm, 2 cm

Sum of the two sides = 3 cm + 2 cm – 5 cm Third side = 6 cm

We have, sum of any two sides < the third sid6 i.e. 5 cm < 6 cm Hence, the triangle is not possible.

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, In ∆OPQ, we have

OP + OQ > PQ

[Sum of any two sides of a triangle is greater than the third side]

(ii) Yes, In ∆OQR, we have OQ + OR > QR

[Sum of any two sides of a triangle is greater than the third side]

(iii) Yes, In ∆OPR, we have OR + OP > RP

[Sum of any two sides of a triangle is greater than the third side]

Question 3.

AM is a median of a triangle ABC.

Is AB + BC + CA > 2AM ?

(Consider the sides of triangles ∆ABM and ∆AMC)

Solution:

Yes, In AABM, we have

AB + BM > AM …(i)

[Sum of any two sides of a triangle is greater than the third side]

In ∆AMC, we have

AC + CM > AM …(ii)

[Sum of any two sides of a triangle is greater than the third side]

Adding eq (i) and (ii), we have

AB + AC + BM + CM > 2AM

AB + AC + BC + > 2AM

AB + BC + CA > 2AM

Hence, proved.

Question 4.

ABCD is a quadrilateral.

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

Solution:

Yes, Given Undefined control sequence \squareABCD in which AC and BD are its diagonals.

In ∆ABC, we have

AB + BC > AC …(i)

[Sum of any two sides is greater than the third side]

In ∆BDC, we have

BC + CD > BD …(ii)

[Sum of any two sides is greater than the third side]

In ∆ADC, we have

CD + DA > AC …(iii)

[Sum of any two sides is greater than the third side]

In ∆DAB, we have

DA + AB > BD …(iv)

[Sum of any two sides is greater than the third side]

Adding eq. (i), (ii), (iii) and (iv), we get

2AB + 2BC + 2CD + 2DA > 2AC + 2BD or AB + BC + CD + DA > AC + BD [Dividing both sides by 2]

Hence, proved.

Question 5.

ABCD is a quadrilateral.

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

Solution:

Yes, we have a quadrilateral ABCD.

In ∆AOB, we have AB < AO + BO …(i)

[Any side of a triangle is less than the sum of other two sides]

In ∆BOC, we have

BC < BO + CO …(ii)

[Any side of a quadrilateral is less than the sum of other two sides]

In ∆COD, we have

CD < CO + DO …(iii)

[Any side of a triangle is less than the sum of other two sides]

In ∆AOD, we have

DA < DO + AO …(iv)

[Any side of a triangle is less than the sum of other two sides]

Adding eq. (i), (ii), (iii) and (iv), we have

AB + BC + CD + DA

∠2AO + ∠BO + ∠CO + ∠DO

∠2(AO + BO + CO + DO)

∠2 [(AO + CO) + (BO + DO)]

∠2(AC + BD)

Thus, AB + BC + CD + DA < 2(AC + BD)

Hence, proved.

Question 6.

The length 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:

Sum of two sides

= 12 cm + 15 cm = 27 cm

Difference of the two sides

= 15 cm – 12 cm = 3 cm

∴ The measure of third side should fall between 3 cm and 27 cm.