**Question 1:**

**Is it possible to have a triangle with the following sides?**

**(a) 2cm, 3cm 5cm**

**(b) 3cm, 6cm, 7cm**

**(c) 6cm, 3cm, 2cm**

*Answer:*

Since the triangle is possible having its sum of the lengths of any two sides would be greater than that length of the third side.

(a) 2cm , 3cm 5cm

2 + 3 > 5 No

2 + 5 > 3 Yes

3 + 5 > 2 Yes

This triangle is not possible.

(b) 3cm, 6cm and 7cm

3 + 6> 7 Yes

6 + 7 > 3 Yes

3 + 7 > 6 Yes

This triangle is possible.

( c) 6cm, 3cm and 2cm

6 + 3 > 2 Yes

6 + 2 > 3 Yes

2 + 3 > 6 No

This triangle is not possible.

** **

**Question 2:**

**Take any point P in the interior of a triangle ABC is:**

**(a) PA + PB > AB?**

**(b) PB + PC > BC?**

**(C ) PC + PA > CA?**

*Answers:*

Join PC, PB and PA

(a) Is PA + PB > AB?

Yes, APB forms a triangle.

(b) Is PB + PC > BC?

Yes, CBP forms a triangle.

( c) Is PC + PA > CA?

Yes, CPA forms a triangle.

** **

**Question 3:**

**PO is a median of a triangle PQR. Is PQ + QR + RP > 2PO? (Consider the sides of a triangle as Δ PQO and Δ POR )**

*Answer:*

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

Therefore,

In

In

Adding equation (1) and (2) we get,

PQ + PO + PR + OR > PO + PO

=> PQ + PR + ( QO + OR ) > 2PO

=> PQ + PR + QR > 2 PO

Hence, it is true.

** **

**Question 4:**

**PQRS is a quadrilateral. Is PQ + QR + RS + RP > PR + QS ?**

*Answer:*

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

Therefore,

In

In

In

In

Adding the equations (1) , (2) , (3) and (4) we get,

PQ + QR + PS + SR + RQ + PS + PQ > PS + PS + SQ + SQ

=>( PQ + PQ ) + ( QR + QR ) + ( PS + PS ) + ( SR + SR ) > 2 PR + 2 SQ

=> 2 PQ + 2 QR + 2 PS + 2 SR > 2 PR + 2 SQ

=> 2 ( PQ + QR + PS + SR) > 2 ( PR + SQ )

=> PQ + QR + PS + SR > PR + SQ

=> PQ + QR + RS + SP > PR + SQ

Hence, it is true.

** **

**Question 5:**

**PQRS is a quadrilateral. Is PQ + QR + RS + SA < 2 (PR + QS)?**

*Answer:*

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

Therefore,

In

In

In

In

Adding the equations (1) , (2) , (3) and (4) We get,

PQ + QR + RS + SP < XP + XQ + XQ + XR + XR + XS + XS + XP

=> PQ + QR + RS + SP < 2XP + 2 XQ + 2 XR + XS

=> PQ + QR + RS + SP < 2[(PX + XR) + (XS + XQ)]

=> PQ + QR + RS + SP < 2[PR + QS]

Hence it is proved.

** **

**Question 6:**

**The lengths of the two sides of a given triangle are 12cm and 15cm respectively. 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 should be greater than the length of the third side.

It is given that the two sides of a triangle are 15cm and 12cm.

Therefore, the third side should be less than 15+ 12 = 27cm

Also that, the third side cannot be less than the difference of the other two sides.

Therefore, the third side has to be greater than 15-12 = 3cm.

Hence, the third side could be the length more than 3cm and less than 27cm.