# Ncert Solutions For Class 7 Maths Ex 6.4

## Ncert Solutions For Class 7 Maths Chapter 6 Ex 6.4

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

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?

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$\Delta$ PQO and Δ$\Delta$ POR )

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 Δ$\Delta$ PQO, PQ + QO > PO – – – – – – – – – – (1)

In Δ$\Delta$ POR, PR + OQ > PO – – – – – – – – – – – (2)

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 ?

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

Therefore,

In Δ$\Delta$ PQR, PQ + QR > PR – – – – – – – – – – (1)

In Δ$\Delta$ PSR, PS + SR > PR – – – – – – – – – (2)

In Δ$\Delta$ SRQ, SR + RQ > SQ – – – – – – – – – (3)

In Δ$\Delta$ PSQ, PS + PQ > SQ – – – – – – – – – – (4)

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)?

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 Δ$\Delta$ PXQ, PQ < XP + XQ – – – – – – – – – – (1)

In Δ$\Delta$ QXR, QR < XQ + XR – – – – – – – – – – – (2)

In Δ$\Delta$ RXS, RS < XR + XS – – – – – – – – – – – (3)

In Δ$\Delta$ PXS, SP < XS + XP – – – – – – – – – – (4)

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?

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.