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

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?

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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 )

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

In Δ 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 ?

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Answer:

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

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Therefore,

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

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

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

In Δ 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)?

Answer:

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

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Therefore,

In Δ PXQ, PQ < XP + XQ – – – – – – – – – – (1)

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

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

In Δ 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?

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.