Detailed NCERT Solutions for Class 7 Maths - Chapter 6 - The Triangle and its properties - Exercise 6.4

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Please find the detailed Solutions for class 7th Maths Chapter 6 – The Triangle and its properties – Exercise 6.4 Solutions, you can check all the solutions from here chapter wise

NCERT Class 7th Mathematics Solutions – Detailed Chapter Wise

NCERT Class 7th Mathematics Chapter 6 – The Triangle and its properties – Exercise 6.4 Solutions

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: Since, the sum of the lengths of any two sides of a triangle is greater than the third side.

(i) 2 cm, 3 cm, 5 cm

2 + 3 > 5 No

3 + 5 > 2 Yes

2+ 5 > 3 Yes

So, this triangles is not possible.

(ii) 3 cm, 6 cm, 7 cm

3 + 6 > 7 Yes

6 + 7 > 3 Yes

3 + 7 > 6 Yes

So, this triangle is possible

(iii) 6 cm, 3 cm, 2 cm

6 + 3 > 2 Yes

6 + 2 > 3 Yes

3 + 2 > 6 No

So, this triangle is not possible.

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?

Sol: Join OP, OP, OQ.

(i) OP + OQ > PQ?

Yes, POQ form a triangle.

(ii) OQ + OR > QR?

Yes, RQO form a triangle.

(iii) OR + OP > RP?

Yes, ROP form a triangle.

3. AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM?
(Consider the sides of triangles
∆ABM and ∆AMC.)

Sol: Since, the sum of the lengths of any two sides of a triangle is greater than the third side.

In ΔABM,

AB + BM > AM —-eq.(i)

In ΔAMC,

AC + MC > AM —-eq.(ii)

Adding eq. (i) and (ii)

AB + BM + AC +MC > AM + AM

AB + AC + (BM+MC) > 2AM

AB + AC + BC > 2AM

Or, AB + BC + CA > 2AM

4. ABCD is a quadrilateral.

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

Sol: Since, the sum of the lengths of any two sides of a triangle is greater than the third side.

Therefore,

In ΔABC, AB + BC > AC —-eq.(i)

In ΔADB AD + AB > DB —-eq.(ii)

In ΔADC AD + DC > AC —-eq.(iii)

In ΔDCB DC + CB > DB —-eq.(iv)

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

AB + BC + AD + AB + AD + DC + DC + CB > AC + DB + AC + DB

AB + AB + BC + BC + DC + DC + AD + AD > AC + AC + DB + DB

2AB + 2BC + 2DC + 2AD > 2AC + 2DB

2( AB + BC + DC + AD ) > 2 (AC + DB)

Dividing both sides by 2, we get

AB + BC + DC + AD > AC + DB

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

5. ABCD is quadrilateral. Is
AB + BC + CD + DA < 2 (AC + BD)?

Sol: Let diagonals AC and BD meets at point O

Since, the sum of the lengths of any two sides of a triangle is greater than the third side.

Therefore,

In ΔAOB AB < OA + OB—-eq.(i)

In ΔBOC BC < OB + OC—-eq.(ii)

In ΔCOD CD < OC + OD —-eq.(iii)

In ΔAOD DA < OD + OA—-eq.(iv)

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

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

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

AB + BC + CD + DA < 2[ (OA + OC) + (OB + OD) ]

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

Hence, it is proved.

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two
measures should the length of the third side fall?

Sol: Since, the sum of the lengths of any two sides of a triangle is greater than the third side.

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

So, the sides should be less than 12 cm + 15 cm = 27 cm

And it should be greater than 15 cm – 12 cm = 3 cm

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