Class 7 Maths NCERT Solutions Chapter 6 - The Triangle and its Properties Exercise 6.4

 Class 7 Maths NCERT Solutions Chapter 6 - The Triangle and its Properties Exercise 6.4

NCERT Solutions for Class 7 Maths Exercise 6.4 Chapter 6 The Triangle and its Properties. The main aim is to help students understand and crack these problems.

NCERT Solutions for Class 7 Maths Chapter 6 wherein problems are solved step by step with 

detailed explanations. Two special triangles equilateral and isosceles and 

the sum of the lengths of two sides of triangles are the topics covered in this exercise 

of NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties.

Question 1: Is it possible to have a triangle with the following sides?

i) Is it possible to have a triangle with the following sides 2 cm, 3 cm, and 5 cm?

Ans: we know that 

The sum of two sides of the triangle is greater than the third side.

2 + 3 = 5 = 5 No 

Hence, The Triangle is not possible. 

ii) Is it possible to have a triangle with the following sides 3 cm, 6 cm, and 7 cm?

Ans: we know that

The sum of the two sides of the triangle is greater than the third side.

3 cm, 6 cm , 7 cm

3 + 6 = 9 > 7 Yes 

7 + 6 = 13 > 3 Yes 

7 + 3 = 10 > 6 Yes 

Hence the triangle is possible.

iii) Is it possible to have a triangle with the following sides 6 cm, 3 cm, and 2 cm?

Ans: we know that 

The sum of the two sides of the triangle is greater than the third side.

6 + 3 = 9 > 2  Yes 

3 + 2 = 5 < 6  No

6 + 2 = 8 > 3  Yes 

Hence, The Triangle is not possible. 

Question 2: Take any point O in the interior of a triangle  PQR . Is 

i) OP + OQ > PQ ? 

Ans: In the given triangle, join OR , OQ, and PO. 

From the diagram, it can be seen that PQR is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, OP + OR > PQ . Yes OP + OR > PQ. 

ii). OQ + OR > QR ? 

Ans: From the diagram, it can be seen that QOP is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So OQ + OR > QR ? Yes OQ + OR > QR

iii) OR + OP > RP? 

Ans: From the diagram, it can be seen that OPR is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So OR + OP > RP , Yes OR + OP > RP

Question 3: AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM ? 

( consider the side of the triangle 🔺️ABM and 🔺️AMC.)

Ans: we know that 

The sum of the length of any two sides is always greater than the third side. 

Now consider the 🔺️ABM.

Here, AB + BM > AM... [ equation 1 ]

Then, consider the 🔺️ACM.

Here, AC + CM > AM... [ equation 2]

By adding equations [1] and [2] we get, 

AB + BM + AC + CM > AM + AM From the figure we have, BC = BM + CM + AB + BC + AC > 2AM 

Hence, the given expression is true.

Question 4 : ABCD is a quadrilateral. 

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

Ans: In a triangle, the sum of the length of the two sides is always greater than the third side. 

Considering ABC

AB + BC > CA [ 1 ]

In BCD 

BC + CD > DB [2] 

In CDA

CD + DA > AC [3] 

In DAB

DA + AB > DB [4]

Adding equation [1] ,[2], [3] and [4] we obtain.

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

2AB + 2BC + 2CD + 2DA > 2AC + 2BD 

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

AB + BC + CD + DA > AC + BD .

Yes, the giving expression is true.

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

Ans: in a triangle, the sum of the length of two sides of the triangle is always greater than the third side.

Considering AOB

OA + OB > AB         [1]

in OBC

OB + OC > BC    [2]

in OCD

OC + OD > CD     [3]

in ODA

OD + OA > DA    [4]

Adding equation [1],[2],[3] and [4] we obtain.

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

2OA + 2OB + 2OD > AB + BC + CD + BA.

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

From the figure 

OA + OC = AC 

OB + OD = BD 

2AC + 2BD > AB + BC + CD + DA

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

Yes, the giving expression is true.

Question 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?

Ans: The sum of the length of any two sides is always greater than the third side. So, the third side's length should be less than the sum of the other two sides.

12 + 15 = 27cm

and the third side can not be less than the difference between the two sides.

  15 - 12 = 3 cm 

So, the length of the third side falls between 3cm to 27cm.

Thank you.