Students can Download Chapter 6 The Triangles and Its Properties Ex 6.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4

Question 1.

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

i) 2 cm, 3 cm, 5 cm.

Solution:

greatest side is 5 cms.

2 + 3 = 5

Here sum of the side is equal to third side.

So it is impossible to construct the triangle by given measures.

Because the property of a triangle says that the sum of the two sides of a triangle is greater than the length of the third side.

ii) 3 cm, 5 cm, 7 cm

Solution:

The greatest side is 7 cm.

3 + 6 = 9 cms > 7

6 + 7 = 13 > 3

7 + 3 = 10 > 6

∴ It is possible to construct the triangle with the sides of 3, 6 and 7 cms.

iii) 6 cm, 3 cm, 2 cm

6 + 3 = 9 > 2

3 + 2 = 5 > 6

2 + 6 = 8 > 3

∴ It is not possible to construct the triangle with the sides of 6, 3 and 2 cms.

Question 2.

Take any point ‘O’ in the interior of a triangle PQR. Is

i) OP + OQ > PQ

Solution:

Yes. Sum of the length of any two sides of a triangle is greater than the length of the 3 side.

ii) OQ + OR > QR

Solution:

Yes, Sum of the length of any two sides of a triangle is greater than the length of the 3^{rd} side.

iii) OR + OP > RP

Solution:

Yes, Sum of the length of any two sides of a triangle is greater than the length of the 3^{rd} side.

Question 3.

AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM ?

(Consider the sides of triangles ∆ ABM and ∆ AMC.)

Solution:

Consider the triangle ABM

In ∆ ABM

AB + BM > AM — 1 (Sum of the length of any 2 sides of a triangle is greater that the 3^{rd} side.)

In ∆ ACM

CM + CA > AM — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

Add 1 and 2

(AB + BM) + (CM + CA) > AM + AM

AB + (BM + CM) + CA > 2AM

AB + BC + CA > 2AM (yes, it is proved.)

Question 4.

ABCD is a quadrilateral.

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

Solution:

Consider the ∆ ABC

AB + BC > CA — 1 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

In ∆ ACD

AD + CD > AC — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

Adding 1 and 2

AB + BC + CD + DA > 2AC — 3

In ∆ ABD

AB + DA > BD — 4 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

In ∆ BCD

BC + CD > BD — 5 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

Adding 4 and 5

AB + DA + BC + CD > 2BD — 6

Adding 3 and 6

∴ AB + BC + CD + DA > AC + BD (yes, it is proved)

Question 5.

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

Solution:

In ∆ AOB OA + OB > AB — 1 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

In ∆ BOC

OB + OC > BC — 2 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

In ∆ COD

OC + CD > CD — 3 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

In ∆ AOD

OA + OD > AD — 4 (Sum of the length of any 2 sides of a triangle is greater than the 3^{rd} side.)

Add all the 4

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

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?

Solution:

Let ‘x’ cm to the length of the 3^{rd} side.

According to the sum of the length of any 2 sides of a triangle is greater than 3^{rd} side.

∴ 12 + 15 >x

27 > x

or x < 27 15 + x > 12

x > 12 – 15 = -3

x + 12 > 15

x > 15 – 12 = 3

∴ x > -3 and x > 3 we should take x > 3

∴ The length of 3^{rd} side of a triangle should be in between 3 cm and 27 cms.