Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.4

1. Which of the following can be the sides of a triangle ?

(a) 8 cm, 10 cm, 18 cm

(b) 6 cm, 4 cm, 8 cm

(c) 35 cm, 38 cm, 40 cm

(d) 3 cm, 4 cm, 10 cm

Solutions:

Drawing of a triangle with given three sides is possible if sum of lengths of any two sides of a triangle is greater than the third side.

(a) Since 8 + 10 = 18

So, 8 cm, 10 cm, 18 cm cannot be the lengths of the sides of the triangle

(b) Since 6 + 4 > 8

4 + 8 >6

8 + 6 > 4

So, 6 cm, 4 cm, 8 cm can be lengths of the sides of a triangle.

(c) Since 35 + 38 > 40

38 + 40 > 35

40 + 35 > 38

So, 35 cm, 38 cm, 40 cm can be lengths of the sides of a triangle.

(d) Since 3 + 4 < 10

So, 3 cm, 4 cm, 10 cm cannot be the lengths of the sides of a triangle.

2. A point O is in interior of a ΔABC use symbols >, < or = to make the following statements true.

Answer:

(a) OA + OB > AB

(b) OB + OC > BC

(c) OA + OC > AC

3. ABCD is a quadrilateral

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

Solution:

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

Proof : In ΔABC ; AB + BC > AC

[∵ sum of lengths of any two sides of a triangle is always greater than the third side] …. (i)

In ΔADC ; CD + DA > AC

[using same reason as above] …. (ii)

In ΔABD ; AB + DA > BD

[using same reason as above] …. (iii)

In ΔBCD ; BC + CD > BD

[using same reason as above] …. (iv)

Adding (i), (ii), (iii) and (iv), we get :

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

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

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

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

Proved.

4. AD is a median of < ABC

Is AB + BC + CA > 2AD ?

Solution:

In ΔABD

AB + BD > AD …(1)

{∵ Sum of the lengths of any two sides of a triangle is greater than the third side}

In ΔACD

CA + DC > AD …(2) (using same reason as above)

Adding (1) and (2), we get

AB + BD + CA + DC > AD + AD

AB + (BD + DC) + CA > 2AD

[D is mid point of BC BD + DC = BC]

Hence AB + BC + CA > 2AD.

5. The length of two sides of a triangle are 4 cm and 6 cm. Between what two measures should the length of the third side fall ?

Solution:

We know that the sum of two sides of a triangle is always greater than the third side.

Therefore, third side has to be less than the sum of the two sides.

The third is thus less than 4 cm + 6 cm = 10 cm

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

Thus the third side has to be more than 6 cm – 4 cm = 2 cm the length of third side should be greater than 2 cm and less than 10 cm.

Hence the length of the third side should fall between 2 cm and 10 cm.