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The Triangles and its Properties

    Two Special Triangles : Equilateral and Isosceles

    Exercise 6.4

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7th class > The Triangles and its Properties > Exercise 6.4

Exercise 6.4

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

a

(i) 2 cm, 3 cm, 5 cm

Solution:

A triangle can be possible only when the of the lengths of any two sides would be than the length of the side.

Take the two sides as 2 cm and 3 cm. Since 2 + 3 = and the third side is also 5 cm.

It possible to have a triangle with the sides 2 cm, 3 cm and 5 cm.

b

(ii) 3 cm, 6 cm, 7 cm

Solution:

First, take the two sides as 3 cm and 6 cm. Since 3 + 6 = and 9 7, the property of the triangle is satisfied.

Now take the two sides as 6 cm and 7 cm. Since 6 + 7 = and 13 3, the property of the triangle is satisfied.

Now, take the two sides as 7 cm and 3 cm. Since 7 + 3 = and 10 6, the property of the triangle is satisfied.

Hence, it possible to have a triangle with the sides 3 cm, 6 cm and 7 cm.

c

(iii) 6 cm, 3 cm, 2 cm

Solution:

Take the two sides as 2 cm and 3 cm. Since 2 + 3 = and 5 is than 6.

It possible to have a triangle with the sides 6 cm, 3 cm and 2 cm.

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

triangle PQR

a

(i) Is OP + OQ > PQ?

Solution:

In the given triangle, join OR, OQ and OP.

triangle PQR

From the diagram, it can be seen that OPQ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be than the length of the third side.

So, yes, OP + OQ > PQ

b

(ii) Is OQ + OR > QR?

Solution:

In the given triangle, join OR, OQ and OP.

triangle PQR

From the diagram, it can be seen that OQR is a 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.

c

(iii) Is OR + OP > RP?

Solution:

In the given triangle, join OR, OQ and OP.

triangle PQR

From the diagram, it can be seen that OPR is a 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.

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

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

triangle ABC

Solution:

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

In ΔABM, AB + BM AM and in ΔAMC, AC + MC AM.

Add both the inequalities and simplify.

AB + BM + AC + MC AM + AM

⇒ AB + AC + (BM + MC) >

Substitute BC for BM + MC.

⇒ AB + AC + > 2AM

Hence, AB + BC + CA > 2AM is true.

4. ABCD is a quadrilateral.Is AB + BC + CD + DA > AC + BD?

quadrilateral

Solution:

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

In △ABC, AB + BC AC.

In △ADC, AD + DC AC.

In △DCB, DC + CB DB.

In △ADB, AD + > DB.

Add all the four inequalities and simplify.

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

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

⇒AB + BC + AD + DC > (AC + DB)

⇒ (AB + BC + AD + DC) > (AC + DB)

Divide both sides by 2 and simplify.

⇒ 22(AB + BC + AD + DC) 22(AC + DB)

⇒ AB + BC + AD + DC AC + DB

Hence, AB + BC + CD + DA > AC + BD is true.

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

Solution:

Draw a quadrilateral ABCD. Join AC and BD.

Quadrilateral

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

In △AOB, AB < OA + .

In △BOC, BC < OB + .

In △COD, DC < OC + .

In △AOD, DA < + OA.

Add all the four inequalities and simplify.

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

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

⇒ AB + BC + CD + DA < [(AO + OC) + (DO + OB)]

Substitute for AO + OC and for DO + OB.

⇒ AB + BC + CD + DA < 2(AC + BD)

Hence, AB + BC + CD + DA < 2(AC + BD) is true.

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:

It is given that two sides of a triangle are 12 cm and 15 cm. The sum of two sides of a triangle is always than the third side of the triangle.

Therefore, the third side of the triangle should be than 12 + 15 = cm.

Also, the third side cannot be than the difference of two sides of a triangle.

Therefore, The third side should be than 15 − 12 = cm.

Therefore, the length of the third side should be between 3 cm and 27 cm.