## MP Board Class 7th Maths Solutions 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

(ii) 3 cm, 6 cm, 7 cm

(iii) 6 cm, 3 cm, 2 cm

Solution:

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

(i) Given that, the sides of the triangle are 2 cm, 3 cm, 5 cm.

It can be observed that,

(2 + 3) cm = 5 cm

However, 5 cm = 5 cm

Hence, this triangle is not possible.

(ii) Given that, the sides of the triangle are

3 cm, 6 cm, 7 cm.

Here, (3 + 6) cm = 9 cm > 7 cm

(6 + 7) cm = 13 cm > 3 cm

(3 + 7) cm = 10 cm > 6 cm

Hence, this triangle is possible.

(iii) Given that, the sides of the triangle are

6 cm, 3 cm, 2 cm.

Here, (6 + 3) cm = 9 cm > 2 cm

However, (3 + 2) cm = 5 cm < 6 cm Hence, this triangle is not possible.

Question 2.

Take any point 0 in the interior of a triangle PQR. Is

(i) OP + OQ > PQ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

Solution:

If O is a point in the interior of a given triangle, then three triangles ∆OPQ, ∆OQR and ∆ORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side.

(i) Yes, as OQR is a triangle with sides OR, OQ and QR.

∴ OQ + OR> QR

(ii) Yes, as OQR is a triangle sides OR, OQ and QR.

∴ OQ + OR > QR

(iii) Yes, as A ORP is a triangle with sides OR, OP and PR.

∴ OR + OP > PR

Question 3.

AM is a median of a triangle ABC.

Is AB + BC + CA > 2AM?

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

Solution:

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

In ∆ABM,

AB + BM > AM …(i)

Similarly, in ∆ACM,

AC + CM > AM …(ii)

Adding (i) and (ii), we obtain

AB + BM + MC +AC > AM + AM

AB + BC + AC > 2AM

Yes, the given expression is true.

Question 4.

ABCD is a quadrilateral.

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

Solution:

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

In ∆ABC; AB + BC > CA …(i)

In ∆BCD; BC + CD > DB …(ii)

In ∆CDA; CD + DA > AC …(iii)

In ∆DAB; DA + AB > DB …(iv)

Adding (i), (ii), (iii) and (iv), 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 given expression is true.

Question 5.

ABCD is a quadrilateral.

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

Solution:

In a triangle, the sum of the lengths of either two sides is always greater than the third side.

Let diagonals AC and BD of the quadrilateral ABCD intersects at O.

In ∆OAB, OA + OB > AB … (i)

In ∆OBC, OB + OC > BC … (ii)

In ∆OCD, OC + CD > CD … (iii)

In ∆ODA, OD + OA > DA … (iv)

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

⇒ 20A + 20B + 20C + 20D > AB + BC + CD + DA

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

⇒ 2(AC) + 2(BD) > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA

Yes, the given 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?

Solution:

In a triangle, the Sum of the lengths of either two side is always greater than the third side.

Lengths of two sides of a triangle are 12 cm and 15 cm.

Let the third side be x cm.

∴ x + 12 > 15 ⇒ x > 3

x + 15 > 12 ⇒ x > – 3 but side length never be negative.

and 12 + 15 > x ⇒ 27 > x

Hence, third side can measure between 3 and 27.