Mathematics – Class 7 – Chapter 6 – The Triangle And Its Properties – Exercise 6.4 – NCERT Exercise Solution

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

(i) 2 cm, 3 cm, 5 cm

Solution:

As we know the sum of any two sides of a triangle must be greater than the third, so it is possible to draw a triangle.

Clearly, we have:

(2 + 3) = 5

5 = 5

Here the sum of any two of these numbers is not greater than the third.

Hence, it is not possible to draw a triangle whose sides are 2 cm, 3 cm, and 5 cm.

(ii) 3 cm, 6 cm, 7 cm

Solution:

As we know the sum of any two sides of a triangle must greater than the third, so it is possible to draw a triangle.

Clearly, we have:

(3 + 6) = 9 > 7

(6 + 7) = 13 > 3

(7 + 3) = 10 > 6

Here the sum of any two of these numbers is greater than the third.

Hence, it is possible to draw a triangle whose sides are 3 cm, 6 cm, and 7 cm.

(iii) 6 cm, 3 cm, 2 cm

Solution:

As we know the sum of any two sides of a triangle must greater than the third, so it is possible to draw a triangle.

Clearly, we have:

(3 + 2) = 5 < 6

Here the sum of any two of these numbers is less than the third.

Hence, it is not possible to draw a triangle whose sides are 6 cm, 3 cm, and 2 cm.

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

(i) OP + OQ > PQ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.4 - NCERT Exercise Solution

Solution:

We take any point O in the interior of a triangle PQR and join OR, OP, OQ.

After joining the point we get three triangles ΔOPQ, ΔOQR, and ΔORP which are shown in the figure below.

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.4 - NCERT Exercise Solution

As we know that the sum of the length of any two sides of a triangle is always greater than the third side.

(i) Yes, ΔOPQ has sides OP, OQ and PQ.

So, OP + OQ > PQ

(ii) Yes, ΔOQR has sides OR, OQ and QR.

So, OQ + OR > QR

(iii) Yes, ΔORP has sides OR, OP and PR.

So, OR + OP > RP

3. AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ΔABM and ΔAMC.)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.4 - NCERT Exercise Solution

Solution:

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

Now we consider the ΔABM,

Here, AB + BM > AM … [equation i]

Now, we consider the ΔACM

Here, AC + CM > AM … [equation ii]

Adding both the equation i.e [i] and [ii] we get,

AB + BM + AC + CM > AM + AM

From the figure we have, BC = BM + CM

AB + BC + AC > 2 AM (we write BC on the place of BM + CM)

Hence, the given expression is true.

4. ABCD is a quadrilateral.

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

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.4 - NCERT Exercise Solution

Solution:

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

Now we consider the ΔABC,

Here, AB + BC > CA … [equation i]

Now, consider the ΔBCD

Here, BC + CD > DB … [equation ii]

Now, consider the ΔCDA

Here, CD + DA > AC … [equation iii]

Now, consider the ΔDAB

Here, DA + AB > DB … [equation iv]

Adding equation [i], [ii], [iii] and [iv] we get,

AB + BC + BC + CD + CD + DA + DA + AB > CA + DB + AC + DB

2AB + 2BC + 2CD + 2DA > 2CA + 2DB

Take out 2 (as common) on both the side,

2(AB + BC + CA + DA) > 2(CA + DB)

AB + BC + CA + DA > CA + DB

Hence, the given expression is true.

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

Solution:

We consider ABCD as quadrilateral and P is the point where the diagonals intersect. As shown in the figure below.

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.4 - NCERT Exercise Solution

We know that the sum of the length of any two sides is always greater than the third side.

Now we consider the ΔPAB,

Here, PA + PB > AB … [equation i]

Then, we consider the ΔPBC

Here, PB + PC > BC … [equation ii]

Now, we consider the ΔPCD

Here, PC + PD > CD … [equation iii]

Now, we consider the ΔPDA

Here, PD + PA > DA … [equation iv]

Adding equation [i], [ii], [iii] and [iv] we get,

PA + PB + PB + PC + PC + PD + PD + PA > AB + BC + CD + DA

2PA + 2PB + 2PC + 2PD > AB + BC + CD + DA

2PA + 2PC + 2PB + 2PD > AB + BC + CD + DA

2(PA + PC) + 2(PB + PD) > AB + BC + CD + DA

From the figure we have, AC = PA + PC and BD = PB + PD

Then,

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

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

Hence, the given expression 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:

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

According to the question, it is given that the two sides of a triangle are 12 cm and 15 cm.

So, the third side length should be less than the sum of the other two sides,

12 + 15 = 27 cm.

Then, it is given that the third side is can not be less than the difference of the two sides, 15 – 12 = 3 cm

Hence, the length of the third side falls between 3 cm and 27 cm.

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