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$\small AB+BC+CD+DA< 2(AC+BD)$            ?

Let the intersection point of the two diagonals be O. As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). So, In  : In  : In  : In  : Now, Adding all four equations we, get which can also be expressed as  Hence this is true.

Is  $\small AB+BC+CD+DA> AC+BD$ ?

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). So, In  : In  : Adding (1) and (2) we get, Hence The given statement 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?

Let ABC be a triangle with AB = 12cm and BC = 15cm Now As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). AB + BC > CA 12 + 15 > CA CA < 27......(1) Also, in a similar way AB + CA  > BC  CA > BC -  AB CA > 15 - 12  CA > 3............(2) Hence from (1) and (2), we can say that the length of third side of the triangle must be...

3.  AM is a median of a triangle ABC.
Is $\small AB+BC+CA> 2AM$ ?
(Consider the sides of triangles $\small \Delta ABM$ and $\small \Delta AMC$.)

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule). So, In  : In  : Adding (1) and (2), we get As we can see M is the point in line BC So, we can say So our equation becomes . Hence it is a True statement.

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

(i)  $\small OP+OQ> PQ$ ?
(ii) $\small OQ+OR> QR$ ?
(iii) $\small OR+OP> RP$?

i) As POQ is a triangle, the sum of any two sides will always be greater than the third side. so  Yes,   . ii) As ROQ is a triangle, the sum of any two sides will always be greater than the third side. so  Yes,   iii) As ROQ is a triangle, the sum of any two sides will always be greater than the third side. so  Yes,

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

As we know, According to the Triangle inequality law, the sum of lengths of any two sides of a triangle would always greater than the length of the third side. So Verifying this inequality by taking all possible combinations, we have, (i) 2 cm, 3 cm, 5 cm 3 + 5 > 2 ----> True 2 + 5 >  3---->  True 2 + 3 > 5 ---->False Hence the triangle is not possible. (ii) 3 cm, 6 cm, 7 cm  3 + 6 > 7 ----->...
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