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Is AB + BC + CA > 2 AM?

(Consider the sides of triangles \(\triangle \)ABM and \(\triangle \)AMC.)

Answer :

We know that,

The sum of the length of any two sides is always greater than the third side.

Now In \(\triangle \)ABM,

Here, AB + BM > AM … [equation i]

Then, In \(\triangle \)ACM

Here, AC + CM > AM … [equation ii]

By adding equation [i] and [ii] we get,

AB + BM + AC + CM > AM + AM

From the figure we have, BC = BM + CM

AB + BC + AC > 2 AM

Hence, the given expression is true.

- Q.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
- Q.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?
- Q.4 ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?
- Q.5 ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
- Q.6 The length of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

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