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

Answer :

In \(\triangle \)ABC, we have

AB + BC > AC …(i)
[Sum of any two sides is greater than the third side]

In \(\triangle \)BDC, we have

BC + CD > BD …(ii)

In \(\triangle \)ADC, we have
CD + DA > AC .... (iii)

In \(\triangle \)DAB, we have
DA + AB > BD …(iv)

Adding eq. (i), (ii), (iii) and (iv), we get

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

Or, AB + BC + CD + DA > AC + BD [Dividing both sides by 2]

Hence, proved

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