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

Answer :

\(\triangle \)AOB, we have AB < AO + BO …(i)
[Any side of a triangle is less than the sum of other two sides]

In \(\triangle \)BOC, we have

BC < BO + CO …(ii)

In \(\triangle \)COD, we have

CD < CO + DO …(iii)

In \(\triangle \)AOD, we have

DA < DO + AO …(iv)

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

AB + BC + CD + DA < 2AO + 2BO + 2CO + 2DO

Or, AB + BC + CD + DA < 2(AO + BO + CO + DO)

Or, AB + BC + CD + DA < 2 [(AO + CO) + (BO + DO)]

Or, AB + BC + CD + DA < 2(AC + BD)

Thus, AB + BC + CD + DA < 2(AC + BD)

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.4 ABCD is a quadrilateral. Is AB + BC + CD + DA > 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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