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Answer:
2) The sum of the length of any two sides 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) .The sum of the length of any two sides is always greater than the third side.
Now consider the ΔABM,
Here, AB + BM > AM … [equation i]
Then, consider the ΔACM
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
4) The sum of the length of any two sides is always greater than the third side.
Now consider the ΔABC,
Here, AB + BC > CA … [equation i]
Then, consider the ΔBCD
Here, BC + CD > DB … [equation ii]
Consider the ΔCDA
Here, CD + DA > AC … [equation iii]
Consider the ΔDAB
Here, DA + AB > DB … [equation iv]
By 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 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) The sum of the length of any two sides is always greater than the third side.
Now consider the ΔPAB,
Here, PA + PB > AB … [equation i]
Then, consider the ΔPBC
Here, PB + PC > BC … [equation ii]
Consider the ΔPCD
Here, PC + PD > CD … [equation iii]
Consider the ΔPDA
Here, PD + PA > DA … [equation iv]
By 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.
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