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298Secondary School Mathematics for Class 913. In the given figure, prove thatCD+DAAB BC(ii) CD + DA + AB + BC > 2AC. |
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Answer» Now by the property of triangles, sum of any two sides > the third side.Therefore, AB + BC > AC Same way CD +DA > ACSimilar way we can get DA + AB > BD and BC + CD > BDNow adding all these inequalities, 2(AB + BC +CD +DA) > 2(AC + BD)Dividing both sides by 2 we get, (AB + BC +CD +DA) > (AC + BD)Thus, the given first identity is proved.For the second identity,2AC > AB + BC Similar way, 2AC > CD + DAAlso, 2BD > DA + AB and 2BD > BC + CDAdding the inequalities, we get 4(AC + BD) > 2(AB + BC +CD +DA)Dividing by 2 on both sides,we arrive at 2(AC + BD) > (AB + BC +CD +DA)or (AB + BC +CD +DA) < 2(AC + BD) Hence second identity is proved. |
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