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Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD - Mathematics

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प्रश्न

Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

योग
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उत्तर

Given in the question, a quadrilateral ABCD.

To proof that AB + BC + CD + DA > AC + BD.

Proof: In triangle ABC,


AB + BC > AC   ...(i) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

In triangle BCD,

BC + CD > BD   ...(ii) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

In triangle CDA,

CD + DA > AC   ...(iii) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

Similarly, in triangle DAB,

AD + AB > BD   ...(iv) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

Now, adding equation (i), (ii), (iii) and (iv), we get

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

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

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

AB + BC + CD + DA > AC + BD

Hence proved.

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अध्याय 7: Triangles - Exercise 7.4 [पृष्ठ ७०]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 7 Triangles
Exercise 7.4 | Q 12. | पृष्ठ ७०

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