Gujarat BoardEnglish MediumSTD 9MathsTriangles4 Marks
Question
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
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Answer
Given ABCD is a quadilateral,
Construction join diagonais AC and BD. To show AB + BC + CD + DA > AC + BD In $\triangle\text{ABC},$ AB + BC > AC [Sum of two sides of a triangle is greater than the third side] ...(i) In $\triangle\text{BCD},$ BC + CD > BD [Sum of two sides of a triangle is greater than the third side] ...(ii) In $\triangle\text{CDA},$ CD + DA > AC [Sum of two sides of a triangle is greater than the third side] ...(iii) In $\triangle\text{DAB},$ DA + AB + > BD [Sum of two sides of a triangle is greater than the third side] ...(iv) On adding Eqs. (i), (ii), (iii) (iv), We get 2(AB + BC + CD + DA) > 2 (AC + BD) ⇒ AB + BC + CD + DA > AC + BD
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