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Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2(BD + AC)

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प्रश्न

Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2(BD + AC)

योग
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उत्तर

Given: ABCD is a quadrilateral.


To show: AB + BC + CD + DA < 2(BD + AC)

Construction: Join diagonals AC and BD.

Proof: In ΔOAB, OA + OB > AB  ...(i) [Sum of two sides of a triangle is greater than the third side]

In ΔOBC, OB + OC > BC   ...(ii) [Sum of two sides of a triangle is greater than the third side]

In ΔOCD, OC + OD > CD  ...(iii) [Sum of two sides of a triangle is greater than the third side]

In ΔODA, OD + OA > DA   ...(iv) [Sum of two sides of a triangle is greater than the third side]

On adding equations (i), (ii), (iii) and (iv), we get

2[(OA + OB + OC + OD] > AB + BC + CD + DA

⇒ 2[(OA + OC) + (OB + OD)] > AB + BC + CD + DA

⇒ 2(AC + BD) > AB + BC + CD + DA  ...[∵ OA + OC = AC and OB + OD = BD]

⇒ AB + BC + CD + DA < 2(BD + AC)

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अध्याय 7: Triangles - Exercise 7.4 [पृष्ठ ७०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
अध्याय 7 Triangles
Exercise 7.4 | Q 11. | पृष्ठ ७०

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