Q3 of 59 Page 194

Let’s prove that the perimeter of a quadrilateral is more than twice of the sum length of any diagonal.

Let us first draw a figure for above question.


ABCD is a quadrilateral. Quadrilateral is any figure with four sides.


AC is one of the diagonals.


The diagonal AC divides the quadrilateral into two triangles.


We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than the third side.


Let us first consider Δ ACD


AD + DC > AC ……….. (i)


Similarly let us consider the other Δ ABC


AB + BC > AC ………… (ii)


Now let us add the equations (i) and (ii)


AD + AC + BC + AB > AC + AC


AD + AC + BC + AB > 2 × AC


Hence it is proved that the perimeter of a quadrilateral is more than twice of the sum length of any diagonal.


More from this chapter

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2

O is any point inside the triangle ABC. Let’s prove that

AB+AC>OB+OC


 2

O is any point inside the triangle ABC. Let’s prove that

AB+BC+AC>OA+OB+OC


 4 A

P is any point inside ABC let’s prove that,

AP+BP>AB


 4 B

P is any point inside ABC let’s prove that,

AB+BC+AC<2(AP+BP+CP)