Q6 of 59 Page 194

Let’s prove that the sum of length of two diagonals is greater than the sum of the length of any two opposite sides of a quadrilateral.

Let us first construct the diagram for the question:


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


AC is one of the diagonals.


Diagonal BD also divides the quadrilateral into two equal triangles.


The diagonal AC divides the quadrilateral into two triangles.


Let the common point of intersection of two diagonals be O.


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 OAB


OA + OB > AB ……….. (i)


Let us first consider Δ OBC


OB + OC > BC ………….. (ii)


Let us first consider Δ OCD


OD + OC > DC ………….. (iii)


Let us first consider Δ ODA


OD + OA > AD ………….. (iv)


Now let us add equations (i) and (iii)


OA + OB + OD + OC > AB + DC


We know that AC = OA + OC and BD = OB + OD


AC + BD > AB + DC


We can also get the same result for other two sides as well by adding equations (ii) and (iv).


OB + OC + OD + OA > AD + BC


AC + BD > AD + BC


Hence it is proved that the sum of length of two diagonals is greater than the sum of the length of any two opposite sides of a quadrilateral.


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