Answer :

Given: ABCD is a rhombus. = {0}.



To prove: Area of OAB = (area of ABCD)


Proof:


O is the mid-point of AC as well as BD.


Further, in rhombus ABCD,


AB BC CD DA


In ∆OAB and ∆OBC,


OA OC


OB OB


AB CB


∆OAB and ∆OBC are congruent by SSS theorem for congruence.


Thus, their areas are equal.


Similarly, ∆OAB, ∆OBC, ∆OCD and ∆ODA are all congruent triangles having equal areas.


∆OAB = ∆OBC = ∆OCD = ∆ODA


Now, ABCD = ∆OAB + ∆OBC + ∆OCD + ∆ODA


ABCD = ∆OAB + ∆OAB + ∆OAB + ∆OAB


ABCD = 4∆OAB



Thus,



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