Question

؈ABCD is a rhombus. AC ∩ BD = {0}. Prove that the area of ΔOAB =1/4 (area of ؈ABCD).

Answer 1

ABCD is a rhombus, AC ∩ BD = {0}. it means, diagonals of rhombus AC , BD is intersected as at O.

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.
ar(∆OAB) = ar(∆OBC) = ar(∆OCD) = ar(∆ODA)
Now, ar(ABCD) = ar(∆OAB) + ar(∆OBC) + ar(∆OCD) + ar(∆ODA)
ar(ABCD) = ar(∆OAB) + ar(∆OAB) + ar(∆OAB) + ar(∆OAB)
ar(ABCD) = 4ar(∆OAB)
ar(∆OAB) = 1/4 × ar(ABCD)