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General
Easy

Question

ABCD is a parallelogram and ' 0 ' is the point of intersection of its diagonals AC and BD. If  the area of straight triangle A O D equals 8 sq.cm , what is the area of the straight triangle BOC ?

hintHint:

we know that in a parallelogram the diagonals bisect each other.
Prove congruence of triangles .Areas of congruent triangles are equal.

The correct answer is: 8 sq.cm


    Ans :- 8 sq.cm
    Explanation :-
    We know Area of straight triangle AOD = 8 sq.cm
    In parallelogram ABCD , diagonals meet at O
    As we know diagonals bisect each other
    AO = OC  ; BO = OD
    And also straight angle AOD equals straight angle COB (vertically opposite angles are equal)

    By SAS ( side-angle -side) congruence
    AOD ≅ COB
    If triangle are congruent then areas of the triangles are equal so
    Area of straight triangle AOD= Area of straight triangle COB
    ∴Area of straight triangle COB = 8 sq. cm
     

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