prove that area of rhombus is half of product of its diagonal.

Asked by shubham deshpande | 6th Jan, 2012, 08:55: PM

Expert Answer:

Area of rhombus ABCD = area of triangle ABD + area of triangle CBD Triangles ABD and CBD are congruent by SSS

Area of rhombus ABCD = 2×(Area of triangle ABD)
AE is perpendicular to DB because the diagonals of a rhombus are perpendicular bisectors of each other.
Area of triangle ABD = DB×AE/2 because a triangle's area is one-half the product of a side and the altitude drawn to that side.
Area of rhombus ABCD = 2×(Area of triangle ABD)
So area of rhombus ABCD = 2×(DB×AE/2) = DB×AE AE = AC/2 because the diagonals of a rhombus are perpendicular bisectors of each other.
So area of rhombus DB×AE = DB×(AC/2) = DB×AC/2

Answered by  | 6th Jan, 2012, 11:52: PM

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