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Answer:
GIVEN: A rhombus ABCD, each side of which is ‘a'. As diagonals of a rhombus are perpendicular to each other. So, AC Perpendicular to BD. Let AC =2x & BD = 2y
PCQD is a rectangle contained by the diagonals of rhombus, as mentioned in the question. And if we consider a rectangle contained by only one of its diagonals, that can be of any size. So, that possibility is ruled out. Here in the first case considered that rectangl is contained by half the diagonals.
PROOF: Area of rhombus ABCD
= 1/2* one diagonal * other diagonal
= 1/2* 2x* 2y = 2xy square unit. ………….(1)
Area of rectangle PCQD
= length * breadth = DP * PC
=> area( rectangle) = DB/2 * AC/2
=> area(rectangle) = x* y = xy square unit ………….(2)
By dividing (1) by (2)
Area(rhombus) / Area( rectangle) = 2xy / xy = 2:1
=> Area of rectangle is half the rhombus contained by its (1/2diagonal).. ●●●●●●●
Now, if we take a rectangle with sides equal to the diagonals of rhombus..
In that case area of rectangle = 2x * 2y = 4xy sq unit ………(3)
If we compare (1) & (3)
area(rhombus)/area( rectangle) = 2xy/4xy = 1/2 1:2
This shows area of rhombus is half the area of rectangle..●●●●●●●