show that the area of a square is equal to half the product of its diagonal
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Let ABCD is a square, each side is x unit. Diagonals AC =BD =y unit.If diagonals intersect at point O. Angle AOB=90° and OA=OB= y/2.
In right angled triangle AOB
OA^2+OB^2=AB^2
(y^2)/4+(y^2)/4=x^2 or x^2=(y^2)/2…………..(1)
Area of square=(side)^2=(x)^2 , [put x^2=(y^2)/2 from eq.(1).]
Area of square =(y^2)/2=(1/2)×y×y=(1/2)AC×BD.
= Half of the product of diagonals. Proved.
In right angled triangle AOB
OA^2+OB^2=AB^2
(y^2)/4+(y^2)/4=x^2 or x^2=(y^2)/2…………..(1)
Area of square=(side)^2=(x)^2 , [put x^2=(y^2)/2 from eq.(1).]
Area of square =(y^2)/2=(1/2)×y×y=(1/2)AC×BD.
= Half of the product of diagonals. Proved.
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