Question

Show that the area of a square is equal to half the product of its diagonals.​

Answer 1

Show that the area of a square is equal to half the product of its diagonals.

Area of square =(y^2)/2=(1/2)×y×y=(1/2)AC×BD. = Half of the product of diagonals. Proved.

Answer 2

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

$\huge\color{purple}{ \colorbox{orange}{\colorbox{white} {Thanks}}}$