Question

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





Pls answer fast​

Answer 1

Step-by-step explanation:

let the side of the square be x

by pythagorean theorem

x²+x²=diagonal ²

because the sides and the diagonal form a right angle triangle with diagonal as hypotenuse

2x²=diagonal ²

(root 2× x)²=diagonal ²

square will get cut

root 2 ×x=diagonal

both diagonals are equal

product of diagonals

root2 ×x ×root2×x=

2x²

2x² is product of diagonals

dividing it by 2

x²

area of square = side²

here side is x

area=x²

x² is obtained in both the ways

hence it is proven

please mark me the brainliest

hope you got it

if it was complicated then I would apologize because it the only way I can prove it because I am in class 7 and I can't elaborate that much but I think it's quite simple to understand it

I got late because I just saw u'r question

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.