Question

prove that the diagonal of the square is √2 times of its side.

Answer 1

Answer:hope this helps u

Step-by-step explanation:

Answer 2

Answer-

Given: A square ABCD with sides AB, BC, CD, DA and Diagonal AC.

To prove: AC= √2 AB

Proof- In Triangle ABC,

AB= BC= let it be s (Sides of square are equal in measure)

Using Pythagorus Theorem where,

Hypotenuse(H)= AC

Perpendicular(P)= BC= s

Base(B)= AB= s

==> H^2= P^2 + B^2

(AC)^2= (BC)^2 + (AB)^2

(AC)^2= s^2 + s^2

(AC)^2= 2s^2

AC= √2s^2

AC= √2 s

AC= √2 AB

Hence, proved.

From this result, we can conclude that

the diagonal of the square is √2 times of its side.