The diagonal of the square is $8\sqrt{2}$ cm. Find the diagonal of another square whose area is triple that of the first square.

$8\sqrt{6}$ cm

We know that,

Area of a square = (Side)²

Diagonal of a square is = Side\(\sqrt {2}\)

The diagonal of the first square is = 8\(\sqrt {2}\)

As given in the question, diagonal = 8\(\sqrt {2}\)= side\(\sqrt {2}\)

= The side of the first square = 8 cm

= The area of the first square = (8)² = 64 cm2

The Area of the Second square is double the first one

= Area of the second one = 64×3

= Side of the 2nd square = \(\sqrt {64 × 3}\)  = 8×\(\sqrt {3}\)

The diagonal 2nd square  = \(\sqrt {2}\)×8×\(\sqrt {3}\) = $8\sqrt{6}$ cm