Let x be the base and side of isosceles \( \Large \triangle ABC \).

By hypothesis,



\( \Large \frac{1}{2}x \times x = 800 \)

=> \( \Large x^{2} = 1600 \)

=> x = 40 m

Therefore, Hypotenuse, BC = 2 x = 40 x/2 m

If s be the side of the maximum square, then

\( \Large s^{2}+\frac{1}{2} \left(x-s\right)s+\frac{1}{2} \left(x-s\right)2=800 \)

=> \( \Large s^{2}+xs-s^{2} = 800 \)

=> \( \Large xs = 800 \)

=> \( \Large s = \frac{800}{40} = 20 m \)

Therefore, Length of diagonal of the square = \( \Large 20 \sqrt{2} m \)