Area of an isosceles right-angled triangle is 800 sq. metres. The greatest possible square has been cut out from it. The length of the diagonal of this square will be


A) \( \Large 10\ \sqrt{2} m\)

B) \( \Large 10\ \sqrt{3} m\)

C) 20 m

D) \( \Large 20\ \sqrt{2} m\)

Correct Answer:
D) \( \Large 20\ \sqrt{2} m\)

Description for Correct answer:

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 = \( \Large 20 \sqrt{2} m \)


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