By using Pythagoras Theorem:-
(Hypotenuse)² = (Base)²+ (Perpendicular)²
Taking, Hypotenuse As H
Base As B
Perpendicular As P
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Solving The Question
H²= B²+ P²
[tex](9 \sqrt{2} ) {}^{2} = x {}^{2} + {x}^{2} \\ \\ \implies \: 81 \times 2 = 2x {}^{2} \\ \\ \implies162 = 2x {}^{2} \\ \\ \implies \frac{162}{2} = x {}^{2} \\ \\ \implies \cancel\frac{162}{2} = x {}^{2} \\ \\ \implies81 = {x}^{2} \\ \\ \implies x = \sqrt{81} \\ \\ \implies \: x = 9[/tex]
Hope This Helps You ❤️
