\#3.) Given an isosceles right triangle with a hypotenuse of \( 7 \sqrt{3} \), determine the length of both legs.

1. Let the length of each leg be \( x \). 2. By the Pythagorean theorem for an isosceles right triangle, the sum of the squares of the legs equals the square of the hypotenuse: \[ x^2 + x^2 = \left(7\sqrt{3}\right)^2 \] 3. Combine like terms: \[ 2x^2 = \left(7\sqrt{3}\right)^2 \] 4. Calculate the square of the hypotenuse: \[ \left(7\sqrt{3}\right)^2 = 7^2 \times 3 = 49 \times 3 = 147 \] So, \[ 2x^2 = 147 \] 5. Solve for \( x^2 \): \[ x^2 = \frac{147}{2} \] 6. Take the square root of both sides: \[ x = \sqrt{\frac{147}{2}} \] 7. Simplify the square root: \[ 147 = 49 \times 3 = 7^2 \times 3 \quad \text{so,} \quad x = \sqrt{\frac{7^2 \times 3}{2}} = \frac{7\sqrt{3}}{\sqrt{2}} \] 8. Rationalize the denominator: \[ \frac{7\sqrt{3}}{\sqrt{2}} = \frac{7\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{7\sqrt{6}}{2} \] Thus, the length of each leg is \( \frac{7\sqrt{6}}{2} \).