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

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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 $$ \frac{7\sqrt{6}}{2} $$.
Each leg of the isosceles right triangle is $$ \frac{7\sqrt{6}}{2} $$ units long.

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

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