1. **Stating the problem:** We are given a right triangle with sides labeled as follows: hypotenuse $5x + 11$, one leg $5x - 4$, and the other leg $3x$. There is also a segment labeled $x + 3$ inside the leg $3x$. We want to verify the Pythagorean theorem and understand the relationships between these expressions. 2. **Recall the Pythagorean theorem:** For a right triangle with legs $a$ and $b$, and hypotenuse $c$, the relationship is: $$c^2 = a^2 + b^2$$ 3. **Assign sides:** Let - Hypotenuse $c = 5x + 11$ - One leg $a = 5x - 4$ - Other leg $b = 3x$ 4. **Check the Pythagorean theorem:** Calculate $a^2 + b^2$ and compare to $c^2$. Calculate $a^2$: $$a^2 = (5x - 4)^2 = 25x^2 - 40x + 16$$ Calculate $b^2$: $$b^2 = (3x)^2 = 9x^2$$ Sum: $$a^2 + b^2 = 25x^2 - 40x + 16 + 9x^2 = 34x^2 - 40x + 16$$ Calculate $c^2$: $$c^2 = (5x + 11)^2 = 25x^2 + 110x + 121$$ 5. **Compare $a^2 + b^2$ and $c^2$:** $$34x^2 - 40x + 16 \stackrel{?}{=} 25x^2 + 110x + 121$$ Bring all terms to one side: $$34x^2 - 40x + 16 - 25x^2 - 110x - 121 = 0$$ Simplify: $$9x^2 - 150x - 105 = 0$$ 6. **Solve the quadratic equation:** $$9x^2 - 150x - 105 = 0$$ Divide both sides by 3: $$\cancel{3} \times 3x^2 - \cancel{3} \times 50x - \cancel{3} \times 35 = 0 \Rightarrow 3x^2 - 50x - 35 = 0$$ Use quadratic formula: $$x = \frac{50 \pm \sqrt{(-50)^2 - 4 \times 3 \times (-35)}}{2 \times 3} = \frac{50 \pm \sqrt{2500 + 420}}{6} = \frac{50 \pm \sqrt{2920}}{6}$$ Simplify $\sqrt{2920}$: $$\sqrt{2920} = \sqrt{4 \times 730} = 2\sqrt{730}$$ So, $$x = \frac{50 \pm 2\sqrt{730}}{6} = \frac{25 \pm \sqrt{730}}{3}$$ 7. **Interpretation:** The value(s) of $x$ that satisfy the Pythagorean theorem for the given sides are: $$x = \frac{25 + \sqrt{730}}{3} \quad \text{or} \quad x = \frac{25 - \sqrt{730}}{3}$$ Since $x$ represents a length, we take the positive root: $$x = \frac{25 + \sqrt{730}}{3}$$ 8. **Perimeter:** The perimeter is given as $3x + 7$ in the problem statement, but the perimeter of the triangle is the sum of all sides: $$P = (5x + 11) + (5x - 4) + 3x = 5x + 11 + 5x - 4 + 3x = (5x + 5x + 3x) + (11 - 4) = 13x + 7$$ So the perimeter is: $$P = 13x + 7$$ **Final answers:** - The value of $x$ satisfying the Pythagorean theorem is: $$x = \frac{25 + \sqrt{730}}{3}$$ - The perimeter of the triangle is: $$P = 13x + 7$$