1. **State the problem:** We have a right triangle where the longer leg is 7 cm longer than the shorter leg, and the hypotenuse is 9 cm longer than the shorter leg. We need to find the lengths of all three sides. 2. **Define variables:** Let the length of the shorter leg be $x$ cm. 3. **Express other sides in terms of $x$:** - Longer leg = $x + 7$ - Hypotenuse = $x + 9$ 4. **Use the Pythagorean theorem:** For a right triangle, $$(\text{shorter leg})^2 + (\text{longer leg})^2 = (\text{hypotenuse})^2$$ So, $$x^2 + (x+7)^2 = (x+9)^2$$ 5. **Expand the squares:** $$x^2 + (x^2 + 14x + 49) = x^2 + 18x + 81$$ 6. **Combine like terms:** $$x^2 + x^2 + 14x + 49 = x^2 + 18x + 81$$ $$2x^2 + 14x + 49 = x^2 + 18x + 81$$ 7. **Bring all terms to one side:** $$2x^2 + 14x + 49 - x^2 - 18x - 81 = 0$$ $$x^2 - 4x - 32 = 0$$ 8. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-4$, $c=-32$. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 1 \times (-32) = 16 + 128 = 144$$ So, $$x = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2}$$ 9. **Find the two possible solutions:** - $$x = \frac{4 + 12}{2} = \frac{16}{2} = 8$$ - $$x = \frac{4 - 12}{2} = \frac{-8}{2} = -4$$ Since length cannot be negative, discard $x = -4$. 10. **Find the lengths:** - Shorter leg = $x = 8$ cm - Longer leg = $8 + 7 = 15$ cm - Hypotenuse = $8 + 9 = 17$ cm **Final answer:** - Shorter leg: 8 cm - Longer leg: 15 cm - Hypotenuse: 17 cm