1. **Problem statement:** A triangle's height is 1 cm longer than its base, and its area is 21 cm². Find the height. 2. **Formula used:** The area of a triangle is given by: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 3. **Set variables:** Let the base be $b$ cm. Then the height is $h = b + 1$ cm. 4. **Write the area equation:** $$21 = \frac{1}{2} \times b \times (b + 1)$$ 5. **Multiply both sides by 2 to clear the fraction:** $$2 \times 21 = b (b + 1)$$ $$42 = b^2 + b$$ 6. **Rewrite as a quadratic equation:** $$b^2 + b - 42 = 0$$ 7. **Solve the quadratic equation using the quadratic formula:** $$b = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-42)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 168}}{2} = \frac{-1 \pm \sqrt{169}}{2}$$ $$b = \frac{-1 \pm 13}{2}$$ 8. **Calculate the two possible values:** - $$b = \frac{-1 + 13}{2} = \frac{12}{2} = 6$$ - $$b = \frac{-1 - 13}{2} = \frac{-14}{2} = -7$$ (not possible since length cannot be negative) 9. **Find the height:** $$h = b + 1 = 6 + 1 = 7$$ **Final answer:** The height is $7$ cm.