1. **State the problem:** We have a right-angled triangle with legs $6x$ cm and $8$ cm, and a rectangle with sides $5$ cm and $4x - 1$ cm. The area of the triangle is 10 cm² greater than the area of the rectangle. We need to find the value of $x$. 2. **Write the formulas for areas:** - Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$ - Area of rectangle = $\text{length} \times \text{width}$ 3. **Express the areas in terms of $x$:** - Triangle area = $\frac{1}{2} \times 6x \times 8 = 24x$ - Rectangle area = $5 \times (4x - 1) = 20x - 5$ 4. **Set up the equation from the problem statement:** $$\text{Triangle area} = \text{Rectangle area} + 10$$ $$24x = (20x - 5) + 10$$ 5. **Simplify the equation:** $$24x = 20x - 5 + 10$$ $$24x = 20x + 5$$ 6. **Isolate $x$:** $$24x - 20x = 5$$ $$\cancel{24x} - \cancel{20x} = 5$$ $$4x = 5$$ 7. **Solve for $x$:** $$x = \frac{5}{4} = 1.25$$ **Final answer:** $$x = 1.25$$ cm