1. **State the problem:** We have a trapezium PQRS with parallel sides PQ and SR. PQ = $2k + 2$ cm, SR = $4k - 1$ cm, height QR = $3k$ cm, and area $L = 150$ cm². We need to express the area $L$ in terms of $k$ and then find the value of $k$. 2. **Formula for the area of a trapezium:** $$L = \frac{(\text{sum of parallel sides}) \times \text{height}}{2}$$ Here, the parallel sides are PQ and SR, and the height is QR. 3. **Write the area in terms of $k$:** $$L = \frac{(2k + 2) + (4k - 1)}{2} \times 3k$$ 4. **Simplify the expression inside the parentheses:** $$(2k + 2) + (4k - 1) = 2k + 2 + 4k - 1 = 6k + 1$$ 5. **Substitute back:** $$L = \frac{6k + 1}{2} \times 3k = \frac{3k(6k + 1)}{2}$$ 6. **Given $L = 150$, set up the equation:** $$150 = \frac{3k(6k + 1)}{2}$$ 7. **Multiply both sides by 2 to clear the denominator:** $$2 \times 150 = 3k(6k + 1)$$ $$300 = 3k(6k + 1)$$ 8. **Divide both sides by 3:** $$\frac{300}{\cancel{3}} = \cancel{3}k(6k + 1) / \cancel{3}$$ $$100 = k(6k + 1)$$ 9. **Expand the right side:** $$100 = 6k^2 + k$$ 10. **Rearrange to form a quadratic equation:** $$6k^2 + k - 100 = 0$$ 11. **Use the quadratic formula:** $$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=6$, $b=1$, $c=-100$. 12. **Calculate the discriminant:** $$\Delta = 1^2 - 4 \times 6 \times (-100) = 1 + 2400 = 2401$$ 13. **Calculate the square root:** $$\sqrt{2401} = 49$$ 14. **Find the two possible values for $k$:** $$k = \frac{-1 \pm 49}{2 \times 6} = \frac{-1 \pm 49}{12}$$ 15. **Calculate each root:** - $$k = \frac{-1 + 49}{12} = \frac{48}{12} = 4$$ - $$k = \frac{-1 - 49}{12} = \frac{-50}{12} = -\frac{25}{6}$$ 16. **Interpret the results:** Since $k$ represents a length factor, it must be positive. Therefore, the valid solution is: $$k = 4$$ **Final answer:** $$\boxed{k = 4}$$