1. Problem Q1: The longer leg of a right triangle is 4 cm more than the shorter leg and the hypotenuse is 20 cm. 2. Let the shorter leg be $x$ cm, so the longer leg is $x+4$ cm. 3. By the Pythagorean theorem: $$x^2 + (x+4)^2 = 20^2$$. 4. Expand the left side: $$x^2 + (x+4)^2 = x^2 + x^2 + 8x + 16 = 2x^2 + 8x + 16$$. 5. Set equal to 400 and simplify: $$2x^2 + 8x + 16 = 400$$. 6. Subtract 400: $$2x^2 + 8x - 384 = 0$$. 7. Divide by 2: $$x^2 + 4x - 192 = 0$$. 8. Factor the quadratic: $$x^2 + 4x - 192 = (x+16)(x-12)$$. 9. Solutions are $x=-16$ or $x=12$, length must be positive so $x=12$ cm. 10. Final answer Q1: a. 12 cm. 11. Problem Q2: A rectangular prism has length 15 cm, width 12 cm and surface area 684 cm^2; find the height $h$. 12. Surface area formula: $$SA = 2(lw + lh + wh)$$. 13. Compute $lw$: $$lw = 15 \times 12 = 180$$. 14. Substitute and divide by 2: $$\frac{SA}{2} = 342 = lw + h(l+w) = 180 + h(15+12)$$. 15. Simplify: $$342 - 180 = 162 = 27h$$. 16. Solve for $h$: $$h = \frac{162}{27} = 6$$ cm. 17. Final answer Q2: c. 6.0 cm. 18. Problem Q3: A regular pentagon-based prism has base side 8 cm and apothem 3 cm; water rises from 3 cm to 5 cm when a stone is immersed. Find the stone's volume. 19. Let $A$ be the area of the pentagonal base; the volume change equals $A(5-3) = 2A$ $\text{cm}^3$. 20. For a regular pentagon base, area is half the perimeter times the apothem: $$A = \frac{1}{2} \times (5 \times 8) \times 3$$. 21. Compute the base area: $$A = \frac{1}{2} \times 40 \times 3 = 60\text{ cm}^2$$. 22. Volume of stone: $$V = 2A = 2 \times 60 = 120\text{ cm}^3$$. 23. Final answer Q3: b. 120 cm^3.