1. **State the problem:** We are given a right-angled triangle AOC with AC = 24 cm, OC = 26 cm, and angle \(\angle AOC = 60^\circ\). Inside it is a children's swimming pool shaped as a quadrant of a circle POA centered at O. We need to find: (a)(i) The radius of the children's swimming pool. (a)(ii) The remaining area of the swimming pool after closing the children's pool area AOB. 2. **Find the radius of the children's swimming pool (a)(i):** - The children's pool is a quadrant of a circle centered at O. - The radius is the length of OA. - Since triangle AOC is right-angled at A, use the Law of Cosines or Pythagoras to find OA. Given: \[ AC = 24, \quad OC = 26, \quad \angle AOC = 60^\circ \] Using Law of Cosines on triangle AOC: \[ OC^2 = OA^2 + AC^2 - 2 \times OA \times AC \times \cos(\angle AOC) \] But we don't know OA, so rearranging is complicated. Instead, since \(\angle AOC = 60^\circ\) and OC = 26, AC = 24, we can find OA by Law of Cosines: \[ OC^2 = OA^2 + AC^2 - 2 \times OA \times AC \times \cos(60^\circ) \] \[ 26^2 = OA^2 + 24^2 - 2 \times OA \times 24 \times \frac{1}{2} \] \[ 676 = OA^2 + 576 - 24 OA \] \[ OA^2 - 24 OA + 576 - 676 = 0 \] \[ OA^2 - 24 OA - 100 = 0 \] Solve quadratic: \[ OA = \frac{24 \pm \sqrt{24^2 - 4 \times 1 \times (-100)}}{2} \] \[ = \frac{24 \pm \sqrt{576 + 400}}{2} = \frac{24 \pm \sqrt{976}}{2} \] \[ \sqrt{976} \approx 31.24 \] \[ OA = \frac{24 \pm 31.24}{2} \] Two solutions: - \(\frac{24 + 31.24}{2} = 27.62\) (too large, more than OC) - \(\frac{24 - 31.24}{2} = -3.62\) (negative, discard) So radius \(r = OA = 27.62\) cm approximately. 3. **Calculate the area of the children's swimming pool (quadrant) (a)(ii):** - Area of full circle: \(\pi r^2\) - Area of quadrant: \(\frac{1}{4} \pi r^2\) Using \(\pi = \frac{22}{7}\) and \(r = 27.62\): \[ \text{Area quadrant} = \frac{1}{4} \times \frac{22}{7} \times (27.62)^2 \] \[ = \frac{1}{4} \times \frac{22}{7} \times 762.5 = \frac{1}{4} \times 2397.86 = 599.47 \text{ cm}^2 \] 4. **Calculate the area of the adult swimming pool (triangle AOC):** - Area of triangle = \(\frac{1}{2} \times AC \times OA \times \sin(\angle AOC)\) \[ = \frac{1}{2} \times 24 \times 27.62 \times \sin 60^\circ \] \[ = 12 \times 27.62 \times \frac{\sqrt{3}}{2} = 12 \times 27.62 \times 0.866 \] \[ = 12 \times 23.91 = 286.92 \text{ cm}^2 \] 5. **Calculate the remaining area after closing the children's pool:** - Remaining area = Area of triangle AOC - Area of quadrant POA \[ = 286.92 - 599.47 = -312.55 \text{ cm}^2 \] This negative value indicates an error in the radius calculation or assumptions. Re-examining step 2, since OC is the longest side (26 cm), and AC = 24 cm, OA must be less than OC. Alternatively, since \(\angle AOC = 60^\circ\), use Law of Cosines to find OA: \[ OA^2 = OC^2 + AC^2 - 2 \times OC \times AC \times \cos(60^\circ) \] \[ = 26^2 + 24^2 - 2 \times 26 \times 24 \times \frac{1}{2} = 676 + 576 - 624 = 628 \] \[ OA = \sqrt{628} = 25.06 \text{ cm} \] This is the correct radius. 6. **Recalculate area of quadrant with corrected radius:** \[ \text{Area quadrant} = \frac{1}{4} \times \frac{22}{7} \times (25.06)^2 = \frac{1}{4} \times \frac{22}{7} \times 628 = \frac{1}{4} \times 1974.86 = 493.72 \text{ cm}^2 \] 7. **Recalculate area of triangle AOC:** \[ = \frac{1}{2} \times 24 \times 25.06 \times \sin 60^\circ = 12 \times 25.06 \times 0.866 = 12 \times 21.69 = 260.28 \text{ cm}^2 \] 8. **Calculate remaining area:** \[ \text{Remaining area} = \text{Area triangle} - \text{Area quadrant} = 260.28 - 493.72 = -233.44 \text{ cm}^2 \] Still negative, meaning the quadrant is larger than the triangle, which is impossible. The problem states the children's pool is a quadrant of a circle centered at O inside the triangle, so the radius must be equal to OA, which is the length from O to A. Since \(\angle AOC = 60^\circ\), and triangle AOC is right angled at A, use Pythagoras to find OA: \[ OA^2 + AC^2 = OC^2 \] \[ OA^2 + 24^2 = 26^2 \] \[ OA^2 = 676 - 576 = 100 \] \[ OA = 10 \text{ cm} \] This is the radius. 9. **Calculate area of quadrant with radius 10 cm:** \[ \text{Area quadrant} = \frac{1}{4} \times \frac{22}{7} \times 10^2 = \frac{1}{4} \times \frac{22}{7} \times 100 = \frac{1}{4} \times 314.29 = 78.57 \text{ cm}^2 \] 10. **Calculate area of triangle AOC:** \[ = \frac{1}{2} \times AC \times OA = \frac{1}{2} \times 24 \times 10 = 120 \text{ cm}^2 \] 11. **Calculate remaining area after closing children's pool:** \[ = 120 - 78.57 = 41.43 \text{ cm}^2 \] **Final answers:** (a)(i) Radius of children's swimming pool = \(10\) cm (a)(ii) Remaining area of swimming pool = \(41.43\) cm²