1. **State the problem:** We have triangle ABC with AB = AC = 15 cm, CP perpendicular to AB, and CP = 9 cm. We need to find the shaded area formed by the arc centered at B passing through C and Q on AB. 2. **Find angle ABC:** Using the right triangle CPC, where CP = 9 cm and AP = x (unknown), and AB = 15 cm. 3. Use Pythagoras theorem in triangle APC: $$AP = \sqrt{AC^2 - CP^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12$$ 4. Since AB = 15 cm and AP = 12 cm, then PB = AB - AP = 15 - 12 = 3 cm. 5. To find angle ABC, use triangle BPC: $$\tan(\angle ABC) = \frac{CP}{PB} = \frac{9}{3} = 3$$ 6. Calculate angle ABC: $$\angle ABC = \arctan(3) \approx 1.249 \text{ radians} \approx 1.25 \text{ radians (3 s.f.)}$$ 7. **Find shaded area:** The shaded area is the segment of the circle centered at B with radius BC = 15 cm, bounded by chord BQ. 8. Length BC = 15 cm (radius), angle ABC = 1.25 radians. 9. Area of sector BQC: $$A_{sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 15^2 \times 1.25 = \frac{1}{2} \times 225 \times 1.25 = 140.625 \text{ cm}^2$$ 10. Length BQ = PB = 3 cm, use triangle BQC to find area of triangle BQC: 11. Height CP = 9 cm, base BQ = 3 cm 12. Area of triangle BQC: $$A_{triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 9 = 13.5 \text{ cm}^2$$ 13. Shaded area (segment) = Area of sector - Area of triangle: $$A_{shaded} = 140.625 - 13.5 = 127.125 \text{ cm}^2$$ **Final answers:** - Angle ABC = $1.25$ radians (3 s.f.) - Shaded area = $127.1$ cm$^2$ (3 s.f.)