1. **State the problem:** We have a circle with center O and points A, B, C, D on the circumference. DB is a diameter, and AC intersects DB at point P. Given lengths are AP = 12 cm, PC = 7 cm, and PB = 6 cm. We need to find the area of the circle. 2. **Use the intersecting chords theorem:** When two chords intersect inside a circle, the products of the segments of each chord are equal. That is, $$AP \times PC = PB \times PD$$ 3. **Calculate PD:** We know AP = 12, PC = 7, PB = 6, so $$12 \times 7 = 6 \times PD$$ $$84 = 6 \times PD$$ Divide both sides by 6: $$\cancel{6} \times PD = \frac{84}{\cancel{6}}$$ $$PD = 14$$ 4. **Find the length of diameter DB:** Since DB is a diameter, and P lies on DB, the length DB = PB + PD = 6 + 14 = 20 cm. 5. **Calculate the radius:** Radius $r = \frac{DB}{2} = \frac{20}{2} = 10$ cm. 6. **Calculate the area of the circle:** $$\text{Area} = \pi r^2 = \pi \times 10^2 = 100\pi$$ 7. **Final answer:** $$\text{Area} \approx 100 \times 3.142 = 314.2 \text{ cm}^2$$ Rounded to 3 significant figures, the area is **314 cm\textsuperscript{2}**.