1. **Problem statement:** We have a square-based pyramid with base ABCD where side AD = 19 cm. Point O is directly above the center of the base. We need to find: a) Length of diagonal AC of the square base. b) Perpendicular height of the pyramid (distance from O to the base). 2. **Finding length AC:** Since ABCD is a square with side length $19$ cm, the diagonal $AC$ can be found using the Pythagorean theorem: $$AC = \sqrt{AD^2 + DC^2}$$ Since $AD = DC = 19$ cm, $$AC = \sqrt{19^2 + 19^2} = \sqrt{2 \times 19^2} = \sqrt{2} \times 19$$ 3. **Calculate $AC$ numerically:** $$AC = 19 \times \sqrt{2} \approx 19 \times 1.414 = 26.87 \text{ cm}$$ Rounded to 3 significant figures: $$AC = 26.9 \text{ cm}$$ 4. **Finding the perpendicular height $h$ of the pyramid:** Point O is above the center of the base. The center of the square base is the midpoint of diagonal $AC$. 5. **Using triangle BCD and angle $\angle BCD = 82^\circ$:** We know $BC = 19$ cm (side of square), and $CD = 19$ cm. 6. **Find length BD using the Law of Cosines in triangle BCD:** $$BD^2 = BC^2 + CD^2 - 2 \times BC \times CD \times \cos(82^\circ)$$ $$BD^2 = 19^2 + 19^2 - 2 \times 19 \times 19 \times \cos(82^\circ)$$ 7. **Calculate $BD^2$:** $$BD^2 = 361 + 361 - 722 \times \cos(82^\circ)$$ $$\cos(82^\circ) \approx 0.1392$$ $$BD^2 = 722 - 722 \times 0.1392 = 722 - 100.56 = 621.44$$ 8. **Calculate $BD$:** $$BD = \sqrt{621.44} \approx 24.93 \text{ cm}$$ 9. **Height calculation:** Since O is above the center of the base, and the height is perpendicular to the base, the height $h$ can be found using right triangle formed by O, center of base, and B or D. 10. **Using Pythagoras theorem in triangle OBD:** $$OB^2 = h^2 + (\frac{BD}{2})^2$$ 11. **Length OB is the slant height from apex O to vertex B. Since O is above center, OB is given or can be found if more info is provided. But since not given, we use the right angle at foot of perpendicular and angle 82° to find height.** 12. **Using angle 82° at C in triangle BCD, height $h$ is perpendicular from O to base, so:** $$h = \frac{BD}{2} \times \tan(82^\circ)$$ 13. **Calculate $h$:** $$\tan(82^\circ) \approx 7.115$$ $$h = \frac{24.93}{2} \times 7.115 = 12.465 \times 7.115 = 88.63 \text{ cm}$$ Rounded to 3 significant figures: $$h = 88.6 \text{ cm}$$ **Final answers:** - a) Length of AC = $26.9$ cm - b) Perpendicular height of pyramid = $88.6$ cm