1. Problem (a): Find the number of planes of symmetry of the cuboid with edges AB = 8 cm, BC = 4 cm, and CR = 5 cm. 2. Explanation: A cuboid has planes of symmetry that divide it into mirror-image halves. The number of planes of symmetry depends on the equality of its edges. 3. Since all edges are different (8 cm, 4 cm, 5 cm), the cuboid has exactly 3 planes of symmetry, each plane bisecting the cuboid through the center parallel to one pair of faces. 4. Answer (a): The cuboid has 3 planes of symmetry. 5. Problem (b): Calculate the angle between the diagonal AR and the plane BCRQ of the cuboid. 6. Explanation: To find the angle between a line and a plane, we use the formula: $$\theta = 90^\circ - \alpha$$ where $\alpha$ is the angle between the line and the normal vector to the plane. 7. Step 1: Assign coordinates to points for calculation. Let A = (0,0,0), B = (8,0,0), C = (8,4,0), R = (8,4,5). 8. Vector AR = R - A = (8,4,5). 9. The plane BCRQ contains points B, C, R, Q. The normal vector to the plane can be found by the cross product of vectors BC and BR. 10. Vector BC = C - B = (0,4,0), Vector BR = R - B = (0,4,5). 11. Normal vector $\vec{n} = \vec{BC} \times \vec{BR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 4 & 0 \\ 0 & 4 & 5 \end{vmatrix} = (20,0,0)$. 12. Normalize $\vec{n}$: $|\vec{n}| = 20$. 13. Calculate the angle $\alpha$ between AR and $\vec{n}$: $$\cos \alpha = \frac{\vec{AR} \cdot \vec{n}}{|\vec{AR}||\vec{n}|} = \frac{(8,4,5) \cdot (20,0,0)}{\sqrt{8^2+4^2+5^2} \times 20} = \frac{160}{\sqrt{64+16+25} \times 20} = \frac{160}{\sqrt{105} \times 20} = \frac{8}{\sqrt{105}}$$ 14. Calculate $\alpha$: $$\alpha = \cos^{-1}\left(\frac{8}{\sqrt{105}}\right) \approx 22.62^\circ$$ 15. Angle between AR and plane BCRQ is: $$\theta = 90^\circ - 22.62^\circ = 67.38^\circ$$ 16. Answer (b): The angle between diagonal AR and plane BCRQ is approximately $67.4^\circ$. 17. Problem (c): Calculate the angle that AV makes with the base of the pyramid with square base ABCD, AB = 11 cm, and AV = 18.6 cm. 18. Explanation: The vertex V is vertically above M, the intersection of diagonals AC and BD. The base is a square with side 11 cm. 19. Step 1: Find length AM. Since ABCD is a square, M is the midpoint of diagonal AC. 20. Length of diagonal AC: $$AC = AB \sqrt{2} = 11 \sqrt{2}$$ 21. Since M is midpoint of AC: $$AM = \frac{AC}{2} = \frac{11 \sqrt{2}}{2} = 5.5 \sqrt{2}$$ 22. Step 2: Triangle AVM is right angled at M because V is vertically above M. 23. Use trigonometry to find angle $\theta$ between AV and base plane: $$\cos \theta = \frac{AM}{AV} = \frac{5.5 \sqrt{2}}{18.6}$$ 24. Calculate: $$\cos \theta \approx \frac{5.5 \times 1.414}{18.6} = \frac{7.777}{18.6} \approx 0.418$$ 25. Find $\theta$: $$\theta = \cos^{-1}(0.418) \approx 65.2^\circ$$ 26. Answer (c): The angle AV makes with the base is approximately $65.2^\circ$.