1. **Problem statement:** We have a triangular prism with given lengths: DE = 22 mm, CD = 15 mm, CF = 9 mm. We need to find: a) The length of AF. b) The size of angle CAF. 2. **Understanding the prism and points:** - Points A, B, C form a right triangle. - Points D, E, F correspond to A, B, C on the opposite face. - DE = 22 mm is the height of the prism (distance between the two triangular faces). - CD = 15 mm and CF = 9 mm are edges on the base face. - AF is the diagonal from A on one face to F on the opposite face. 3. **Find coordinates for points to use distance and angle formulas:** Assume: - Point C at origin (0,0,0). - Point D at (15,0,0) since CD = 15 mm along x-axis. - Point F at (0,9,0) since CF = 9 mm along y-axis. - Point E at (15,0,22) since DE = 22 mm vertical from D. - Point A corresponds to E shifted by vector from C to A. Since A corresponds to E, and A is above C by height 22 mm, and A is at (0,0,22). 4. **Coordinates:** - A = (0,0,22) - F = (0,9,0) 5. **Calculate length AF:** $$AF = \sqrt{(0-0)^2 + (9-0)^2 + (0-22)^2} = \sqrt{0 + 81 + 484} = \sqrt{565}$$ 6. **Simplify length AF:** $$AF = \sqrt{565} \approx 23.75 \text{ mm}$$ 7. **Calculate angle CAF:** Vectors: - $ \vec{CA} = A - C = (0,0,22) - (0,0,0) = (0,0,22) $ - $ \vec{CF} = F - C = (0,9,0) - (0,0,0) = (0,9,0) $ Angle between $ \vec{CA} $ and $ \vec{CF} $ is given by: $$\cos \theta = \frac{\vec{CA} \cdot \vec{CF}}{|\vec{CA}| |\vec{CF}|}$$ Dot product: $$\vec{CA} \cdot \vec{CF} = 0*0 + 0*9 + 22*0 = 0$$ Magnitudes: $$|\vec{CA}| = 22, \quad |\vec{CF}| = 9$$ So: $$\cos \theta = \frac{0}{22 \times 9} = 0$$ Therefore: $$\theta = \cos^{-1}(0) = 90^\circ$$ 8. **Final answers:** a) Length of AF = 23.75 mm b) Angle CAF = 90.00 degrees