1. **Problem statement:** Given a regular square pyramid PADIM with slant height $PR=10$ and base diagonal $12\sqrt{2}$, find the following: (a) Length $ID$ (b) Altitude of the pyramid (c) Length $RD$ (d) Length $PD$ (lateral edge) --- 2. **Step (a): Find $ID$** - The base is a square with diagonal $12\sqrt{2}$. - For a square with side length $s$, diagonal $d = s\sqrt{2}$. - So, $12\sqrt{2} = s\sqrt{2} \implies s = 12$. - $ID$ is a side of the square base, so $ID = 12$. --- 3. **Step (b): Find the altitude of the pyramid** - The slant height $PR=10$ is the height of the triangular face from apex $P$ to midpoint $R$ of base edge $ID$. - Since $ID=12$, midpoint $R$ divides $ID$ into two segments of length 6. - The altitude $h$ is the perpendicular height from $P$ to the base plane. - Triangle $PRD$ is right-angled at $R$ with $PR=10$ (slant height) and $RD=6$ (half base side). - Using Pythagoras theorem in triangle $PRD$: $$h = \sqrt{PR^2 - RD^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$$ - So, the altitude $h = 8$. --- 4. **Step (c): Find $RD$** - $RD$ is half the base side length, so $RD = 6$. --- 5. **Step (d): Find $PD$ (lateral edge length)** - $PD$ is the edge from apex $P$ to base vertex $D$. - Consider right triangle $PDR$ where $PR=10$, $RD=6$, and $PD$ is the hypotenuse. - Using Pythagoras theorem: $$PD = \sqrt{PR^2 + RD^2} = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} = 2\sqrt{34}$$ --- **Final answers:** - $ID = 12$ - Altitude $= 8$ - $RD = 6$ - $PD = 2\sqrt{34}$