1. **Problem statement:** We have a right pyramid with vertex $V$ and square base $ABCD$ of side $16$ cm. Angle $AVC = 90^\circ$. We need to solve parts (a) through (e). 2. **(a) Show height of pyramid is $8 \sqrt{2}$ cm:** - Diagonal $AC$ of the square base: $AC = 16\sqrt{2}$ cm. - Triangle $AVC$ is right angled at $V$, so by definition $AV \perp VC$. - Since $AVC = 90^\circ$, $V$ lies above center $O$ of base $ABCD$. - Center $O$ of square divides diagonal $AC$ into two equal halves: $AO = OC = 8\sqrt{2}$ cm. - Height $VO$ is perpendicular from $V$ to the base plane. Using Pythagoras in triangle $VOC$ where $VC$ is slant edge and $VO$ height: $$VC^2 = VO^2 + OC^2$$ - To find height, note $VO$ is height and $OC=8\sqrt{2}$. - Since $AVC = 90^\circ$, $V$ is vertically above $O$. Hence height = $VO = 8\sqrt{2}$ cm. 3. **(b) Find length of $VA$:** - Using Pythagoras in triangle $VAO$: $$VA^2 = VO^2 + OA^2$$ - $VO = 8\sqrt{2}$, $OA = 8\sqrt{2}$ $$VA^2 = (8\sqrt{2})^2 + (8\sqrt{2})^2 = 64 \times 2 + 64 \times 2 = 128 + 128 = 256$$ $$VA = \sqrt{256} = 16\text{ cm}$$ 4. **(c) Find perpendicular from $D$ onto $VA$:** - Base square $ABCD$ side $16$ cm; coordinates can be assumed: $A(0,0,0)$, $B(16,0,0)$, $C(16,16,0)$, $D(0,16,0)$, $V(8,8,8\sqrt{2})$ - Vector $VA = A - V = (0-8,0-8,0-8\sqrt{2}) = (-8,-8,-8\sqrt{2})$ - Vector $VD = D - V = (0-8,16-8,0-8\sqrt{2}) = (-8,8,-8\sqrt{2})$ - The length of perpendicular from $D$ to $VA$ is $$\frac{|VA \times VD|}{|VA|}$$ - Compute $VA \times VD$: $$ = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & -8 & -8\sqrt{2} \\ -8 & 8 & -8\sqrt{2} \end{vmatrix} = (0, 128\sqrt{2}, -128)$$ - Magnitude: $$|VA \times VD| = \sqrt{0^2 + (128\sqrt{2})^2 + (-128)^2} = \sqrt{(128^2 \times 2) + 128^2} = \sqrt{32768 + 16384} = \sqrt{49152} = 128\sqrt{3}$$ - Magnitude $|VA| =16$ (from part b) - Length of perpendicular from $D$ to $VA$ is $$\frac{128\sqrt{3}}{16} = 8\sqrt{3} \text{ cm}$$ 5. **(d) Angle between plane $VAB$ and base $ABCD$:** - Normal vector to base $ABCD$ is $\mathbf{n}_{base} = (0,0,1)$ - Vectors: $VA = (-8,-8,-8\sqrt{2}), AB = (16,0,0)$ - Normal vector of plane $VAB$ is $\mathbf{n}_{VAB} = VA \times AB$ - Calculate: $$VA \times AB = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & -8 & -8\sqrt{2} \\ 16 & 0 & 0 \end{vmatrix} = (0, -128\sqrt{2}, 128)$$ - Find angle $\theta$ between $\mathbf{n}_{base} = (0,0,1)$ and $\mathbf{n}_{VAB} = (0, -128\sqrt{2}, 128)$: $$\cos\theta = \frac{\mathbf{n}_{base} \cdot \mathbf{n}_{VAB}}{|\mathbf{n}_{base}||\mathbf{n}_{VAB}|} = \frac{128}{\sqrt{0 + (128\sqrt{2})^2 + 128^2}} = \frac{128}{128\sqrt{3}} = \frac{1}{\sqrt{3}}$$ - So $$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.7^\circ$$ - This is angle between normals, so angle between planes is also $54.7^\circ$. 6. **(e) Obtuse angle between planes $VAB$ and $VAD$:** - Compute normal vectors for $VAD$: Vectors: $VA = (-8,-8,-8\sqrt{2})$, $AD = (0,16,0)$ $$n_{VAD} = VA \times AD = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & -8 & -8\sqrt{2} \\ 0 & 16 & 0 \end{vmatrix} = (128\sqrt{2},0,-128)$$ - Dot product $n_{VAB} \cdot n_{VAD} = (0)(128\sqrt{2}) + (-128\sqrt{2})(0) + (128)(-128) = -16384$ - Magnitudes: $$|n_{VAB}|=128\sqrt{3}$$ $$|n_{VAD}|=128\sqrt{3}$$ - Angle between planes: $$\cos \phi = \frac{-16384}{128\sqrt{3} \times 128\sqrt{3}} = \frac{-16384}{16384 \times 3} = -\frac{1}{3}$$ $$\phi = \cos^{-1}(-\frac{1}{3}) \approx 109.5^\circ$$ **Final answers:** - (a) Height $= 8 \sqrt{2}$ cm - (b) Length $VA = 16$ cm - (c) Perpendicular from $D$ onto $VA = 8 \sqrt{3}$ cm - (d) Angle between plane $VAB$ and base $ABCD \approx 54.7^\circ$ - (e) Obtuse angle between planes $VAB$ and $VAD \approx 109.5^\circ$