1. **Problem statement:** We have a square-based pyramid ABCDE with volume 180 cm³. M is the center of the base ABCD and lies vertically below E. Given $\vec{EM} = 7.5$ cm, find: (a) the height $ME$ of the pyramid, (b) the length $AE$, (c) the angle $EAM$, (d) the angle between planes $BCE$ and $ABCD$. 2. **Formula for volume of a pyramid:** $$V = \frac{1}{3} \times \text{area of base} \times \text{height}$$ Since the base is square with side length $a$, area of base $= a^2$. 3. **Find the height $ME$ (part a):** Given volume $V = 180$ cm³ and height $h = ME$, we have: $$180 = \frac{1}{3} a^2 h$$ Rearranged: $$h = \frac{3 \times 180}{a^2} = \frac{540}{a^2}$$ 4. **Find side length $a$ of the base:** Point $M$ is the center of the square base, so $EM$ is the height $h$ of the pyramid, given as 7.5 cm. 5. **Relate $a$ and $EM$ using the diagonal:** Since $M$ is the center of the square base, the distance from $M$ to any vertex (like $A$) is half the diagonal of the square base. Diagonal of square base $= a\sqrt{2}$, so $$AM = \frac{a\sqrt{2}}{2}$$ Given $AM = 6$ cm, $$6 = \frac{a\sqrt{2}}{2} \implies a = \frac{12}{\sqrt{2}} = 6\sqrt{2}$$ 6. **Calculate height $h = ME$ using volume formula:** $$h = \frac{540}{a^2} = \frac{540}{(6\sqrt{2})^2} = \frac{540}{36 \times 2} = \frac{540}{72} = 7.5 \text{ cm}$$ This matches the given $EM$, confirming the height. 7. **Find length $AE$ (part b):** $AE$ is the distance from vertex $A$ on the base to apex $E$. Using Pythagoras in triangle $AME$: $$AE = \sqrt{AM^2 + ME^2} = \sqrt{6^2 + 7.5^2} = \sqrt{36 + 56.25} = \sqrt{92.25} = 9.6 \text{ cm}$$ 8. **Find angle $EAM$ (part c):** Angle $EAM$ is at vertex $A$ between points $E$ and $M$. Vectors: $$\overrightarrow{AE} = \overrightarrow{E} - \overrightarrow{A}$$ $$\overrightarrow{AM} = \overrightarrow{M} - \overrightarrow{A}$$ Since $M$ is center of base and $A$ is a vertex, $AM$ is along the base. Using cosine formula: $$\cos \theta = \frac{AE^2 + AM^2 - ME^2}{2 \times AE \times AM}$$ Plug in values: $$\cos \theta = \frac{9.6^2 + 6^2 - 7.5^2}{2 \times 9.6 \times 6} = \frac{92.25 + 36 - 56.25}{115.2} = \frac{72}{115.2} = 0.625$$ So, $$\theta = \cos^{-1}(0.625) \approx 51.3^\circ$$ 9. **Find angle between planes $BCE$ and $ABCD$ (part d):** Plane $ABCD$ is the base, horizontal. Plane $BCE$ contains points $B$, $C$ (base vertices), and apex $E$. The angle between planes is the angle between their normal vectors. Normal vector to base $ABCD$ is vertical. Vector $BC$ lies on base, vector $BE$ goes from base to apex. Calculate vector $BC$ and $BE$: $$BC = C - B = (a,0,0) - (0,a,0) = (a, -a, 0)$$ $$BE = E - B = (\frac{a}{2}, \frac{a}{2}, h) - (0,a,0) = (\frac{a}{2}, -\frac{a}{2}, h)$$ Normal to plane $BCE$ is: $$\vec{n} = BC \times BE$$ Calculate cross product: $$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & -a & 0 \\ \frac{a}{2} & -\frac{a}{2} & h \end{vmatrix} = \mathbf{i}(-a \times h - 0) - \mathbf{j}(a \times h - 0) + \mathbf{k}(a \times -\frac{a}{2} - (-a) \times \frac{a}{2})$$ $$= -a h \mathbf{i} - a h \mathbf{j} + 0 \mathbf{k} = (-a h, -a h, 0)$$ Normal to base $ABCD$ is vertical: $$\vec{n_0} = (0,0,1)$$ Angle $\phi$ between planes: $$\cos \phi = \frac{|\vec{n} \cdot \vec{n_0}|}{|\vec{n}| |\vec{n_0}|} = \frac{|0|}{\sqrt{(-a h)^2 + (-a h)^2} \times 1} = 0$$ So, $$\phi = 90^\circ$$ **Final answers:** (a) Height $ME = 7.5$ cm (b) Length $AE = 9.6$ cm (c) Angle $EAM \approx 51.3^\circ$ (d) Angle between planes $BCE$ and $ABCD = 90^\circ$