Solution
We are given a right pyramid with a square base $ABCD$, vertex $E$, and the following information:
- The base $ABCD$ is horizontal with $AB = BC = 12\ \text{cm}$.
- The diagonals of the base intersect at point $O$, and $E$ is vertically above $O$.
- The angle between $EA$ and the plane $ABCD$ is $30^\circ$.
(a) Find the exact value of $h$, the height of the pyramid
The diagonal of the square base is:
$$ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 12^2} = \sqrt{288} = 12\sqrt{2}\ \text{cm} $$
Since $E$ is vertically above $O$, the line $EO$ is perpendicular to the base $ABCD$. The point $O$ is the midpoint of the diagonals $AC$ and $BD$.
In the triangle $\triangle EAO$:
- $AO = \frac{1}{2}AC = \frac{12\sqrt{2}}{2} = 6\sqrt{2}\ \text{cm}$,
- The angle $\angle EAO = 30^\circ$,
- The height $h = EO$ is opposite $\angle EAO$, and $AO$ is adjacent.
Using the tangent of $\angle EAO$:
$$ \tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{AO} $$
$$ \tan(30^\circ) = \frac{h}{6\sqrt{2}} $$
$$ h = 6\sqrt{2}\cdot\tan(30^\circ) $$
Since $\tan(30^\circ)=\frac{1}{\sqrt{3}}$:
$$ h = 6\sqrt{2}\cdot\frac{1}{\sqrt{3}} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}\ \text{cm} $$
Final Answer:
$$ h = 2\sqrt{6}\ \text{cm} $$
(b) Calculate the size of $\angle EFO$ to the nearest degree
The point $F$ lies on $AD$ such that $AF : FD = 1 : 4$. The total length of $AD$ is $12\ \text{cm}$, so:
$$ AF = \frac{1}{5}\cdot12 = \frac{12}{5} = 2.4\ \text{cm}, \quad FD = \frac{4}{5}\cdot12 = 9.6\ \text{cm} $$
Let $M$ be the midpoint of $AD$. Then:
$$ AM = 6 $$
and
$$ FM = 6 - 2.4 = 3.6 $$
Thus,
$$ OF = \sqrt{6^2 + 3.6^2} = \frac{6\sqrt{34}}{5} $$
Let $\angle EFO = \theta$. Then:
$$ \tan\theta = \frac{2\sqrt{6}} {\frac{6\sqrt{34}}{5}} $$
$$ \theta = 35^\circ $$
Final Answer:
$$ \angle EFO = 35^\circ $$
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