Question

SMNF is a regular triangular pyramid. SO (height)=6cm. Measure of the SEO angle is 60 degrees. Find: MF, apothem SE, total area of pyramid, volume of pyramid and the area of the SME triangle.

https://docs.google.com/drawings/d/1lB1C5bVx0ffa-QMdn5_jty58NssietYAFYhtdQDPLfY/edit?usp=sharing

Answer 1

ANSWER ON QUESTION #66883 – MATH – GEOMETRY

QUESTION

SMNF is a regular triangular pyramid. SO (height) = 6 cm. Measure of the SEO angle is 60 degrees ($\angle SEO = 60{}^\circ$). Find: MF, apothem SE, total area of pyramid, volume of pyramid and the area of the SME triangle.

SOLUTION

1) Since $SMNF$ is a regular triangular pyramid, the base of the height falls in the centroid of the triangle $\Delta MNF$.

2) Since $SMNF$ is a regular triangular pyramid, the triangle $\Delta MNF$ is regular. This means that $ME$ is a height, median, bisector.

$$\left\{ \begin{array}{l} SO \perp ME \\ ME \perp NF \end{array} \right\} \to SE \perp NF$$

by the theorem of three perpendiculars.

Conclusion, $SE$ is an apothem.

3) Consider a triangle $\Delta SEO$:

$SO \perp OE$, hence the triangle $\Delta SEO$ is right.

$$\left\{ \begin{array}{l} \angle SEO = 60{}^\circ \\ SO = 6 \end{array} \right\} \to \tan \angle SEO = \frac{SO}{OE} \to OE = \frac{SO}{\tan \angle SEO} = \frac{SO}{\tan 60{}^\circ} = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3};$$ $$\sin \angle SEO = \frac{SO}{SE} \to SE = \frac{SO}{\sin \angle SEO} = \frac{SO}{\sin 60{}^\circ} = \frac{6}{\frac{\sqrt{3}}{2}} = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}.$$ $$SE = 4\sqrt{3} \text{ cm is an apothem}$$ $$OE = 2\sqrt{3} \text{ cm}$$

4) Consider a regular triangle $\Delta MNF$.

$O$ is the center of mass of a triangle. As we know, the center of mass divides the median in the ratio 2 to 1 counting from the top of the triangle.

In this case,

$$\frac{MO}{OE} = \frac{2}{1} \rightarrow MO = 2OE \rightarrow ME = MO + OE = 2OE + OE = 3OE = 3 \cdot 2\sqrt{3} = 6\sqrt{3}$$ $$ME = 6\sqrt{3} \text{ cm is a height of a regular triangle}$$

We write down the formula for the height of a regular triangle

$$h = \frac{a\sqrt{3}}{2}, \quad a \text{ is the length of triangle side}$$

In this case,

$$ME = \frac{MF\sqrt{3}}{2} \rightarrow MF = \frac{2ME}{\sqrt{3}} = \frac{2 \cdot 6\sqrt{3}}{\sqrt{3}} = 12$$ $$MF = 12 \text{ cm is a side of the regular pyramid}$$

The area of the base is

$$A_1 = A_{\Delta MNF} = \frac{MF^2\sqrt{3}}{4} = \frac{12^2\sqrt{3}}{4} = 36\sqrt{3}$$

By the definition, the volume of the pyramid is

$$V = \frac{1}{3} S_1 \cdot h = \frac{1}{3} \cdot 36\sqrt{3} \cdot 6 = 72\sqrt{3}$$ $$V = 72\sqrt{3} \text{ cm}^3$$

By the definition,

$$A_{\Delta SME} = \frac{1}{2} \cdot SO \cdot ME = \frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3} \text{ cm}^2$$ $$A_{\Delta SME} = 18\sqrt{3} \text{ cm}^2$$

By the definition, the total area is

$$A _ {t o t a l} = A _ {b a s e} + \frac {1}{2} \cdot P \cdot L,$$

where

$A_{base} = A_{1}$ is the area of the base,

$P$ is the base perimeter,

$L$ is an apothem.

In this case,

$$\begin{array}{l} A _ {t o t a l} = 3 6 \sqrt {3} + \frac {1}{2} \cdot 3 M F \cdot S E = 3 6 \sqrt {3} + \frac {1}{2} \cdot 3 \cdot 1 2 \cdot 4 \sqrt {3} = \\ = 5 4 \sqrt {3} + 7 2 \sqrt {3} = 1 2 6 \sqrt {3} \\ \end{array}$$ $$\boxed {A _ {t o t a l} = 1 2 6 \sqrt {3} c m ^ {2}}$$

**ANSWER:**

$$M F = 1 2 c m$$ $$S E = 4 \sqrt {3} c m$$ $$A _ {t o t a l} = 1 2 6 \sqrt {3} c m ^ {2}$$ $$V = 7 2 \sqrt {3} c m ^ {3}$$ $$A _ {\Delta S M E} = 1 8 \sqrt {3} c m ^ {2}$$

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