Dann gilt im Dreieck $AMSAMS$ mit dem Satz des Pythagoras:

$h^2=(3a{)}^2-{\left({\displaystyle \frac{d}{2}}\right)}^{2}h^2=(3a)^2-\backslash left(\backslash dfrac\{d\}\{2\}\backslash right)^2$

$h^2=9a^2-{\left({\displaystyle \frac{\sqrt{2}a}{2}}\right)}^{2}h^2=9a^2-\backslash left(\backslash dfrac\{\backslash sqrt\{2\}a\}\{2\}\backslash right)^2$

$\Rightarrow h=\sqrt{9a^2-{\displaystyle \frac{1}{2}}{a}^2}={\displaystyle \frac{\sqrt{34}}{2}}a\backslash Rightarrow\backslash ;h=\backslash sqrt\{9a^2-\backslash dfrac\{1\}\{2\}a^2\}=\backslash dfrac\{\backslash sqrt\{34\}\}\{2\}a$

Im gezeichneten Dreieck ist die Dreiecksgrundseite die halbe Dreiecksseite $aa$.

Im Dreieck gilt dann mit dem Satz des Pythagoras:

$h_{\mathrm{\Delta }}=\sqrt{(3a{)}^2-{\left({\displaystyle \frac{a}{2}}\right)}^2}=\sqrt{9a^2-{\displaystyle \frac{a^2}{4}}}h\_\{\backslash Delta\}=\backslash sqrt\{(3a)^2-\backslash left(\backslash dfrac\{a\}\{2\}\backslash right)^2\}=\backslash sqrt\{9a^2-\backslash dfrac\{a^2\}\{4\}\}$

$h_{\mathrm{\Delta }}={\displaystyle \frac{\sqrt{35}}{2}}ah\_\{\backslash Delta\}=\backslash dfrac\{\backslash sqrt\{35\}\}\{2\}a$

Für den Oberflächeninhalt folgt dann:

$O=G+4\cdot A_{\mathrm{\Delta }}=G+4\cdot \frac{1}{2}\cdot a\cdot h_{\mathrm{\Delta }}O=G+4\backslash cdot\; A\_\{\backslash Delta\}=G+4\backslash cdot\backslash frac\{1\}\{2\}\backslash cdot\; a\backslash cdot\; h\_\{\backslash Delta\}$

$O=a^2+4\cdot {\displaystyle \frac{1}{2}}\cdot a\cdot {\displaystyle \frac{\sqrt{35}}{2}}\cdot aO=a^2+4\backslash cdot\backslash dfrac\{1\}\{2\}\backslash cdot\; a\backslash cdot\; \backslash dfrac\{\backslash sqrt\{35\}\}\{2\}\backslash cdot\; a$

$O=a^2+\sqrt{35}\cdot a^2=(1+\sqrt{35})\cdot a^{2}O=a^2+\backslash sqrt\{35\}\backslash cdot\; a^2=(1+\backslash sqrt\{35\})\backslash cdot\; a^2$

$O\approx \mathrm{6,916}{a}^{2}O\backslash approx6\{,\}916\; a^2$

Die Länge der Dreiecksseite $[AB][AB]$ ist die Diagonalenlänge $d=\sqrt{2}ad=\backslash sqrt\{2\}a$.

Die Höhe $hh$ in diesem Dreieck ist die halbe Pyramidenhöhe.

$\Rightarrow h={\displaystyle \frac{1}{2}}{\displaystyle \frac{\sqrt{34}}{2}}a={\displaystyle \frac{\sqrt{34}}{4}}a\backslash Rightarrow\backslash ;h=\backslash dfrac\{1\}\{2\}\backslash dfrac\{\backslash sqrt\{34\}\}\{2\}a=\backslash dfrac\{\backslash sqrt\{34\}\}\{4\}a$

Die Längen der beiden Seiten $[AC][AC]$ und $[BC][BC]$ sind gleich groß. Die Länge wird wieder mit dem Satz des Pythagoras berechnet:

$\overline{AC}=\overline{BC}=\sqrt{h^2+{\left({\displaystyle \frac{d}{2}}\right)}^2}=\sqrt{{\left({\displaystyle \frac{\sqrt{34}}{4}}a\right)}^2+{\left({\displaystyle \frac{\sqrt{2}a}{2}}\right)}^2}\backslash overline\{AC\}=\backslash overline\{BC\}=\backslash sqrt\{h^2+\backslash left(\backslash dfrac\{d\}\{2\}\backslash right)^2\}=\backslash sqrt\{\backslash left(\backslash dfrac\{\backslash sqrt\{34\}\}\{4\}a\backslash right)^2+\backslash left(\backslash dfrac\{\backslash sqrt\{2\}a\}\{2\}\backslash right)^2\}$

Im linken Dreieck gilt:

$\tan \alpha ={\displaystyle \frac{h}{\frac{d}{2}}}={\displaystyle \frac{2\cdot h}{d}}\backslash tan\backslash alpha=\backslash dfrac\{h\}\{\backslash frac\{d\}\{2\}\}=\backslash dfrac\{2\backslash cdot\; h\}\{d\}$

$\tan \alpha ={\displaystyle \frac{2\cdot {\displaystyle \frac{\sqrt{34}}{4}}a}{\sqrt{2}a}}\backslash tan\backslash alpha=\backslash dfrac\{2\backslash cdot\; \backslash dfrac\{\backslash sqrt\{34\}\}\{4\}a\}\{\backslash sqrt\{2\}a\}$

$\tan \alpha ={\displaystyle \frac{\sqrt{34}}{2\cdot \sqrt{2}}}\backslash tan\backslash alpha=\backslash dfrac\{\backslash sqrt\{34\}\}\{2\backslash cdot\; \backslash sqrt\{2\}\}$

$\Rightarrow \alpha ={\tan }^{-1}\left({\displaystyle \frac{\sqrt{34}}{2\cdot \sqrt{2}}}\right)\approx {\mathrm{64,12}}^{\circ }\backslash Rightarrow\backslash ;\backslash alpha=\backslash tan^\{-1\}\backslash left(\backslash dfrac\{\backslash sqrt\{34\}\}\{2\backslash cdot\; \backslash sqrt\{2\}\}\backslash right)\backslash approx\; 64\{,\}12^\backslash circ$

Der Winkel $ABCABC$ ist gleich dem Winkel $\alpha \backslash alpha$.

Der Winkel $ACB={180}^{\circ }-2\cdot \alpha ={180}^{\circ }-{\mathrm{128,24}}^{\circ }={\mathrm{51,56}}^{\circ }ACB=180^\backslash circ-2\backslash cdot\; \backslash alpha=180^\backslash circ-128\{,\}24^\backslash circ=51\{,\}56^\backslash circ$.