heureka

In triangle ABC, $${\mathtt{AB}} = {\sqrt{{\mathtt{2}}}}$$ ,point D is on side BC, BD=2*DC,cosDAC= $${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}$$ ,cosC= $${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}$$ ,AC+BC=?

$$\small{ \begin{array}{l} \angle DAC = \arccos(\frac{3\cdot\sqrt{10}}{10}) = 18.4349488229\ensurement{^{\circ}}\\\\ \angle C = \arccos(\frac{2\cdot\sqrt{5}}{5}) = 26.5650511771\ensurement{^{\circ}}\\\\ \angle DAC + \angle C = 45 \ensurement{^{\circ}} \\\\ \angle ADC = 180 \ensurement{^{\circ}} -45 \ensurement{^{\circ}} = 135\ensurement{^{\circ}}\\\\ \angle BDA = 180 \ensurement{^{\circ}} -135 \ensurement{^{\circ}} = 45\ensurement{^{\circ}}\\\\ \end{array} }$$

Now we define:

DC = x,   BD = 2x,

H = foot (of a perpendicular) from A on line BC between B and D. ( $$\small{\angle BDA = 45\ensurement{^{\circ}}}$$  !!!)

h = AH

u = BH,  v = HD,  DB = u+v = 2x

$$\small{ \begin{array}{l} H = \text{foot (of a perpendicular) from }~ A \text{ to line $\overline{BC}$ } \\ h = \overline{AH}\qquad u = \overline{BH}\qquad v = \overline{HD}\qquad x = \overline{DC}\qquad \overline{BC} = 3x \end{array} }\\\\ \small{ \begin{array}{rcl} \tan{(45\ensurement{^{\circ}} )}=1 &=& \frac{h}{v} \\ h &=& v\\\\ \tan{(C)} = 0.5 &=& \frac{h}{v+x} \\ 0.5 &=& \frac{v}{v+x} \\ v+x &=& 2v\\ v &=& x\\\\ \overline{BD} = u+v &=& 2x \\ u&=& 2x-v \\ u &=& 2x -x \\ u &=& x\\\\ u^2+h^2 &=& \sqrt{2}^2 \quad | \quad u= 2x-v\\ (2x-v)^2 + v^2 &=& 2 \quad | \quad v=x\\ (2x-x)^2 + x^2 &=& 2 \\ x^2+x^2 &=& 2\\ 2x^2 &=& 2 \\ x^2 &=& 1\\ x &=& 1\\\\ \overline{BC} = 3x = 3\cdot 1 = 3\\\\ \frac {\overline{AC}} { \sin{(135\ensurement{^{\circ}})} }&=&\frac{x}{\sin{(DAC)}}\\\\ \frac {\overline{AC}} {\sin{(135\ensurement{^{\circ}} )}} &=&\frac{1}{ \sin{(DAC)} }\\\\ \overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{(DAC)} } \\\\ \end{array} }\\\\$$

$$\small{ \begin{array}{rcl} \overline{AC} &=& \frac {\sin{(135\ensurement{^{\circ}} )}} { \sin{ ( 18.4349488229 \ensurement{^{\circ}} )} } \\\\ \overline{AC} &=& 2.23606797750 \\\\ \overline{AC}+\overline{BC}& =& 2.23606797750 +3\\\\ \mathbf{ \overline{AC}+\overline{BC}} &\mathbf{ =}& \mathbf{5.23606797750} &\\ \hline \end{array} }$$

25^x=5^x+7

$$\begin{array}{rcl} 25^x &=& 5^x+7 \\ 5^{2x} &=& 5^x+7 \\ 5^{2x} - 5^x-7 &=& 0 \qquad |\qquad u = 5^x \\ u^2 -u-7 &=& 0\\ u_{1,2} &=& \frac{1}{2}\cdot [ 1\pm \sqrt{ 1-4\cdot (-7) } ] \\ u_{1,2} &=& \frac{1}{2}\cdot [ 1\pm \sqrt{ 29 } ] \\\\ u_1 &=& \frac{1+\sqrt{29}}{2} \qquad u_2 = \frac{1-\sqrt{29}}{2}\\\\ 5^x&=& u \qquad |\qquad \ln() \\ x\ln{(5)} &=& \ln{(u)} \\ x &=& \frac{\ln{(u)}} {\ln{(5)}}\\\\ x_1 &=& \frac{ \ln{ ( \frac{1+\sqrt{29}}{2} ) } } {\ln{(5)}}\\ x_1 &=& 0.72126430677\\\\ x_2 &=& \frac{ \ln{ ( \frac{1-\sqrt{29}}{2} ) } } {\ln{(5)}} \qquad \text{no solution}\\ \end{array}$$

$$\begin{array}{rcl} {\text{\bf{Stunde 4: }} \\ \sum \limits_{k=0}^{\infty } \frac{3}{4^k} &=& \dfrac{3}{4^0} +\dfrac{3}{4^1} +\dfrac{3}{4^2} +\dfrac{3}{4^3} +\dfrac{3}{4^4} +\dots \\\\ &=& 3\cdot \left( \dfrac{1}{4^0} \right ) +3\cdot \left( \dfrac{1}{4^1} \right ) +3\cdot \left( \dfrac{1}{4^2} \right ) +3\cdot \left( \dfrac{1}{4^3} \right ) +3\cdot \left( \dfrac{1}{4^4} \right ) +\dots \\\\ &=& 3 +3\cdot \left( \dfrac{1}{4} \right )^1 +3\cdot \left( \dfrac{1}{4} \right )^2 +3\cdot \left( \dfrac{1}{4} \right )^3 +3\cdot \left( \dfrac{1}{4} \right )^4 +\dots \\\\ \end{array}$$

Dies ist eine unendliche geometrische Reihe mit  $$a=3$$   und

  $$\small{\text{$ q = \dfrac{1}{4} $}}$$

Die Summe ist

  $$\small{\text{$s=\dfrac{a}{1-q} = \dfrac{3}{1-\frac{1}{4} } = \dfrac{3}{\frac{3}{4} } =4 $}}\\\\ \begin{array}{rcl} \sum \limits_{k=0}^{\infty } \frac{3}{4^k} =4 \end{array}$$

integrate ∫(2cos3x + 3sinx) / sin^3x dx

$$\small{\begin{array}{l|lll} \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx \quad & \quad \cos{(3x)}= \cos{(x)}\cos{(2x)} -\sin(x)\sin{(2x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}[1-2\sin^2{(x)}]-\sin{(x)}\cdot 2\sin{(x)}\cos{(x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}[1-2\sin^2{(x)}]-2\sin^2{(x)}\cos{(x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}-2\cos{(x)}\sin^2{(x)}-2\sin^2{(x)}\cos{(x)}\\ \quad &\quad \cos{(3x)} = \cos{(x)}-4\cos{(x)}\sin^2{(x)}\\ =\int \dfrac{2[\cos{(x)}-4\cos{(x)}\sin^2{(x)}] + 3\sin{(x)} }{ \sin^3{(x)} }\ dx & \\ =\int \dfrac{2\cos{(x)}-8\cos{(x)}\sin^2{(x)} + 3\sin{(x)} }{ \sin^3{(x)} }\ dx & \\ = 2 \int \dfrac{ \cos{(x)} } { \sin^3{(x)} }\ dx -8 \int \dfrac{ \cos{(x)}\sin^2{(x)} } { \sin^3{(x)} }\ dx +3 \int \dfrac{ \sin{(x)} } { \sin^3{(x)} }\ dx & \\ = 2 \int \dfrac{ 1 } { \sin^2{(x)} }\cdot\cot{(x)}\ dx -8 \int \dfrac{ \cos{(x)} } { \sin{(x)} }\ dx +3 \int \dfrac{ 1 } { \sin^2{(x)} }\ dx &\\ \quad & \quad \text{Formula:} \\ \quad & \quad \boxed{ \int \frac{f'(x)}{f(x)}=\ln{(f(x))} ~~ \int \frac{\cos{(x)}}{\sin{(x)}}\ dx = \ln {( \sin{(x)} )} } \\ \quad & \quad \boxed{ (\cot{(x)})' = -\frac{1}{\sin^2{(x)}} ~~\int \frac{1}{\sin^2{(x)}}\ dx = -\cot{(x)}}\\ \quad & \quad \boxed{ \int f'(x) \cdot [f(x)]^1 = \frac{ [f(x)]^2}{2} ~~\int \dfrac{ 1 } { \sin^2{(x)} }\cdot\cot{(x)}\ dx = -\frac{\cot^2{(x)}}{2} }\\ = 2 (-\frac{\cot^2{x}}{2}) - 8 \ln {( \sin{(x)} )} + 3 (-\cot{(x)} )&\\\\ = -\cot^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1&\\ \end{array}}\\\\\\$$

$$\small{\begin{array}{rcl} \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& -\cot^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1 \quad | \quad -\cot^2{(x)}=1-\csc^2{(x)}\\\\ \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& 1-\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c_1 \\\\ \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& -\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + (c_1+1) \quad | \quad c = c_1+1 \\\\ \int \dfrac{2\cos{(3x)} + 3\sin{(x)} } { \sin^3{(x)} }\ dx &=& -\csc^2{(x)} - 8\ln {( \sin{(x)} )} - 3\cot{(x)} + c \\\\ \end{array}}$$

How many numbers between 1000 and 2000 leave a remainder of 3 when divided by 11 ?

$$\small{ \begin{array}{rcl} n_1 \cdot 11 + 3 &\ge& 1000 \\ n_1 &\ge& \dfrac{1000-3}{11} = 90.6363636364 \\ n_1 &=& 91 \\\\ n_2 \cdot 11 + 3 &\le& 2000 \\ n_2 &\le& \dfrac{2000-3}{11} = 181.545454545 \\ n_2 &=& 181 \\\\ n&=& n_2-n_1+1\\ n&=& 181-91+1\\ \mathbf{n}&\mathbf{=}& \mathbf{91}\\\\ \hline \end{array} }$$

How many numbers between 1000 and 2000 leave a remainder of 3 when divided by 11 ?

$$\small{ \begin{array}{rcl} 1. & 1004 & \text{remainder~} 3 \\ 2. & 1015 & \text{remainder~} 3\\ 3. & 1026 & \text{remainder~} 3\\ 4. & 1037 & \text{remainder~} 3\\ 5. & 1048 & \text{remainder~} 3\\ 6. & 1059 & \text{remainder~} 3\\ 7. & 1070 & \text{remainder~} 3 \\ 8. & 1081 & \text{remainder~} 3\\ 9. & 1092 & \text{remainder~} 3\\ 10. & 1103 & \text{remainder~} 3\\ 11. & 1114 & \text{remainder~} 3\\ 12. & 1125 & \text{remainder~} 3\\ 13. & 1136 & \text{remainder~} 3\\ 14. & 1147 & \text{remainder~} 3\\ 15. & 1158 & \text{remainder~} 3\\ 16. & 1169 & \text{remainder~} 3\\ 17. & 1180 & \text{remainder~} 3\\ 18. & 1191 & \text{remainder~} 3\\ 19. & 1202 & \text{remainder~} 3\\ 20. & 1213 & \text{remainder~} 3\\ 21. & 1224 & \text{remainder~} 3\\ 22. & 1235 & \text{remainder~} 3\\ 23. & 1246 & \text{remainder~} 3\\ 24. & 1257 & \text{remainder~} 3\\ 25. & 1268 & \text{remainder~} 3\\ 26. & 1279 & \text{remainder~} 3\\ 27. & 1290 & \text{remainder~} 3\\ 28. & 1301 & \text{remainder~} 3\\ 29. & 1312 & \text{remainder~} 3\\ 30. & 1323 & \text{remainder~} 3\\ 31. & 1334 & \text{remainder~} 3\\ 32. & 1345 & \text{remainder~} 3\\ 33. & 1356 & \text{remainder~} 3\\ 34. & 1367 & \text{remainder~} 3\\ 35. & 1378 & \text{remainder~} 3\\ 36. & 1389 & \text{remainder~} 3\\ 37. & 1400 & \text{remainder~} 3\\ 38. & 1411 & \text{remainder~} 3\\ 39. & 1422 & \text{remainder~} 3\\ 40. & 1433 & \text{remainder~} 3\\ 41. & 1444 & \text{remainder~} 3\\ 42. & 1455 & \text{remainder~} 3\\ 43. & 1466 & \text{remainder~} 3\\ 44. & 1477 & \text{remainder~} 3\\ 45. & 1488 & \text{remainder~} 3\\ 46. & 1499 & \text{remainder~} 3\\ 47. & 1510 & \text{remainder~} 3\\ 48. & 1521 & \text{remainder~} 3\\ 49. & 1532 & \text{remainder~} 3\\ 50. & 1543 & \text{remainder~} 3\\ 51. & 1554 & \text{remainder~} 3\\ 52. & 1565 & \text{remainder~} 3\\ 53. & 1576 & \text{remainder~} 3\\ 54. & 1587 & \text{remainder~} 3\\ 55. & 1598 & \text{remainder~} 3\\ 56. & 1609 & \text{remainder~} 3\\ 57. & 1620 & \text{remainder~} 3\\ 58. & 1631 & \text{remainder~} 3\\ 59. & 1642 & \text{remainder~} 3\\ 60. & 1653 & \text{remainder~} 3\\ 61. & 1664 & \text{remainder~} 3\\ 62. & 1675 & \text{remainder~} 3\\ 63. & 1686 & \text{remainder~} 3\\ 64. & 1697 & \text{remainder~} 3\\ 65. & 1708 & \text{remainder~} 3\\ 66. & 1719 & \text{remainder~} 3\\ 67. & 1730 & \text{remainder~} 3\\ 68. & 1741 & \text{remainder~} 3\\ 69. & 1752 & \text{remainder~} 3\\ 70. & 1763 & \text{remainder~} 3\\ 71. & 1774 & \text{remainder~} 3\\ 72. & 1785 & \text{remainder~} 3\\ 73. & 1796 & \text{remainder~} 3\\ 74. & 1807 & \text{remainder~} 3\\ 75. & 1818 & \text{remainder~} 3\\ 76. & 1829 & \text{remainder~} 3\\ 77. & 1840 & \text{remainder~} 3\\ 78. & 1851 & \text{remainder~} 3\\ 79. & 1862 & \text{remainder~} 3\\ 80. & 1873 & \text{remainder~} 3\\ 81. & 1884 & \text{remainder~} 3\\ 82. & 1895 & \text{remainder~} 3\\ 83. & 1906 & \text{remainder~} 3\\ 84. & 1917 & \text{remainder~} 3\\ 85. & 1928 & \text{remainder~} 3\\ 86. & 1939 & \text{remainder~} 3\\ 87. & 1950 & \text{remainder~} 3\\ 88. & 1961 & \text{remainder~} 3\\ 89. & 1972 & \text{remainder~} 3\\ 90. & 1983 & \text{remainder~} 3\\ 91. & 1994 & \text{remainder~} 3\\ \end{array} }$$

$$\small{ \begin{array}{lcl} {\text{\bf{Stunde 12: }} \\ C_{16} = 12_{10} \qquad \text{Hex} \Rightarrow \text{dezimal}\\\\ {\text{\bf{Stunde 1: }} \\ \text{Die Fourier Transformation der Delta Funktion: }=1\\\\ {\text{\bf{Stunde 2: }} \\ \sqrt[8]{256}=\sqrt[8]{2^8}=2\\\\ {\text{\bf{Stunde 3: }} \\ \int \limits_{0}^{2\pi} \int \limits_{0}^{\sqrt{\dfrac{3}{\pi}}} r \ dr \ d\varphi =\int \limits_{0}^{2\pi} \begin{bmatrix} \frac{r^2}{2}\end{bmatrix}_{0}^{ \sqrt{ \frac{3}{\pi} } } \ d\varphi =\int \limits_{0}^{2\pi} \frac{3}{2\pi} \ d\varphi =\frac{3}{2\pi}\int \limits_{0}^{2\pi} \ d\varphi =\frac{3}{2\pi} \begin{bmatrix} \varphi \end{bmatrix}_{0}^{ 2\pi } = \frac{3}{2\pi} \cdot 2\pi = 3\\\\ {\text{\bf{Stunde 4: wurde bereits geraten.}} }\\\\ {\text{\bf{Stunde 5: }} \\ 0101_2 = (1\cdot 2 + 0)\cdot 2 + 1 = 5_{10} \qquad \text{dual} \Rightarrow \text{dezimal}\\\\ {\text{\bf{Stunde 6: }} \\ 3! = 1\cdot2\cdot3 = 6 \\\\ {\text{\bf{Stunde 7: }} \\ \binom76 = \frac{7!}{6!(7-6)!} = 7 \\\\ {\text{\bf{Stunde 8: }} \\ e^x = \lim \limits_{n\rightarrow \infty }(1+\frac{x}{n})^n \\ \ln[ \lim \limits_{n\rightarrow \infty }(1+\frac{8}{n})^n ] = \ln{(e^8)} = 8 \\\\ {\text{\bf{Stunde 9: }} \\ \begin{vmatrix} 10 & 1 \\ 1 & 1 \end{vmatrix} = 10\cdot 1 - 1\cdot 1 = 10-1=9 \qquad \text{Determinante}\\\\ {\text{\bf{Stunde 10: wurde bereits geraten.}} }\\\\ {\text{\bf{Stunde 11: }} \\ 23_4 = 2\cdot 4 + 3 = 11_{10} \qquad 4_{\text{er-System}}} \Rightarrow \text{dezimal}\\\\ \end{array} }$$

$$\small{ \begin{array}{l} \text{For a positive integer }$n$ \text{, } $\varphi(n)$ denotes the number of positive integers\\ \text{less than or equal to } $n$ \text{ that are relatively prime to } $n$.\text{ What is }$\varphi(2^{100})$? \end{array} }$$

$$\small{ \begin{array}{rcl} \text{Prime Factorization: }$2^{100} &=& \textcolor[rgb]{1,0,0}{2}^{100}$\\\\ $\varphi( 2^{100} ) &=& 2^{100}\cdot (1-\frac{1}{\textcolor[rgb]{1,0,0}{2}})\\\\ \varphi(2^{100}) &=& 2^{100}\cdot\frac12\\\\ \mathbf{ \varphi(2^{100})} &\mathbf{=}& \mathbf{2^{99}}=633~825~300~114~114~700~748~351~602~688$\\ \end{array} }$$

$$\small{ \begin{array}{l} \text{For a positive integer }$n$ \text{, } $\varphi(n)$ denotes the number of positive integers\\ \text{less than or equal to } $n$ \text{ that are relatively prime to } $n$.\text{ What is }$\varphi(12)$? \end{array} }$$

$$\small{ \begin{array}{rcl} \text{Prime Factorization: }$12 &=& \textcolor[rgb]{1,0,0}{2}^2\cdot \textcolor[rgb]{0,150,0}{3}$\\\\ $\varphi(12) &=& 12\cdot (1-\frac{1}{\textcolor[rgb]{1,0,0}{2}})\cdot (1-\frac{1}{\textcolor[rgb]{0,150,0}{3}})\\ \varphi(12) &=& 12\cdot\frac12\cdot \frac23\\ \mathbf{ \varphi(12)} &\mathbf{=}& \mathbf{4}$\\ \end{array} }$$

Drei bauklötze (würfel, zylinder, kegel) wurden aufeinander gestapelt. Der Kegel past genau auf den zylinder. Der durchmesser des Zylinders ist so groß wie seine Höhe. Die Kantenlänge des Würfels ist doppelt so groß, wie der Durchmesser desZylinders. der Kegel ist 3mal so hoch wie der Zylinder. Zusammen haben sie ein Volumen von 114,27cmhoch3. Welche Kantenlänge hat der Würfel und wie groß ist die oberfläche des zusammengesetzten Körpers ?

$$\small{\text{$ \begin{array}{|lcr|} \hline d=\mathrm{Durchmesser~Zylinder} \\ \hline &&\\ h_{\mathrm{Kegel}} &=& 3d \\ &&\\ h_{\mathrm{Zylinder}} &=& d \\ &&\\ a_{\mathrm{Wuerfel}} &=& 2d \\ &&\\ \hline \end{array} \begin{array}{|lclclcl|} \hline &&&&&&\\ V_{\mathrm{Kegel}} &=& \frac{1}{3}\pi (\frac{d}{2})^2\cdot h_{\mathrm{Kegel}} &=& \frac{1}{3}\pi (\frac{d}{2})^2\cdot 3d = \pi (\frac{d}{2})^2\cdot d &=& \pi \dfrac{d^3}{4}\\ &&&&&&\\ V_{\mathrm{Zylinder}} &=& \pi (\frac{d}{2})^2\cdot h_{\mathrm{Zylinder}} &=& \pi (\frac{d}{2})^2\cdot d &=& \pi \dfrac{d^3}{4}\\ &&&&&&\\ V_{\mathrm{Wuerfel}} &=& a^3 &=& (2d)^3 &=& 8d^3\\ &&&&&&\\ \hline \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} 114,27\mathrm{~cm^3} =V_{Zusammen} &=& V_{\mathrm{Kegel}} + V_{\mathrm{Zylinder}} + V_{\mathrm{Wuerfel}} \\\\ V &=& \pi \frac{d^3}{4} + \pi \frac{d^3}{4} + 8d^3\\\\ V &=& 2\pi \frac{d^3}{4} + 8d^3 \\\\ V &=& d^3 ( \frac{\pi}{2} + 8 )\\\\ d^3 &=& \dfrac{V}{ \frac{\pi}{2} + 8 } \\\\ \mathbf{d} &\mathbf{=}& \mathbf{ \sqrt[3]{ \dfrac{V}{ \frac{\pi}{2} + 8 } } } \\\\ d & = & \sqrt[3]{ \dfrac{114,27\mathrm{~cm^3} }{ \frac{\pi}{2} + 8 } }\\\\ \mathbf{d} &\mathbf{=}& \mathbf{ 2,285571004 \mathrm{~cm} }\\\\ a_{\mathrm{Wuerfel}} &=& 2d \\\\ \mathbf{a_{\mathrm{Wuerfel}} } &\mathbf{=}& \mathbf{ 4,57114200801\mathrm{~cm} }\\\\ \end{array} $}}\\\\$$

Berechnung der Oberfläche O :

$$\small{\text{$ \begin{array}{|lclcrcl|} \hline &&&&&&\\ M_{\mathrm{Kegel}} &=& \pi \cdot \frac{d}{2}\cdot s && s^2 &=& (3d)^2+ (\frac{d}{2})^2 \\ &&&&&&\\ &&&& s^2 &=& 9d^2+\frac{d^2}{4} \\ &&&&&&\\ &&&& s^2 &=& d^2 (9+\frac{1}{4}) \\ &&&&&&\\ &&&& s^2 &=& d^2 \cdot \frac{37}{4} \\ &&&&&&\\ &&&& s^2 &=& (\frac{d}{2})^2 \cdot 37 \\ &&&&&&\\ &&&& s &=& \frac{d}{2}\cdot\sqrt{37} \\ &&&&&&\\ M_{\mathrm{Kegel}} &=& \pi \cdot \frac{d}{2}\cdot \frac{d}{2}\cdot\sqrt{37} &&&=& \pi \cdot \frac{d^2}{4}\cdot\sqrt{37} \\ &&&&&&\\ M_{\mathrm{Zylinder}} &=& \pi d\cdot h_{\mathrm{Zylinder}} &=& \pi d\cdot d &=& \pi d^2\\ &&&&&&\\ O_{\mathrm{Wuerfel}} &=& 6a^2 &=& 6\cdot(2d)^2 &=& 24d^2\\ &&&&&&\\ A_{\mathrm{Kreis}} &=& \pi \cdot (\frac{d}{2})^2 &&&=& \pi\cdot \frac{d^2}{4} \\ &&&&&&\\ \hline \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} O_{Zusammen} &=& M_{\mathrm{Kegel}} + M_{\mathrm{Zylinder}} + O_{\mathrm{Wuerfel}} - A_{\mathrm{Kreis}} \\\\ &=& \pi \cdot \frac{d^2}{4}\cdot\sqrt{37} + \pi d^2 + 24d^2 - \pi\cdot \frac{d^2}{4}\\\\ &=& \frac{d^2}{4} ( \pi \cdot\sqrt{37} +4\pi + 96 -\pi ) \\\\ &=& \frac{d^2}{4} ( \pi \cdot\sqrt{37} +3\pi + 96 ) \\\\ &=& \frac{d^2}{4} [ \pi \cdot (\sqrt{37} +3 )+ 96 ] \\\\ &=& 1,30595870359\cdot 124,534340039 \\\\ \mathbf{ O_{\mathrm{Zusammen}} } &\mathbf{=}& \mathbf{162,636705270\mathrm{~cm}^2} \end{array} $}}$$