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 :

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

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