Flächeninhalt, Umfang

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Flächeninhalt, Umfang

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Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Guest 18.06.2015

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8⁺⁶ Answers

#5

+26396

+5

Beste Antwort

Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Gegebene Geradengleichungen der Form y = mx + b oder y - mx = b

$\small{\text{$ \begin{array}{llll} (1) & y = -x+1 & m_1 = -1 & b_1 = 1 \\ (2) & y = -\frac{1}{2}x-1 & m_2 = -\frac{1}{2} & b_2 = -1 \\ (3) & y = 2x+4 & m_3 = 2 & b_3 = 4 \\ \end{array} $}}$

I. Berechnung der Schnittpunkte:

1. Schnitt $\small{\text{$S_1$}}$ Gerade (1) mit Gerade (2 ):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_1 \cdot x_s &=& b_1 \\ 1\cdot y_s - m_2 \cdot x_s &=& b_2 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_1 & -m_1\\ b_2 &-m_2 \end{vmatrix} } {\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -(-1)\\ -1 & -(-\frac{1}{2}) \end{vmatrix} } {\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ -1 & \frac{1}{2} \end{vmatrix} } {\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix} } =\dfrac { 1\cdot \frac{1}{2} - (-1)\cdot 1 } { 1\cdot \frac{1}{2} - 1\cdot 1 } =\dfrac { \frac{1}{2} + 1 } { \frac{1}{2} - 1 } =\dfrac { \frac{3}{2} } { -\frac{1}{2} } = -3\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_1\\ 1 & b_2 \end{vmatrix} } {\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix} } {\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix} } {\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix} } =\dfrac { 1\cdot (-1) - 1\cdot 1 } { 1\cdot \frac{1}{2} - 1\cdot 1 } =\dfrac { -1-1 } { \frac{1}{2} - 1 } =\dfrac { -2 } { -\frac{1}{2} } = 4 \end{array} $}}$

2. Schnitt $\small{\text{$S_2$}}$ Gerade (2) mit Gerade (3):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_2 \cdot x_s &=& b_2 \\ 1\cdot y_s - m_3 \cdot x_s &=& b_3 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_2 & -m_2\\ b_3 &-m_3 \end{vmatrix} } {\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix} } =\dfrac {\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix} } {\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix} } =\dfrac {\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix} } {\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix} } =\dfrac { -1\cdot 2 - 4 \cdot(- \frac{1}{2}) } { 1\cdot (- 2) - 1 \cdot \frac{1}{2} } =\dfrac { -2+2} { -2 - \frac{1}{2} } =\dfrac { 0 } { -\frac{5}{2} } = 0\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_2\\ 1 & b_3 \end{vmatrix} } {\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} } {\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} } {\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix} } =\dfrac { 1\cdot 4 - 1\cdot (- 1) } { 1\cdot (- 2) - 1 \cdot \frac{1}{2} } =\dfrac { 4+1 } { -2 - \frac{1}{2} } =\dfrac { 5 } { -\frac{5}{2} } = -2 \end{array} $}}$

3. Schnitt $\small{\text{$S_3$}}$ Gerade (3) mit Gerade (1):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_3 \cdot x_s &=& b_3 \\ 1\cdot y_s - m_1 \cdot x_s &=& b_1 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_3 & -m_3\\ b_1 &-m_1 \end{vmatrix} } {\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix} } =\dfrac {\begin{vmatrix} 4 & -2\\ 1 & -(-1)\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix} } =\dfrac {\begin{vmatrix} 4 & -2\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix} } =\dfrac { 4\cdot 1 - 1 \cdot (-2) } { 1\cdot 1 - 1 \cdot (-2) } =\dfrac { 4+2} { 1+2 } =\dfrac { 6 } { 3 } = 2\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_3\\ 1 & b_1 \end{vmatrix} } {\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix} } =\dfrac { 1\cdot 1 - 1\cdot 4 } { 1\cdot 1 - 1 \cdot (-2) } =\dfrac { 1-4 } { 1+2 } =\dfrac { -3 } { 3 } = -1 \end{array} $}}$

Die 3 Schnittpunkte zusammengefasst:

$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array} $}}$

Die Strecke von $\small{\text{$S_1 $ nach $ S_2$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_1S_2} =\sqrt{ [(4)-(-2)]^2 + [0-(-3)]^2 } =\sqrt{ 6^2 + 3^2 } =\sqrt{ 36 + 9 } =\sqrt{ 45 } =6,70820393250 \end{array} $}}$

Die Strecke von $\small{\text{$S_2 $ nach $ S_3$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [(-1)-(-2)]^2 + (2-0)^2 } =\sqrt{ (-1+2)^2 + 2^2 } =\sqrt{ 1^2 + 4 } =\sqrt{ 5 } =2.23606797750 \end{array} $}}$

Die Strecke von $\small{\text{$S_3 $ nach $ S_1$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [4-(-1)]^2 + (-3-2)^2 } =\sqrt{ (4+1)^2 + (-5)^2 } =\sqrt{ 5^2 + 5^2 } =\sqrt{ 50 } =7.07106781187 \end{array} $}}$

Der Umfang der eingeschlossenen Fläche ( Dreieck) beträgt: $\small{\text{$\overline{S_1S_2}+\overline{S_2S_3}+\overline{S_2S_3} =6,70820393250+2,23606797750+7,07106781187 =16,0153397219 $}}$

Die Fläche des Dreiecks nach Heron:

$\small{\text{$ \boxed{~~ A =\sqrt{ s\cdot(s-a)\cdot(s-b)\cdot(s-c)} \qquad s = (a + b + c)/2 ~~} $}}\\\\ \small{\text{$ \begin{array}{lcl} a= \overline{S_1S_2} = 6,70820393250\\ b= \overline{S_2S_3} = 2,23606797750\\ c= \overline{S_3S_1} = 7,07106781187 \\\\ s = \dfrac{16,0153397219}{2}= 8,00766986093\\ \end{array} $}}\\\\\\ \small{\text{$ \begin{array}{lcl} A =\sqrt{ 8,00766986093\cdot(8,00766986093-6,70820393250)\cdot(8,00766986093-2,23606797750)\cdot(8,00766986093-7,07106781187)}\\ A =\sqrt{ 8,00766986093\cdot 1,29946592843 \cdot 5,77160188343\cdot 0,93660204906}\\ A =\sqrt{ 56.2499999994 }\\ A=7,5 \end{array} $}}\\\\$

Der Flächeninhalt der eingeschlossenen Fläche ( Dreieck) beträgt 7,5

Probe der Fläche:

$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array} $}}\\\\ \small{\text{$ \boxed{ ~~ A = \dfrac{1}{2}\cdot \begin{vmatrix} 1 & 1 & 1 \\ x_{s_1} & x_{s_2} & x_{s_3} \\ y_{s_1} & y_{s_2} & y_{s_3} \\ \end{vmatrix} ~~ } $}}\\\\ \small{\text{$ \begin{array}{rcl} A &=&\dfrac{1}{2}\cdot \begin{vmatrix} 1 & 1 & 1 \\ 4 & -2 & -1 \\ -3 & 0 & 2 \\ \end{vmatrix} \\\\ &=&\frac{1}{2} \cdot [ 1\cdot (-2) \cdot 2 + 4\cdot 0 \cdot 1 + (-3)\cdot 1 \cdot (-1) - (-3)\cdot (-2) \cdot 1 - 0\cdot (-1) \cdot 1 - 4\cdot 1 \cdot 2 ]\\\\ &=&\frac{1}{2} \cdot [ 1\cdot (-2) \cdot 2 + (-3)\cdot 1 \cdot (-1) - (-3)\cdot (-2) \cdot 1 - 4\cdot 1 \cdot 2 ]\\\\ &=&\frac{1}{2} \cdot [ -4 + 3 - 6 \cdot 1 - 8]\\\\ &=&\frac{1}{2} \cdot [ -15 ]\\\\ &=& - 7,5 \\\\ |A| &=& 7,5 \qquad \mathrm{okay} \end{array} $}}$

heureka 19.06.2015

#6

+12530

0

Omi67 19.06.2015

0 Benutzer online

heureka · Answer 1 · 2015-06-19T09:31:26

Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Gegebene Geradengleichungen der Form y = mx + b oder y - mx = b

$\small{\text{$ \begin{array}{llll} (1) & y = -x+1 & m_1 = -1 & b_1 = 1 \\ (2) & y = -\frac{1}{2}x-1 & m_2 = -\frac{1}{2} & b_2 = -1 \\ (3) & y = 2x+4 & m_3 = 2 & b_3 = 4 \\ \end{array} $}}$

I. Berechnung der Schnittpunkte:

1. Schnitt $\small{\text{$S_1$}}$ Gerade (1) mit Gerade (2 ):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_1 \cdot x_s &=& b_1 \\ 1\cdot y_s - m_2 \cdot x_s &=& b_2 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_1 & -m_1\\ b_2 &-m_2 \end{vmatrix} } {\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -(-1)\\ -1 & -(-\frac{1}{2}) \end{vmatrix} } {\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ -1 & \frac{1}{2} \end{vmatrix} } {\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix} } =\dfrac { 1\cdot \frac{1}{2} - (-1)\cdot 1 } { 1\cdot \frac{1}{2} - 1\cdot 1 } =\dfrac { \frac{1}{2} + 1 } { \frac{1}{2} - 1 } =\dfrac { \frac{3}{2} } { -\frac{1}{2} } = -3\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_1\\ 1 & b_2 \end{vmatrix} } {\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix} } {\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix} } {\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix} } =\dfrac { 1\cdot (-1) - 1\cdot 1 } { 1\cdot \frac{1}{2} - 1\cdot 1 } =\dfrac { -1-1 } { \frac{1}{2} - 1 } =\dfrac { -2 } { -\frac{1}{2} } = 4 \end{array} $}}$

2. Schnitt $\small{\text{$S_2$}}$ Gerade (2) mit Gerade (3):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_2 \cdot x_s &=& b_2 \\ 1\cdot y_s - m_3 \cdot x_s &=& b_3 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_2 & -m_2\\ b_3 &-m_3 \end{vmatrix} } {\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix} } =\dfrac {\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix} } {\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix} } =\dfrac {\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix} } {\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix} } =\dfrac { -1\cdot 2 - 4 \cdot(- \frac{1}{2}) } { 1\cdot (- 2) - 1 \cdot \frac{1}{2} } =\dfrac { -2+2} { -2 - \frac{1}{2} } =\dfrac { 0 } { -\frac{5}{2} } = 0\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_2\\ 1 & b_3 \end{vmatrix} } {\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} } {\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} } {\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix} } =\dfrac { 1\cdot 4 - 1\cdot (- 1) } { 1\cdot (- 2) - 1 \cdot \frac{1}{2} } =\dfrac { 4+1 } { -2 - \frac{1}{2} } =\dfrac { 5 } { -\frac{5}{2} } = -2 \end{array} $}}$

3. Schnitt $\small{\text{$S_3$}}$ Gerade (3) mit Gerade (1):

$\small{\text{$ \begin{array}{rcl} 1\cdot y_s - m_3 \cdot x_s &=& b_3 \\ 1\cdot y_s - m_1 \cdot x_s &=& b_1 \\ \\ \hline \\ y_s = \dfrac {\begin{vmatrix} b_3 & -m_3\\ b_1 &-m_1 \end{vmatrix} } {\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix} } =\dfrac {\begin{vmatrix} 4 & -2\\ 1 & -(-1)\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix} } =\dfrac {\begin{vmatrix} 4 & -2\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix} } =\dfrac { 4\cdot 1 - 1 \cdot (-2) } { 1\cdot 1 - 1 \cdot (-2) } =\dfrac { 4+2} { 1+2 } =\dfrac { 6 } { 3 } = 2\\\\ x_s = \dfrac {\begin{vmatrix} 1 & b_3\\ 1 & b_1 \end{vmatrix} } {\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix} } =\dfrac {\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix} } {\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix} } =\dfrac { 1\cdot 1 - 1\cdot 4 } { 1\cdot 1 - 1 \cdot (-2) } =\dfrac { 1-4 } { 1+2 } =\dfrac { -3 } { 3 } = -1 \end{array} $}}$

Die 3 Schnittpunkte zusammengefasst:

$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array} $}}$

Die Strecke von $\small{\text{$S_1 $ nach $ S_2$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_1S_2} =\sqrt{ [(4)-(-2)]^2 + [0-(-3)]^2 } =\sqrt{ 6^2 + 3^2 } =\sqrt{ 36 + 9 } =\sqrt{ 45 } =6,70820393250 \end{array} $}}$

Die Strecke von $\small{\text{$S_2 $ nach $ S_3$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [(-1)-(-2)]^2 + (2-0)^2 } =\sqrt{ (-1+2)^2 + 2^2 } =\sqrt{ 1^2 + 4 } =\sqrt{ 5 } =2.23606797750 \end{array} $}}$

Die Strecke von $\small{\text{$S_3 $ nach $ S_1$}}$ :

$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [4-(-1)]^2 + (-3-2)^2 } =\sqrt{ (4+1)^2 + (-5)^2 } =\sqrt{ 5^2 + 5^2 } =\sqrt{ 50 } =7.07106781187 \end{array} $}}$

Der Umfang der eingeschlossenen Fläche ( Dreieck) beträgt: $\small{\text{$\overline{S_1S_2}+\overline{S_2S_3}+\overline{S_2S_3} =6,70820393250+2,23606797750+7,07106781187 =16,0153397219 $}}$

Die Fläche des Dreiecks nach Heron:

$\small{\text{$ \boxed{~~ A =\sqrt{ s\cdot(s-a)\cdot(s-b)\cdot(s-c)} \qquad s = (a + b + c)/2 ~~} $}}\\\\ \small{\text{$ \begin{array}{lcl} a= \overline{S_1S_2} = 6,70820393250\\ b= \overline{S_2S_3} = 2,23606797750\\ c= \overline{S_3S_1} = 7,07106781187 \\\\ s = \dfrac{16,0153397219}{2}= 8,00766986093\\ \end{array} $}}\\\\\\ \small{\text{$ \begin{array}{lcl} A =\sqrt{ 8,00766986093\cdot(8,00766986093-6,70820393250)\cdot(8,00766986093-2,23606797750)\cdot(8,00766986093-7,07106781187)}\\ A =\sqrt{ 8,00766986093\cdot 1,29946592843 \cdot 5,77160188343\cdot 0,93660204906}\\ A =\sqrt{ 56.2499999994 }\\ A=7,5 \end{array} $}}\\\\$

Der Flächeninhalt der eingeschlossenen Fläche ( Dreieck) beträgt 7,5

Probe der Fläche:

$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array} $}}\\\\ \small{\text{$ \boxed{ ~~ A = \dfrac{1}{2}\cdot \begin{vmatrix} 1 & 1 & 1 \\ x_{s_1} & x_{s_2} & x_{s_3} \\ y_{s_1} & y_{s_2} & y_{s_3} \\ \end{vmatrix} ~~ } $}}\\\\ \small{\text{$ \begin{array}{rcl} A &=&\dfrac{1}{2}\cdot \begin{vmatrix} 1 & 1 & 1 \\ 4 & -2 & -1 \\ -3 & 0 & 2 \\ \end{vmatrix} \\\\ &=&\frac{1}{2} \cdot [ 1\cdot (-2) \cdot 2 + 4\cdot 0 \cdot 1 + (-3)\cdot 1 \cdot (-1) - (-3)\cdot (-2) \cdot 1 - 0\cdot (-1) \cdot 1 - 4\cdot 1 \cdot 2 ]\\\\ &=&\frac{1}{2} \cdot [ 1\cdot (-2) \cdot 2 + (-3)\cdot 1 \cdot (-1) - (-3)\cdot (-2) \cdot 1 - 4\cdot 1 \cdot 2 ]\\\\ &=&\frac{1}{2} \cdot [ -4 + 3 - 6 \cdot 1 - 8]\\\\ &=&\frac{1}{2} \cdot [ -15 ]\\\\ &=& - 7,5 \\\\ |A| &=& 7,5 \qquad \mathrm{okay} \end{array} $}}$

Gast · Answer 2 · 2015-06-19T05:36:44

y ₍₁₎ = -x+1

y ₍₂₎ =-1/2*x-1

y ₍₃₎ = 2*x+4

Erstmal würde ich die Schnittpunkte berechnen:

$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\ {\mathtt{y}}={\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\ {\mathtt{x}} = {\frac{{\mathtt{4}}}{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}\\ {\mathtt{x}} = {\mathtt{1.333\: \!333\: \!333\: \!333\: \!333\: \!3}}\\ \end{array} \right\}$

$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\\ {\mathtt{y}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{{\mathtt{8}}}{{\mathtt{3}}}}\\ {\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{10}}}{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{2.666\: \!666\: \!666\: \!666\: \!666\: \!7}}\\ {\mathtt{x}} = -{\mathtt{3.333\: \!333\: \!333\: \!333\: \!333\: \!3}}\\ \end{array} \right\}$

$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\\ {\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{2}}\\ {\mathtt{x}} = -{\mathtt{1}}\\ \end{array} \right\}$

Lasst mich nachdenken...

Gast · Answer 3 · 2015-06-19T06:09:15

Nun zum Umfang:

P_(y1/y2)=(4/3;-1/3)

P_(y2/y3)=(-10/3;-8/3)

P_(y3/y1)=(-1/2)

${\mathtt{Strecke}} = {\sqrt{{\left[{\mathtt{x2}}{\mathtt{\,-\,}}{\mathtt{x1}}\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{y2}}{\mathtt{\,-\,}}{\mathtt{y1}}\right]}^{{\mathtt{2}}}}}$

Umfang = Strecke P_(y1/y2)P_(y2/y3)+ Strecke P_(y2/y3)P_(y3/y1)+ Strecke P_(y3/y1)P_(y1/y2)=

${\mathtt{Umfang}} = {\sqrt{{\left[\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right){\mathtt{\,-\,}}\left({\mathtt{2}}\right)\right]}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\left[{\mathtt{\,-\,}}\left({\mathtt{1}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[\left({\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{Umfang}} = {\mathtt{13.734\: \!815\: \!540\: \!536\: \!239\: \!6}}$

Schaut bitte mal drüber ob alles stimmt.

Gast · Answer 4 · 2015-06-19T06:48:39

Nun zur Fläche:

${\mathtt{Flaeche}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\sqrt{{\left[{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{c}}}^{{\mathtt{2}}}\right]}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left[{{\mathtt{a}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{c}}}^{{\mathtt{4}}}\right]}}$

a = Strecke P_(y1/y2)P_(y2/y3)= ${\sqrt{{\left[\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}}$

b = Strecke P_(y2/y3)P_(y3/y1)= ${\sqrt{{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right){\mathtt{\,-\,}}\left({\mathtt{2}}\right)\right]}^{{\mathtt{2}}}}}$

C = Strecke P_(y3/y1)P_(y1/y2)= ${\sqrt{{\left[{\mathtt{\,-\,}}\left({\mathtt{1}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[\left({\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}}$

Gast · Answer 5 · 2015-06-19T06:53:57

Ach die Formel will einfach nicht, sie sollte so aussehen:

Flaeche = 1/4*sqrt([(sqrt([(4/3)-(-10/3)]^2+[(-1/3)-(-8/3)]^2)^2)+(sqrt([(-10/3)-(-1)]^2+[(-8/3)-(2)]^2)^2)+(sqrt([(-1)-(4/3)]^2+[(2)-(-1/3)]^2)^2)]^2-2[(sqrt([(4/3)-(-10/3)]^2+[(-1/3)-(-8/3)]^2)^4)+(sqrt([(-10/3)-(-1)]^2+[(-8/3)-(2)]^2)^4)+(sqrt([(-1)-(4/3)]^2+[(2)-(-1/3)]^2)^4)])

Gast · Answer 6 · 2015-06-19T10:40:48

Danke Heureka für die Korrektur. (Habe ein Minuszeichen in der zweiten Gleichung übersehen.)

In der Klasse 10 ober 11 rechnet man übrigens so:

${\mathtt{F}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left(\left[{\mathtt{x1}}{\mathtt{\,\times\,}}\left({\mathtt{y2}}{\mathtt{\,-\,}}{\mathtt{y3}}\right)\right]{\mathtt{\,\small\textbf+\,}}\left[{\mathtt{x2}}{\mathtt{\,\times\,}}\left({\mathtt{y3}}{\mathtt{\,-\,}}{\mathtt{y1}}\right)\right]{\mathtt{\,\small\textbf+\,}}\left[{\mathtt{x3}}{\mathtt{\,\times\,}}\left({\mathtt{y1}}{\mathtt{\,-\,}}{\mathtt{y2}}\right)\right]\right)$

${\mathtt{Flaeche}} = {\left|{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left(\left[{\mathtt{4}}{\mathtt{\,\times\,}}\left({\mathtt{0}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right]{\mathtt{\,-\,}}\left[{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)\right]{\mathtt{\,-\,}}\left[{\mathtt{1}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0}}\right)\right]\right)\right|} \Rightarrow {\mathtt{Flaeche}} = {\mathtt{7.5}}$

heureka · Answer 7 · 2015-06-22T05:25:15

Die Fläche A eines Dreiecks mit Determinante berechnen:

Gegeben sind drei Punkte :

$\small{\text{$ S_1(x_1|y_1) = (4|-3) \qquad S_2 (x_2|y_2)=(-2|0) \qquad S_3(x_3|y_3)=(-1|2) $}}$

$\small{\text{$ \begin{array}{|l|l|l|} \hline & x & y \\ \hline S_1 & x_1=4 & y_1=-3 \\ S_2 & x_2=-2 & y_2= 0 \\ S_3 & x_3=-1 & y_3=2 \\ \hline \end{array} $}}\\\\$

$\small{\text{$ A = \dfrac{1}{2}\cdot \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} $ }}\\\\$

Auflösung nach der 1. Spalte:

$\small{\text{$ \begin{array}{lll} A &=& \dfrac{1}{2}\cdot \begin{vmatrix} \textcolor[rgb]{1,0,0}{x_1} & y_1 & 1 \\ \textcolor[rgb]{1,0,0}{x_2} & y_2 & 1 \\ \textcolor[rgb]{1,0,0}{x_3} & y_3 & 1 \\ \end{vmatrix} =\frac12\cdot \left( \textcolor[rgb]{1,0,0}{x_1}\cdot \begin{vmatrix} y_2 & 1 \\ y_3 & 1 \\ \end{vmatrix} -\textcolor[rgb]{1,0,0}{x_2}\cdot \begin{vmatrix} y_1 & 1 \\ y_3 & 1 \\ \end{vmatrix} +\textcolor[rgb]{1,0,0}{x_3}\cdot \begin{vmatrix} y_1 & 1 \\ y_2 & 1 \\ \end{vmatrix} \right)\\\\ &=& \frac12\cdot \left(~~ [~x_1\cdot (y_2-y_3)~]-[~x_2\cdot (y_1-y_3)~]+[~x_3\cdot (y_1-y_2)~] ~~\right)\\\\ &=& \mathbf{ \frac12\cdot \left(~~ [~x_1\cdot (y_2-y_3)~]+[~x_2\cdot (y_3-y_1)~]+[~x_3\cdot (y_1-y_2)~] ~~\right) } \\ \\ &=& \frac12\cdot \left(~~ [~4\cdot (0-2)~]-[~2\cdot (2-(-3))~]-[~1\cdot (-3-0)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ [~4\cdot (-2)~]-[~2\cdot (2+3))~]-[~1\cdot (-3)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ -8-10+3 ~~\right)\\\\ &=& -7,5\\\\ A&=& 7,5 \end{array} $ }}\\\\$

Auflösung nach der 2. Spalte:

$\small{\text{$ \begin{array}{lll} A &=& \dfrac{1}{2}\cdot \begin{vmatrix} x_1 & \textcolor[rgb]{1,0,0}{y_1} & 1 \\ x_2 & \textcolor[rgb]{1,0,0}{y_2} & 1 \\ x_3 & \textcolor[rgb]{1,0,0}{y_3} & 1 \\ \end{vmatrix} =\frac12\cdot \left( -\textcolor[rgb]{1,0,0}{y_1}\cdot \begin{vmatrix} x_2 & 1 \\ x_3 & 1 \\ \end{vmatrix} +\textcolor[rgb]{1,0,0}{y_2}\cdot \begin{vmatrix} x_1 & 1 \\ x_3 & 1 \\ \end{vmatrix} -\textcolor[rgb]{1,0,0}{y_3}\cdot \begin{vmatrix} x_1 & 1 \\ x_2 & 1 \\ \end{vmatrix} \right)\\\\ &=& \frac12\cdot \left(~~ -[~y_1\cdot (x_2-x_3)~] +[~y_2\cdot (x_1-x_3)~] -[~y_3\cdot (x_1-x_2)~] ~~\right)\\\\ &=& \mathbf{ \frac12\cdot \left(~~ [~y_1\cdot (x_3-x_2)~] +[~y_2\cdot (x_1-x_3)~] +[~y_3\cdot (x_2-x_1)~] ~~\right) }\\\\ &=& \frac12\cdot \left(~~ [~-3\cdot (-1-(-2))~] +[~0\cdot (4-(-1))~] +[~2\cdot (-2-4)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ [~-3\cdot 1~] +[~2\cdot (-6)~] ~~\right)\\\\ &=& \frac12\cdot \left(~~ -3 - 12 ~~\right)\\\\ &=& \frac12\cdot \left(~~ -15~~\right)\\\\ &=& -7,5\\\\ A&=& 7,5 \end{array} $ }}\\\\$

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