NMMRUS 90 Loesung

Wie breit muss der Streifen sein?

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Also, das Feld hat die Länge L und die Breite B - die gesuchte Breite des Streifens sei x. Die ungemähte Fläche (die in der Mitte über bleibt) ist L-2x lang und B-2x breit. Die Fläche in der Mitte soll genau halb so groß sein, wie die Fläche des gesamten Feldes.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {{L B} \over 2} = (L - 2x)\cdot (B - 2x)}

Bevor wir weiterrechnen - eine kleine (sicher einsichtige) Bedingung:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B > 2x}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L > 2x}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {{L B} \over 2} = L B -2x\cdot (L + B) + 4x^2}

Weil ich mir die "große" Formel für die quadratische Gleichung nicht merken kann dividiere ich jetzt durch 4.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 - {{L + B}\over 2}\cdot x + {{L B \over 8}} = 0}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1,2} = {{L + B}\over 4} \pm \sqrt{({{L + B}\over 4})^2 - {{L B \over 8}}}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1,2} = {{L + B}\over 4} \pm \sqrt{{L^2+2 L B+B^2 -2 L B}\over 16}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1,2} = {{L + B}\over 4} \pm {\sqrt{L^2+B^2} \over 4}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1,2} = {{L + B \pm \sqrt{L^2+B^2}} \over 4}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D = \sqrt{L^2+B^2}} ist die Diagonale des Rechtecks. Aber müssen wir sie abziehen - oder dazuzählen - oder ist es gar egal - also beides richtig? Aus der Bedingung B > 2x folgt, dass wir die Diagonale abziehen müssen, weil (ohne Beschränkung der Allgemeinheit Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L \ge B} - die Länge hieße nicht Länge, wenn sie nicht länger wäre):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x = {{L+B+D}\over 2} < B}

Da Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L \ge B} und Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D \ge B} ist, ist

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {{L+B+D}\over 2} \ge {{B+B+B}\over 2}}

Und zusammen

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {{B+B+B}\over 2} < B}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {3\over 2}\cdot B < B}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {3\over 2} < 1}

Ups - das kann nicht sein => die Diagonale muss abgezogen werden.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = {{L+B-D}\over 4}}