II gimnazjum

Matematyka z plusem 2

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Rozwiązanie 1

Obw=60cm p=10 cm a - bok rombu q - druga przekątna 4a=60cm   ∣:4

h - wysokość trapezu h2+(224−6​)2=152

a) a=7 b=4 c=2 d2=42+52   d2=16+25   d2=41

a - długość przyprostokątnej  a2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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32=h2+12   9=h2+1

132=H2+52   169=H2+25   H2=144

a) a2=32+62   a2=9+36   a2=45

∣AB∣2=(7−(−2))2+(−7−(−9))2   ∣AB∣2=92+22   ∣AB∣2=81+4

|AB| = a  |AC| = b  |BC| = c    a2=(4−(−3))2+(1−(−2))2   a2=72+32   a2=49+9

∣AB∣2=25+4=29

a - bok kwadratu  d - przekątna kwadratu 4a=12 cm a=3 cm

a - bok kwadratu  a2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​=3 cm   a=2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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a - bok kwadratu  a to wysokość trójkąta ABC o boku 2 a=223<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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