II gimnazjum

Matematyka z plusem 2

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/24646/CHx4xi5uIMZ0u/wHEQ2bVg%3D%3D_346b870cbbfff0e536a3e8feaf63d9ee378e735821db049310d2f1e885f380a2.png"> P1​=180=(3a)2   9a2=180   a2=20   P2​=a2=20   Z twierdzenia Pitagorasa: (3a)2=a2+b2   180=20+b2   b2=160   P3​=b2=160 <section class="answer"><h4 class="answer-heading">Odpowiedź:</h4>Pola pozostałych kwadratów są równe 20 i 160.</section>

Rozwiązanie 1

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31488/RfyU%2BE8%2B4r8mSHxjtOgsZw%3D%3D_86638dd06b43af2d0126a7c999aed670154795ddaf8d3cd19c56211ff76d9a7e.jpg">
Mamy trzy kwadraty o bokach a, b, c.

h - wysokość budynku h2+92=152

a=6 b - druga przyprostokątna c - przeciwprostokątna   Pierwszy trójkąt: b=2⋅6=12   c2=62+122

Pierwszy trójkąt:  <img style="vertical-align: middle;" src="../../../math/formula?%5Cdisplaystyle%7Ba%7D%3D%7B10%7D%5C%20%7Bc%7D%7Bm%7D"> <img style="vertical-align: middle;" src="../../../math/formula?%5Cdisplaystyle%7Bb%7D%3D%7B20%7D%5C%20%7Bc%7D%7Bm%7D"> c = ?  c2=a2+b2   c2=102+202   c2=100+400

Żeby zaznaczyć te punkty w układzie współrzędnych należy skonstruować odcinek o długości √2 i √3. √2 - to przeciwprostokątna trójkąta prostokątnego o przyprostokątnych 1 i 1 √3 - to przyprostokątna trójkąta prostokątnego o przyprostokątnej 1 i przeciwprostokątnej 2. A=(2,3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.28em' viewBox='0 0 400000 1296' preserveAspectRatio='xMinYMin slice'><path d='M263,681c0.7,0,18,39.7,52,119
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c340,-704.7,510.7,-1060.3,512,-1067
l0 -0
c4.7,-7.3,11,-11,19,-11
H40000v40H1012.3
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c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1
s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26
c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z
M1001 80h400000v40h-400000z'/></svg>​)

a) a=33<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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c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​   b=22<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
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​   c=32<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
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​   a2=(33<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
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​)2=27

Sprawdzamy czy przekątne przecinają sie pod kątem prostym, czyli czy trójkąt o bokach 5, 12, 13 jest prostokątny.

a) 22+32=4+9=13&lt;16=42

A: Sprawdzamy czy trójkąt o bokach 13, 12 i 17-12=5 jest prostokątny 122+52=144+25=169=132

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