III gimnazjum

Matematyka z plusem 3

a) Obliczamy długość krótszego boku prostokąta:  42+x2=52

Rozwiązanie 1

Dla przypomnienia - długości boków w trójkątach prostokątnych o kątach 90°, 60° 30° oraz 90°, 45°, 45°

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28951/AsJ8VaWfD1Emrt7anTEt%2BQ%3D%3D_8c38ffe20272880ccdab7f46753dd1b7d039fce4845b40f2f48b2fcd9aec60fa.jpg"> 32+h2=52 9+h2=25

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28952/vG8CTKV1F3FU/5b7PyhvVw%3D%3D_70e3f40bff34507be91228fa736313826903da44d6e885d1968f8ef0a12f51f3.jpg" width="388" height="197"> W rombie przekątne dzielą się na połowy pod katem prostym.  x2+52=72 x2+25=49

h  - długość drugiej wysokości P=6 cm⋅4 cm=24 cm2

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28954/Iu9CX1qtvqJ3bdypH%2BqL1Q%3D%3D_24b94f4a942b8f056cba01509998f1414366d02d681361646249c64597bac34c.jpg">   a) 12+32=∣AB∣2             1+9=∣AB∣2             ∣AB∣2=10             ∣AB∣=10<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>​   62+22=∣BC∣2   36+4=∣BC∣2

Plac wygląda tak:  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28958/8/o1nC0WbQoJ49MPOzkMgw%3D%3D_2fe5ebe92dabb2138adb958baa444195a0f1eb3862c692da52fc3a11810748c7.jpg" width="455" height="231">   Dla przypomnienia, długości boków w trójkątach prostokątnych o kątach 90°, 60° 30° oraz 90° 45°, 45° <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/23434/ITQ753RZp6YkJ5kQnuel4A%3D%3D_4f4b1642c86f5f780a039da5c1c51f977f4f4be540c875f0471f70f002d78a7f.jpg" width="768" height="221">   10=2b   b=10:2=5   b3<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>​=53<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>​