2 szkoły ponadpodstawowej

II liceum

II technikum

MATeMAtyka 2. Zakres podstawowy. Po gimnazjum

a) O=4a 2πr=4a r=2π4a​

Rozwiązanie 1

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31724/fy4InQYTcczLg9HoDg/xRw%3D%3D_a6dca159dd5d3ea79c93d759f9b5e43fc6679419f42eba698efd7b146a92758f.png" width="315" height="318"> a=1 r=21​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
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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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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l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ r=21​⋅1⋅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
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>​=21​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
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>​

Musimy obliczyć pole kwadratu i odjąć od niego pole czterech okręgów. P1​=162=256​

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31727/hcRxUb19hge12BE4uUNC7Q%3D%3D_4a4a5a6e5f0fd2c20209cbfd4b4eab64f4c283d07e86715c25b1763a9fb83550.png" width="509" height="455"> Obliczamy pole odcinka koła (żółta część): PI​=P1​−P2​ Kąt środkowy wyznaczony przez tę cieciwę ma miarę 90o( trójkąt o bokach a, a, a√2 to trójkąt prostokątny o kątach 90o,45o,45o) P1​=360o90o​π42=41​⋅16π=4π

a)  P=360o90o​⋅π122=41​⋅144π=36π​​

Dorysowujemy sobie kąt wpisany ACB oparty na tym samym łuku co kąt środkowy AOB. Kąt wpisany oparty na tym samym łuku co kąt środkowy ma miarę dwa razy od niego mniejszą.     <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31738/yTWAXIdJnchHtlDyPLXx9Q%3D%3D_7157acdc85647b5b8e13881c7cb9295ec6a0c75b4553196ad35e648cca6927b0.png"> Dla trójkąta ABC Układamy zależność trygonometryczną: sin(21​α)=1053<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
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​

a) <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31735/0GsUJDWENSjJH3baHMIxzA%3D%3D_ad818a2b61bf829a09bd66bb86d66bda4bdb3158765716888dff5873d0290baa.png"> Zauważamy, że trójkąt OBA wyznaczony przez cięciwę i promienie jest równoboczny, zatem: ∣∠AOB∣=60o P=360o60o​⋅π122=61​⋅144π=24π​​