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MATeMAtyka 2. Zakres podstawowy. Po gimnazjum

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
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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
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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
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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
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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​​

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MATeMAtyka 2. Zakres podstawowy. Po gimnazjum