III gimnazjum

Matematyka wokół nas 3

Boki trójkąta ABC: 4,8,10 Boki trójkąta A1​B1​C1​:x,10,y

Rozwiązanie 1

k=ObABC​ObA1​B1​C1​​​=8+12+168+12+16−6​=3630​=65​

W trójkacie K'L'M' przyprostokatne mają długość: 24 cm i 7 cm. Obliczmy długość przeciwprostokątnej (korzystamy z tw. Pitagorasa): a2=242+72   a2=576+49

∣AB∣2=(6,5)2−(2,5)2 ∣AB∣=42,25−6,25<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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M834 80h400000v40h-400000z'/></svg>​=6 k=2,53​=1,2

∣BE∣=102−62<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>​ ∣BE∣=100−36<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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ ∣BE∣=64<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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<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25096/xOAYjAvXyCfaU%2Bg/zWfLcA%3D%3D_bce06b3d940dad00fa97a4d29fa8aaf13cc74316998cdaf669a00b7362f76b2a.jpg" width="640" height="276"> ∣∠DOC∣=∣∠AOB∣

Wiemy, że różnica długości przyprostokątnych drugiego trójkąta jest równa 4 cm oraz że skala podobieństwa to 1/4. Zatem możemy obliczyć, ile wynosi różnica długości przyprostokątnych pierwszego trójkąta:  41​⋅4 cm=1 cm     Teraz o długościach przyprostokątnych pierwszego trójkąta wiemy już, że ich suma wynosi 3 cm, a różnica wynosi 1 cm. Zatem te przyprostokątne muszą mieć długości 1 cm i 2 cm. Teraz możemy obliczyć długości przyprostokątnych drugiego trójkąta (są one 4 razy dłuższe niż w pierwszym trójkącie) 4⋅1 cm=4 cm

Rysunek: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/47557/sZobdRwr271gHum1/f5syg%3D%3D_466d9f9e5141a09dd26bcfb153e0a191d581222d95947c6061306027fdfa58f4.jpg" width="452" height="250"> Trójkąt ABC: ∣AB∣=2c   ∣BC∣=2a   ∣AC∣=2b

Rysunek pomocniczy: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/47553/sFuCIz/F/u84Xn1NNv%2B/7A%3D%3D_ce395e2b8a7151da1e7cdd6174f6ae2a5bf9ed708de4a4dd13e5d6c45920ef98.jpg" width="295" height="238"> Wiemy, że: ∣AB∣=16 cm   ∣AC∣=20 cm   ∣BC∣=12 cm