[latex]Rysunki pogladowe w zalacznikach.\\7.\x=8-4=4\\Z tw. Pitagorasa:\\h^2+4^2=(4+8)^2\\h^2+16=12^2\\h^2=144-16\\h^2=128\\h=sqrt{128}\\h=sqrt{64cdot2}\\h=8sqrt2[/latex] [latex]Figura sklada sie z dwoch trapezow prostokatnych:\\P_f=2cdotfrac{8+4}{2}cdot8sqrt2=12cdot8sqrt2=96sqrt2[/latex] [latex]6.\|EF|=frac{60sqrt3}{2}=30sqrt3 (cm)\\P_{Delta EAB}=frac{60^2sqrt3}{4}=frac{3600sqrt3}{4}=900sqrt3 (cm^2)\\h=|EF|-30; h=30sqrt3-30=30(sqrt3-1) (cm)\\Delta EABsimDelta ENM-trojkaty podobne\\skala podobienstwa:k=frac{30(sqrt3-1)}{30sqrt3}=frac{sqrt3-1}{sqrt3}\\stosunek pol figur podobnych jest rowny kwadratowi skali\podobienstwa, zatem:[/latex] [latex]P_{Delta ENM}=900sqrt3cdotleft(frac{sqrt3-1}{sqrt3} ight)^2=900sqrt3cdotfrac{3-2sqrt3+1}{3}=300sqrt3cdot(4-2sqrt3)\\=1200sqrt3-1800=600(2sqrt3-3) (cm^2)[/latex] [latex]9.\a=17cm\e;f-przekatne\\e-f=14cm o e=f+14\\Bok rombu i polowy przekatnych tworza trojkat prostokatny.\Na mocy tw. Pitagorasa:\\a^2=(frac{e}{2})^2+(frac{f}{2})^2\\left(frac{f+14}{2} ight)^2+left(frac{f}{2} ight)^2=17^2\\frac{f^2+28f+196}{4}+frac{f^2}{4}=289 /cdot4[/latex] [latex]f^2+28f+196+f^2=1156\\2f^2+28f+196-1156=0\\2f^2+28f-960=0 /:2\\f^2+14f-480=0\\Delta=14^2-4cdot1cdot(-480)=196+1920=2116; sqrtDelta=sqrt{2116}=46\\f_1=frac{-14-46}{2cdot1} < 0; f_2=frac{-14+46}{2cdot1}=16 (cm)\\e=14+16=30 (cm)\\P=frac{ef}{2}; P=frac{16cdot30}{2}=240 (cm^2)[/latex] [latex]10.\tgalpha=frac{3}{4}; tgalpha=frac{h}{x}\\frac{h}{x}=frac{3}{4} o h=frac{3}{4}x\\Z tw. Pitagorasa:\\x^2+h^2=10^2\\x^2+left(frac{3x}{4} ight)^2=100\\x^2+frac{9x^2}{16}=100 /cdot16\\16x^2+9x^2=1600\\25x^2=1600 /:25\\x^2=64\\x=sqrt{64}\\x=8 (cm)[/latex] [latex]Obw=40cm; Obw=2cdot10+2x+2b\\20+2cdot8+2b=40\\2b=40-20-16\\2b=4 /:2\\b=2 (cm)\\a=b+2x\\a=2+2cdot8=2+16=18 (cm)[/latex] [latex]11.\d_1-sa to dwie dlugosci wysokosci trojkata rownobocznego\\d_1=2cdotfrac{asqrt3}{2}=asqrt3\\d_2=2a\\Suma przekatnych:\\S=_2+2d_1=2a+2asqrt3=2a+asqrt{2^2cdot3}=2a+asqrt{12}\\Obwod:\Obw=6a=2a+4a=2a+asqrt{4^2}=2a+asqrt{16}\\Obw=2a+asqrt{16} > S=2a+asqrt{12}[/latex]
