Question

Vă rog repede dau coroană

Answer 1

   

$\displaystyle\bf\\1)\\a)\\Unghiurile~~A,~B,~C,~D~~dp~~cu~~2,~4,~6,~8\\\textbf{Suma unghiurilor unui patrulater este egala cu }~~360^o\\2+4+6+8=20\\\\k=\frac{360}{20}=18\\A=2\times18=36^o\\B=4\times18=72^o\\C=6\times18=108^o\\D=8\times18=144^o\\Cel~mai~mic~unghi~este:~~\boxed{\bf~A=36^o}\\\\b)\\A=12~cm^2\\L=\sqrt{A}=\sqrt{12}=2\sqrt{3}~cm\\P=4\times L=4\times 2\sqrt{3}=\boxed{\bf8\sqrt{3}~cm}$

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$\displaystyle\bf\\c)\\P=2\times L+ 2\times l=2\times 6+ 2\times \Big(\frac{2}{3}\times 6\Big)=\\\\=2\times 6+ 2\times 4=12+8=\boxed{\bf20~cm}$

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$\displaystyle\bf\\d)\\\textbf{Daca un romb are un unghi de }~60^o,\\\textbf{atunci diagonala mica imparte rombul in 2 triunghiuri echilaterale.}\\<A = 60^o~~\implies~~BD~este~diagonala~mica.\\BD=12~cm\\\textbf{Ceallalta diagonala este dublul inaltimii triunghiului echilateral.}\\\\AC=2\times\frac{12\sqrt{3}}{2}=12\sqrt{3}~cm\\\\A=\frac{BD\times AC}{2}=\frac{12\times12\sqrt{3}}{2}=\boxed{\bf72\sqrt{3}~cm^2}$

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$\displaystyle\bf\\e)\\\textbf{Diagonalele paralelogramului impart paralelogramul }\\\textbf{in 4 triunghiuri echivalente.}\\\textbf{Triunghiurile echivalente sunt triunghiuri }\\\textbf{de aceeasi arie chiar daca sunt de forme diferite.}\\Aria_{\Delta BOC}=Aria_{\Delta COD}=Aria_{\Delta DOA}=Aria_{\Delta AOB}=12~cm^2\\\\\implies~Aria~par. ABCD=4\times~Aria_{\Delta AOB}=4\times12~cm^2=\boxed{\bf48~cm^2}$

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$\displaystyle\bf\\f)\\h_{trapez}=8~cm\\linia~mijlocie~(lm)=12~cm\\\\lm=\frac{AB+CD}{2}~unde~AB~||~CD\\\\Aria_{tr.~ABCD}=h\times\frac{AB+CD}{2}=h\times lm=8\times12=\boxed{\bf96~cm^2}$