Question

va rog repede punctul a doar! ​

Answer 1

   

$\displaystyle\bf\\Se~da:\\\\\frac{CD}{AB}=\frac{2}{5}~~~~~\implies~CD=2k~cm~~si~~AB=5k~cm\\\\Din~desen~se~vede:\\AC\perp BD\\\\h=DD'=DD'=21~cm\\Se~cere:AB=?~~si~~CD=?\\\\Rezolvare:\\\\Un~trapez~ortodiagonal~are~linia~mijlocie~egala~cu~inaltimea.\\\\L_m=h\\h=21~cm\\\\L_m=\frac{AB+CD}{2}=\frac{5k+2k}{2}=\frac{7k}{2}\\\\\frac{7k}{2}=21\\\\7k=21\times2\\\\7k=42\\\\k=\frac{42}{7}\\\\k=6\\\\ AB=5k=5\times6\\\boxed{\bf AB=30}\\\\CD=2k=2\times6\\\boxed{\bf CD=12}$

 

 

Answer 2

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$\it \dfrac{CD}{AB}=\dfrac{2}{5}=0,4 \Rightarrow CD=0,4\cdot AB\ \ \ \ \ (*)\\ \\ \\ ABCD\ -\ trapez\ isoscel\ ortodiagonal \Rightarrow \ell_m=h \Rightarrow \dfrac{AB+CD}{2}=21|_{\cdot2}\Rightarrow\\ \\ \\ \Rightarrow AB+CD=42\ \stackrel{(*)}{\Longrightarrow}\ AB+0,4\cdot AB=42|_{\cdot10} \Rightarrow 10AB+4AB=420 \Rightarrow \\ \\ \\ \Rightarrow 14AB=420|_{:14} \Rightarrow AB=30\ cm$