Question

Patrulaterul ABCD are diagonalele [AC] si [BD] perpendiculare. Dacă AB=5, BC=4√5,
CD=10, AC=11, sa se determine perimetrul patrulaterului si lungimea diagonalei BD.

Answer 1

 

$\displaystyle\bf\\\textbf{Se da patrulaterul ABCD cu diaagonalele }~AC\perp BD.\\AB=5\\BC=4\sqrt{5}\\CD=10\\AC=11\\Vezi~desenul~atasat.\\Se!cere:\\a)~Perimetrul~patrulaterului\\b)~lungimea~diagonalei~BD$

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$\displaystyle\bf\\Rezolvare:\\\\AC\perp BD~~si~~AC~\bigcap~BD=\{O\}\\\\\textbf{Se formeaza triunghiurile dreptunghice: }\\\\\Delta AOB,~\Delta BOC,~\Delta COD~si~\Delta DOA\\\\\textbf{Calculam catetele: }~OA,~OB,~OC,~OD\\\\ \textbf{Cu teorema lui Pitagora vom scrie un sistem de ecuatii.}\\\\OA^2 + OB^2 = 25\\OB^2 + OC^2 = 80\\\\\textbf{Descompunem numerele 25 si 80 in suma de 2 patrate perfecte.}\\\\25 = 9 + 16\\80 = 16 + 64$

.

$\displaystyle\bf\\Rezulta~ca:\\OA^2=9\\OB^2=16\\OC^2=64\\\\OA=\sqrt{9}=3\\OB=\sqrt{16}=4\\OC=\sqrt{64}=8\\\\Verificare:Stim~ca~AC=11~~(din~enunt)\\AC=OA+OC=3+8=11`~~Corect\\\\CD = 10~~si~~OC = 8\\\\OD=\sqrt{CD^2 - OC^2}=\sqrt{10^2 - 8^2}=\sqrt{100 - 64}=\sqrt{36}=6~cm\\\\OA=3~~si~~OD=6\\\\AD=\sqrt{OA^2 + OD^2}=\sqrt{3^2 + 6^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}~cm\\\\a)~~P=AB+BC+CD+AD=\\~\boxed{\bf~P=5+4\sqrt{5}+10+3\sqrt{5}=15+7\sqrt{5}~cm}\\\\\\b)~~\boxed{\bf~BD = OB + OD = 4 + 6 = 10~cm}$