Răspuns: [tex]\bf A_{ABCD}=400~cm^{2}[/tex]
Explicație pas cu pas:
Salutare!
[tex]\bf Ducem~ d(D;AB)=M[/tex]
[tex]\bf d(C;AB)=N[/tex]
[tex]\bf In~\triangle{AMD}-dreptunghic~ avem:[/tex]
[tex]\bf \angle{AMD}=90^{\circ}[/tex]
[tex]\bf \angle{DAM}=45^{\circ}[/tex]
[tex]\bf \angle{MDA}=45^{\circ}[/tex]
[tex]\bf \implies \triangle{AMD}~este ~dreptunghic ~isoscel[/tex]
[tex]\bf \implies AM=DM=x[/tex]
[tex]\bf In~ \triangle{AMD}~aplicam~teorema~lui ~Pitagora[/tex]
[tex]\bf{AD^{2}=AM^{2}+DM^{2}[/tex]
[tex]\bf x^{2}+x^{2}=20^{2}[/tex]
[tex]\bf 2\cdot x^2=400~~~\bigg|:2[/tex]
[tex]\bf x^2=200\implies x = 10\sqrt{2}~cm[/tex]
[tex]\bf AM = DM = 10\sqrt{2}~cm[/tex]
[tex]\bf Analog~ si~ in~ cazul ~\triangle{BNC}-dreptunghic~[/tex]
[tex]\bf \implies T~diag.~ca~\angle{DCM}=\angle{NCM}=\dfrac{90}{2} =45^{\circ}[/tex]
[tex]\bf DN \cap CM=\{O\}[/tex]
[tex]\bf In~\triangle{DOC}~ avem~doua~unghiuri~de~45^{\circ}\implies \angle{O}=90^{\circ}[/tex]
[tex]\bf d_1 \bot d_2 \implies MDCN-patrat[/tex]
[tex]\bf AM=MD=DC=MN=CN=NB=10\sqrt{2}~cm[/tex]
[tex]\bf AB=AM+MN+NB[/tex]
[tex]\bf AB=3 \cdot 10\sqrt{2}[/tex]
[tex]\bf AB=30\sqrt{2}[/tex]
[tex]\bf A_{ABCD}=\dfrac{(Baza~mare+baza~mica)\cdot Inaltimea}{2}[/tex]
[tex]\bf A_{ABCD} =\dfrac{(AB+CD)\cdot DM}{2}[/tex]
[tex]\bf A_{ABCD}=\dfrac{(30\sqrt{2}+10\sqrt{2})\cdot 10\sqrt{2} }{2}[/tex]
[tex]\bf A_{ABCD}=\dfrac{(30\sqrt{2}+10\sqrt{2})\cdot \not10\sqrt{2} }{\not2}[/tex]
[tex]\bf A_{ABCD}=(30\sqrt{2}+10\sqrt{2})\cdot 5\sqrt{2}[/tex]
[tex]\bf A_{ABCD}=40\sqrt{2}\cdot 5\sqrt{2}[/tex]
[tex]\bf A_{ABCD}=200\cdot 52[/tex]
[tex]\boxed{\bf A_{ABCD}=400~cm^{2}}[/tex]
#copaceibrainly
