[tex]a)\\ \\ tr.ABD, m(A)=90, T.P. : BD^{2}=AD^{2}+AB^{2}=\\(4*6)^{2}+(4*8)^{2}=\\16*36+16*64=\\16(36+64)=16*100\\
BD= \sqrt{16*100} =4*10=40 (cm\\
b)\\ \\
tg(ACB)= \frac{AB} {BC} =\frac{32} {24}=\frac{8} {6}= \frac{4} {3}\\c)\\ \\tr. ABD, m(A)=90, m(E)=90, T.Inaltimii: AE=\frac{AB*AD} {BD}=\frac{32*24} {40} \\
\frac{96} {5} = 19,2 (cm)
\\DB int. AC= {O}\\
DO=AO=OC=\frac{AC} {2}=20 (cm)\\
\\cos DAB= AD/BD=24/40=6/10=3/5\\
cos DAB= DE/AD\\
DE/AD=3/5 DE/24=3/5\\DE=14,5 (cm)\\
DE=FB=BD-EF
[/tex][tex]EF=BD-DE-FB=40-14,5-14,5=40-29=11\\
EO=OF=EF/2=11/2 (cm)[/tex][tex]A tr. AEF=\frac{AE*EF}{2}=\frac{19,2*11}{2}=10,56 (cm^{2})\\
tr. AEF= tr. EFC (criteriul L.U.L) \\
A. AECF=2*A.AEF=2*10,56=21,12\\
tr. AEC= tr.AFC (AECF paralelogram)\\A. AECF=2*A.AEC\\
A.AEC=\frac{A.AECF}{2}=10,56 (cm^{2}[/tex]
