Question

Fie dreptunghiul ABCD cu AB=32 cm, BC=24 cm. Calculati:
a) lungimea diagonalei BD.
b) tg unghiului ACB
c) aria triunghiului AEC stiind ca AE perpendicular pe DB. (AE_|_DB)
Dau coroana

Answer 1

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