Question

1. Pe cercul C(O,r), se considera, in aceasta ordine, punctele A,B,C,D. Notam cu M,N,P,Q mijloacele arcelor AB,BC,CD,DA. Aratati ca MP _|_ NQ.

2.Coarda [AB] subintinde un arc de 120 grade in cercul C(O,4cM). Paralela prin A la OB intersecteaza cercul a doua oara in C. Aflati aria patrulaterului AOBC.

Answer 1

Notam intersectia dintre MP si NQ cu T.
Se stie ca m(QTM)=m(PTM)=[m(MQ)+m(PN)]/2=(1/2m(AB)+1/2m(AD)+1/2m(BC)+1/2(CD))/2=
1/2(m(AB)+m(BC)+m(CD)+m(DA))/2=360/4=90
Obs: Trebuie facuta o figura ca sa fie clar.
 2)Se demonstreaza ca AOBC este romb avnd latura de 4 cm egala cu raza si unghiul ascutit de 60 grade.Asadar aria lui va fi 
2*4²√3/2=16√2

Answer 2

$1)~Notam~cu~S~intersectia~dreptelor~MP~si~QN. \\ \\ m( \angle MOQ)= \frac{m( \stackrel{\frown}{MQ} )+ m(\stackrel{\frown}{PN})}{2}= \frac{ m(\stackrel{\frown}{AM})+m( \stackrel{\frown}{AQ})+m( \stackrel{\frown}{CN})+m(\stackrel{\frown}{CP}) }{2}= \\ \\ = \frac{ \frac{m(\stackrel{\frown}{AB)}}{2}+ \frac{m(\stackrel{\frown}{BC})}{2}+ \frac{m(\stackrel{\frown}{CD})}{2} + \frac{m(\stackrel{\frown}{AD})}{2} }{2}= \frac{360 \textdegree}{4}=90 \textdegree \Rightarrow ~ MP \perp QN .$

$2)~m(\stackrel{\frown}{AB})=120 \textdegree \Rightarrow m( \angle AOB)=120 \textdegree. \\ \\ AO=BO~(raze) \Rightarrow \Delta AOB-isoscel~de~baza~[AB]. \\ \\ m( \angle OAB)=m(\angle OBA)= \frac{180 \textdegree - m( \angle AOB)}{2}= \frac{180 \textdegree - 120 \textdegree}{2}=30 \textdegree. \\ \\ AC ~||~BO \Rightarrow m( \angle CAB)=m( \angle ABO) =30 \textdegree~(alterne~interne).$

$m(\angle ACB)= \frac{ 360 \textdegree - m(\stackrel{\frown}{AB})}{2}= \frac{360 \textdegree -120 \textdegree}{2}= 120 \textdegree. \\ \\ m( \angle ACB) = m( \angle AOB) si~AO~||~BC \Rightarrow AOBC~romb~ cu \\ \\ m( \angle CAO)= 2 \cdot m( \angle OAB)=60 \textdegree. \\ \\ Deci~ \Delta~AOC~si~ \Delta BOC-echilaterale~cu~latura~de~4~cm. \\ \\ A_{ABCD}=2 \cdot A_{AOC}=2 \cdot \frac{4^2 \sqrt{3}}{4}=8 \sqrt{3}~(cm^2).$