Question

bună. am nevoie urgent de ajutor vă rog frumos.......​

Answer 1

Răspuns:

2√2., al

la AB

si 2, la CD

Explicație pas cu pas:

AB??CD, ABNCD trapez isoscel; deci arc AD ptop tot cu 5

8k+5k+6k+5k=360

24k ...360 grade

k=15 grade

arc AB=6*15=90 grade tr AOB dreptunghic isoscel cateta 4cm...deci distanta pana la AB=ipotenuz\a/2=2rad2

mas arc DC=8*15=120 grade

deci mas unghi ODC=30 grade dedi dist =Raz/2=4/2=2cm

probleam GREA!!

Answer 2

$\it AB||CD \Rightarrow m(\stackrel\frown{DA})=m(\stackrel\frown{BC})\\ \\ \{m(\stackrel\frown{AB}),\ \ m(\stackrel\frown{BC}),\ \ m(\stackrel\frown{CD})\}\ d.\ p.\ \{6,\ \ 5,\ \ 8\} \Rightarrow\begin{cases}\ \it m(\stackrel\frown{AB})=6k\\ \\ \it m(\stackrel\frown{BC})=5k\\ \\ \it m(\stackrel\frown{CD})=8k\end{cases}\\ \\ \\ m(\stackrel\frown{DA})=m(\stackrel\frown{BC})=5k$

$\it 6k+5k+8k+5k=360^o \Rightarrow 24k=360^o|_{:24} \Rightarrow k=15^o \Rightarrow \\ \\ \Rightarrow m(\stackrel\frown{AB})=6\cdot15^o=90^o;\ \ \ m(\stackrel\frown{CD})=8\cdot15^o=120^o$

$\it m(\widehat{AOB})=m(\stackrel\frown{AB})=90^o \Rightarrow \Delta AOB-dreptunghic\ isoscel, OA=OB=4\ cm,\\ \\ BC=4\sqrt2\ cm;\ \ \ d(O,\ \ AB)=\dfrac{OA\cdot OB}{AB}=\dfrac{4\cdot4}{4\sqrt2}=\dfrac{4}{\sqrt2}=\dfrac{4\sqrt2}{2}=2\sqrt2\ cm$

$\it m(\widehat{COD})=m(\stackrel\frown{CD})=120^o \Rightarrow m(\widehat{OCD})=m(\widehat{ODC})=30^o\\ \\ Th.\ \angle30^o \Rightarrow d(O,\ \ CD)=OC:2=4:2=2\ cm$