Question

Pe cercul de centru o si rade de 12 cm se considera punctele A,B,C in aceasta ordine astfel incat AB=12 radical din 2 cm si AC=12 radical din 3 cm. Determinati aria sectorului circular determinat de arcul BC.

Answer 1

$AO^{2}+BO^{2}=12 ^{2} +12^{2}=2*12^{2}=244~(cm^2). \\ AB^2=(12 \sqrt{2} )^{2}=244~(cm^2) \\ \\ Deci~AO^2+BO^2=AB^2 \Rightarrow m(AOB)=90grade~(T.R.Pitagora).$

$In~triunghiul~\Delta AOC,~isoscel~de~baza~[AC],~luam~M-mijlocul \\ segmentului~[AC] \Rightarrow OM \perp AC. \\ AM=MC= \frac{AC}{2}= 6\sqrt{3} ~(cm). \\ \\ T.Pitagora~in~\Delta AOM:~OM ^{2}+AM ^{2} = AO^{2} \Rightarrow \\ \RightarrowOM= \sqrt{AO^2-AM^{2}} = \sqrt{144-108}= \sqrt{36}=6~(cm). \\ \\ sin(MAO)= \frac{MO}{AO}= \frac{6}{12}= \frac{1}{2} \Rightarrow m(MAO)=30 \textdegree. \\ m(AOC)=180 \textdegree -2*30 \textdegree =120 \textdegree.$

$m(\ \textless \ BOC)=m(AOC)-m(AOB)=120 \textdegree - 90 \textdegree=30 \texdegree. \\ \\ A_{sectorBOC} = \pi R^{2}*\frac{ m(BOC)}{360 \textdegree}= \pi *144* \frac{30 \textdegree}{360 \textdegree}= 12 \pi ~(cm^2).$