Question

Urgent!!!!!Exercitiul 6!Multumesc anticipat!

Answer 1

$a)~ A_{necolorata} = \frac{90 \textdegree}{360 \textdegree}* \pi R^{2}= \frac{1}{4}* \pi *100=25 \pi ~(cm^{2}). \\ A_{colorata}= A_{patrat} - A_{necolorata}=100-25 \pi ~(cm^2). \\ \\ b)~Triunghiul~este~echilateral~(isoscel~cu~un~unghi~de~60 \textdegree). \\ A_{\Delta}= \frac{ l^{2} \sqrt{3} }{4} = \frac{144 \sqrt{3} }{4}=36 \sqrt{3}~(cm^2). \\ A_{sector} = \frac{60 \textdegree}{360 \textdegree} * \pi R ^{2} = \frac{1}{6}* \pi *144= 24 \pi ~(cm^2).$
$A_{s.c.} = A_{sector}- A_{\Delta} = 24 \pi -36 \sqrt{3}~ (cm^2). \\ \\ s.c.=segment~circular$