Question

A medida do raio da circunferência que contém os vértices A, B e C;
A(4,6) B(-2,-2) C(5,-1)

Answer 1

$\mathrm{Os\ pontos\ A=(4,6),\ B=(-2,-2)\ e\ C=(5,-1)}\ \textrm{determi-}\\ \mathrm{nam\ um\ tri\hat{a}ngulo\ de\ lados\ \mathbf{a, b}\ e\ \mathbf{c}\ e\ c\ \acute{a}rea\ \mathbf{S},\ tal}\ \textrm{que:}\\\\ \mathrm{a=d_{BC}=\sqrt{(-1+2)^2+(5+2)^2}=\sqrt{50}=5\sqrt{2}\ u.c.}\\\\ \mathrm{b=d_{AC}=\sqrt{(-1-6)^2+(5-4)^2}=\sqrt{50}=5\sqrt{2}\ u.c.}\\\\ \mathrm{c=d_{AB}=\sqrt{(-2-6)^2+(-2-4)^2}=\sqrt{100}=10\ u.c.}$
$\mathrm{Perceba\ que\ \mathbf{c^2=a^2+b^2},\ e\ o\ tri\hat{a}ngulo\ \acute{e}\ ret\hat{a}ngulo.}\ \textrm{En-}\\ \mathrm{t\~ao,\ a\ \acute{a}rea\ \mathbf{S}\ ser\acute{a}}:\\\\ \mathrm{S=\dfrac{ab}{2}=\dfrac{(5\sqrt{2})(5\sqrt{2})}{2}=\dfrac{50}{2}=25\ u.a.}$
$\mathrm{A\ \acute{a}rea\ \mathbf{S}\ tamb\acute{e}m\ pode\ ser\ escrita\ em\ fun\c{c}\~ao\ dos\ lados\ e}\\ \mathrm{do\ raio\ \mathbf{R}\ da\ circunfer\hat{e}ncia\ circunscrita:}\\\\ \mathrm{S=\dfrac{abc}{4R}\ \to\ 25=\dfrac{(5\sqrt{2})(5\sqrt{2})(10)}{4R}\ \to\ R=\dfrac{(50)(10)}{4(25)}\ \to}\\\\ \mathrm{\to\ \boxed{\mathbf{R=5\ u.c.}}}$