Odpowiedź:
I AB I ² = ( - 2 -6)² + ( 3 + 1)² = 80 = 16*5
I AB I = 4√5
===========
I BC I² = ( - 3 +2)² + ( - 4 - 3 )² = 50 = 25*2
I BC I = 5√2
==========
I AC I² = ( - 3 - 6 )² + ( - 4 + 1)² = 90 = 9*10
I AC I = 3√10
oraz
--->
AB = [ - 2 - 6, 3 - ( - 1) ] = [ -8 , 4 ]
AC = [ - 3 - 6, - 4 - (-1) ] = [ - 9, - 3 ]
Pole ΔABC
---> --->
P = 0,5 *I det ( AB , AC ) I = 0,5* I -8*( - 3) - 4*(- 9) I = 0,5*60 = 30
Korzystamy z wzoru :
P = [tex]\frac{a*b*c}{4 R}[/tex]
więc
R = [tex]\frac{a*b*c}{4 P} = \frac{4\sqrt{5} *5\sqrt{2}*3\sqrt{10} }{4*30} = \frac{600}{120}= 5[/tex]
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Verified answer
Odpowiedź:
I AB I ² = ( - 2 -6)² + ( 3 + 1)² = 80 = 16*5
I AB I = 4√5
===========
I BC I² = ( - 3 +2)² + ( - 4 - 3 )² = 50 = 25*2
I BC I = 5√2
==========
I AC I² = ( - 3 - 6 )² + ( - 4 + 1)² = 90 = 9*10
I AC I = 3√10
===========
oraz
--->
AB = [ - 2 - 6, 3 - ( - 1) ] = [ -8 , 4 ]
--->
AC = [ - 3 - 6, - 4 - (-1) ] = [ - 9, - 3 ]
Pole ΔABC
---> --->
P = 0,5 *I det ( AB , AC ) I = 0,5* I -8*( - 3) - 4*(- 9) I = 0,5*60 = 30
Korzystamy z wzoru :
P = [tex]\frac{a*b*c}{4 R}[/tex]
więc
R = [tex]\frac{a*b*c}{4 P} = \frac{4\sqrt{5} *5\sqrt{2}*3\sqrt{10} }{4*30} = \frac{600}{120}= 5[/tex]
================================
Szczegółowe wyjaśnienie: