A triangle has corners at #(3 ,5 )#, #(4 ,9 )#, and #(4 ,2 )#. What is the area of the triangle's circumscribed circle?

1 Answer

Area circumscribed circle #=(85pi)/2=133.5176878" "#square units

Explanation:

For any plane triangle, the #color(red)("Sine Law")# states that

#a/sin A=b/sin B=c/sin C#

The radius #R# of the triangle's circumscribed circle is given

#R=a/(2*sin A)=b/(2*sin B)=c/(2*sin C)#

If we let the vertices as #A(3, 5)# and #B(4, 9)# and #C(4, 2)# then we have

#R=(BC)/(2*sin A)=(AC)/(2*sin B)=(AB)/(2*sin C)#

We only need to use

#R=(BC)/(2*sin A)#

Angle A can be determined through the slopes
#m_(AB)=(9-5)/(4-3)=4# and #m_(AC)=(5-2)/(3-4)=-3#

Angle #A=tan^-1 ((m_(AB)-m_(AC))/(1+m_(AB)*m_(AC)))#

Angle #A=tan^-1 ((4-(-3))/(1+4*(-3)))#

Angle #A=tan^-1 (7/(-11))=147.5288077^@#

We can now compute for the radius #R#

#R=(BC)/(2*sin A)=7/(2*sin 147.5288077^@)#

#R=6.519202405#

Compute the area of the triangle's circumscribed circle

Area #=piR^2#

Area #=pi(6.519202405)^2#

Area #=42.5 pi#
Area #=133.5176878" "#square units

God bless....I hope the explanation is useful.