A triangle has corners at (7 ,3 ), (2 ,5 ), and (1 ,4 ). What is the area of the triangle's circumscribed circle?

1 Answer

Area A=1073/98 *pi=34.39723385" "#square units

Explanation:

Let A(1, 4) , B(7, 3), C(2, 5)

One solution is to determine the orthocenter H(x_H, y_H), then solve for the radius R

R=sqrt((x_A-x_H)^2+(y_A-y_H)^2)

the circumcenter is the intersection of the line segment bisectors of the side AB and side AC which have the midpoints at (4, 7/2) and (3/2, 9/2) respectively.

the equation of the line segment bisector at AB is

y-7/2=-1/m_(AB)*(x-4)" "where " "m_(AB) = -1/6

y-7/2=6*(x-4)

simplified
12x-2y=41

the equation of the line segment bisector at AC is

y-9/2=-1/m_(AC)*(x-3/2)" "where m_(AC)=1

y-9/2=-1*(x-3/2)

simplified

x+y=6

simultaneous solution using
x+y=6 and 12x-2y=41 yields

H(x_H, y_H)=(53/14, 31/14)

Solve for R now, using any vertex of the triangle ...
let us choose A(1, 4)

R=sqrt((1-53/14)^2+(4-31/14)^2)

R=sqrt(2146)/14

it follows ,....the area of the circle with center (53/14, 31/14) radius R (see the orange line segment)

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A=pi*R^2
A=pi*(sqrt(2146)/14)^2
A=1073/98 *pi=34.39723385" "#square units

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