Question

A triangle has corners at `(8 ,1 )`, `(4 ,5 )`, and `(6 ,7 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area of triangle's circumscribed circle is `31.42`sq.unit

Explanation:

The three corners are `A (8,1) B (4,5) and C (6,7)`

Distance between two points `(x_1,y_1) and (x_2,y_2)` is

`D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2`

Side `AB= sqrt ((8-4)^2+(1-5)^2)=sqrt(32) ~~ 5.66`unit

Side `BC= sqrt ((4-6)^2+(5-7)^2)=sqrt(8) ~~2.83`unit

Side `CA= sqrt ((6-8)^2+(7-1)^2)=sqrt(40) ~~ 6.32`unit

Area of Triangle is `A_t = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|`

`A_t = |1/2(8(5−7)+4(7−1)+6(1−5))|` or

`A_t = |1/2(-16+24-24)| = 1/2*16 =8.0` sq.unit.

Radius of circumscribed circle is `R=(AB*BC*CA)/(4*A_t)` or

`R=(sqrt(32)*sqrt(8)*sqrt(40))/(4*8) ~~ 3.16`

Area of triangle's circumscribed circle is

`A_c=pi*R^2=pi*3.16^2~~31.42` sq.unit [Ans]

Answer 2

Area of circumcircle `A_C = pi R^2 = pi (3.1623)^2 = color (blue)(31.4164)`

Explanation:

Refer figure abov.

B is mid point of AB

Coordinates of `B = (8+4)/2, (1+5)/2 = ((6,3)`

Similarly coordinates of `F = (4+6)/2, (5+7)/2 = (5,6)`

Slope of AD `m1 = ((5-1) / (4-8)) = -1`

Slope of perpendicular line through B `m_B= -1/m1 = 1`

Eqn of perpendicular line through B is

`y - 3 = 1 *(x - 6)`

`y - x = -3` `color(red)(Eqn (1))`

Slope of CD `m2 = (7-5) / (6-4) = 1`

Slope of perpendicular line through F `m_F = -1/m2 = -1`

Eqn of perpendicular line through F is

`y - 6 = -1 * (x - 5)`

`y + x = 11` `color(red)(Eqn (2)`

Solving Eqns (1), (2) we get the coordinates of circumcenter O.

`y=4, x = 7` `O(7, 4)`

Radius of circumcircle is distance of O from the vertices A, D or C.

I.e `R = OA = OB = OC`

`OA = R = sqrt((8-7)^2 + (1-4)^2) = 3.1623`

Area of circumcircle `A_C = pi R^2 = pi (3.1623)^2 = color (blue)(31.4164)`