Question

A triangle has corners at `(9 ,4 )`, `(3 ,9 )`, and `(5 ,8 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area of circum circle `A_c ^2 = color(purple)(479.0937)`

Explanation:

The steps to find the circum circle area of a triangle:

1. Find and Calculate the midpoint of given coordinates or midpoints (AB, AC, BC)

2. Calculate the slope of the particular line

3.By using the midpoint and the slope, find out the equation of line (y-y1) = m (x-x1)

1. Find out the other line of equation in the similar manner

2. Solve the two bisector equation by finding out the intersection point

3. Calculate the intersection point will be the circum center of the given triangle

4. Find the length of radius (r) of the circum circle, being the distance between circum center and one of the vertices.

5. Calculate the area of the circum circle using formula `pi r^2`

Mid Point of BC `D = (3+5)/2, (9+8)/2 = (4, 17/2)`

Slope of BC `m_(BC) = (9-8) / (3-5) = -1/2`

Slope of PD `m_(PD) = - 1 /(- 1/2) = 2`

Equation of perpendicular bisector PD is

`y - (17/2) = 2 (x - 4)`

`2y - 17 = 4x - 16`

`2y - 4x = 1`color(red)(Eqn (1))#

Mid Point of CA `E = (9+5)/2, (8+4)/2 = (7,6)`

Slope of CA `m_(CA) = (8-4) / (5-9) = -1`

Slope of PE `m_(PE) = - 1 / -1 = 1`

Equation of perpendicular bisector PE is

`y - 6 = 1 (x - 7)`

`y - x = -1` `color(red)(Eqn (2))`

Mid Point of AB `F = (9+3)/2, (4+9)/2 = (6, 13/2)`

Slope of AB `m_(AB) = (9-4) / (3-9) = -5/6`

Slope of PF `m_(PF) = - 1 / (-5/6) = 6/5`

Equation of perpendicular bisector PF is

`y - (13/2) = (6/5) (x - 6)`

`10y - 65 = 12x - 72`

`10y - 12x = -7` `color(red)(Eqn (3))`

Sloving Equations (1) & (2), we get the coordinates of circum center P

`color(green)(P (-3/2, -5/2)`

Radius of circum cirlce `r = PA = PB = PC`

`:. PA = r = sqrt(((-3/2)-9)^2 + ((-5/2)-4)^2) = color(blue)(12.3491)`

Verification :

`PB = r = sqrt((-3/2)-3)^2 + ((-5/2)-9)^2 = 12.3491`

Area of circum circle `A_c = pi r^2 = pi * 12.3491^2 = color(purple)(479.0937)`