Question

A triangle has corners at `(9 ,7 )`, `(2 ,1 )`, and `(5 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`"Details are shown below. Please check my math."`

Explanation:

• The corner coordinates of the ABC triangle are on the circumference circle.
• the first step is to find the edge lengths of triangle a, b, c.
• We can find the distance between two known coordinates by using the following formula.

`P_1(x_1,y_1)" , "P_2(x_2,y_2)`

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

1. The length of a side:
   `a=sqrt((5-2)^2+(2-1)^2)=sqrt(3^2+1^2)=sqrt(9+1)=sqrt(10)" units"`

2. The length of b side:
   `b=sqrt((9-5)^2+(7-2)^2)=sqrt(4^2+5^2)=sqrt(16+25)=sqrt(41)" units"`

3. The length of c side:
   `c=sqrt((9-2)^2+(7-1)^2)=sqrt(7^2+6^2)=sqrt(49+36)=sqrt(85)" units"`

-In the second step, we can calculate the area of the triangle known as corner coordinates.

`A(x_1,y_1)" , "B(x_2,y_2)" , "C=(x_3,y_3)`
`A(9,7)" , "B(2,1)" , "C(5,2)`

`"triangle's area="1/2*|9*1+2*2+5*7-2*7-5*1-9*2 |`

`"triangle's area="1/2*|9+4+35-14-5-18 |`

`"triangle's area="1/2*|9+4+35-14-5-18 |=5.5" units"^2`

• now we can use the formula given below.

`area(ABC)=(a*b*c)/(4*r)`

`5.5=(sqrt (10)*sqrt(41)*sqrt(85))/(4*r)`

`22r=sqrt(10*41*85)`

`r=(sqrt(10*41*85))/22`

`r=(sqrt(34850))/22`

`r=8.49" units"`

• the area of the triangle's circumscribed circle:

`area=pi*r^2`

`area=3.14*(8.49)^2`

`area=226.33" units"^2`