Question

A triangle has corners at `(5 ,1 )`, `(7 ,9 )`, and `(4 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area of circumscribed circle of the triangle is `61.95` sq.unit.

Explanation:

`A(5,1),B(7,9),C(4,2)`

Side `AB= sqrt ((5-7)^2+(1-9)^2)=sqrt(68) ~~8.25`unit

Side `BC= sqrt ((7-4)^2+(9-2)^2)=sqrt(58) ~~7.62`unit

Side `CA= sqrt ((4-5)^2+(2-1)^2)=sqrt(2) ~~ 1.41`unit

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

`A_t = |1/2(5(9−2)+7(2−1)+4(1−9))|` or

`A_t = |1/2(35+7-32)| = 1/2*10 =5.0` sq.unit.

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

`R=(sqrt(68)*sqrt(58)*sqrt(2))/(4*5) ~~ 4.44`

Area of circumscribed circle is `A_c=pi*R^2=pi*4.44^2~~61.95`

sq.unit [Ans]

Answer 2

`493/25pi`

Explanation:

The idea is to find the centre of the circle, then use `A=pir^2`

We will make use of the circle theorem:

The perpendicular from the radius to a chord bisects the chord

This means that we can find the centre from where the perpendicular bisectors of two of the lines intersect. From here, we can find the square of the radius, and the area.

I'll give the points some names.
Let `A=(4,2), B=(5,1)` and `C=(7,9)`

First, let's sketch the information we're given.

We will begin by finding the equation of the perpendicular bisector of AB.

`"midpoint"_"AB"=((4+5)/2,(1+2)/2)`
`=(9/2,3/2)`

`"grad"_"AB"=(2-1)/(4-5)`
`=-1`

`:. "grad"_"perp"=1`

`:. "eqn of perp:"`
`y-3/2=x-9/2`
`y=x-3`

Now, we will do the same for BC.

`"midpoint"_"BC"=((5+7)/2,(9+1)/2)`
`=(6,5)`

`"grad"_"BC"=(9-1)/(7-5)`
`=4`

`"grad"_"perp"=-1/4`

`:. "eqn of perp":`

`y-5=-1/4(x-6)`
`y-5=-1/4x+3/2`
`y=-1/4x+13/2`

The centre of the circle will be the point at which these two lines intersect, so we solve these equations simultaneously.

`y=-1/4x+13/2`
`y=x-3`

`-1/4x+13/2=x-3`
`19/2=5/4x`
`x=38/5`

`y=(38/5)-3`
`y=23/5`

So the centre of the centre is `(38/5,23/5)`

From this, we can work out the radius. In fact, it's easier to work out `r^2`, since we can put this into the area formula straight away.

`r=sqrt((x-x_1)^2+(y-y_1)^2)`

becomes

`r^2=(x-x_1)^2+(y-y_1)^2`

We'll take the centre as `(x,y)` and any other point, say B, for `(x_1,y_1)`

`r^2=(38/5-5)^2+(23/5-1)^2`
`=493/25`

Since `A=pir^2`,

`A=493/25pi`