Find the circumcentre of A(0,0) B(3,5) and C(4,2).

ANSWER

Let A(4,6), B(0,4), C(6,2) be the vertices of the given △ABC.

Let P(x,y) be the circumcentre of △ABC. Then,

PA=PB=PC⟹PA

2

=PB

2

=PC

2

Now, PA

2

=PB

2

(x−4)

2

+(y−6)

2

=(x−0)

2

+(y−4)

2

x

2

+y

2

−8x−12y+52=x

2

+y

2

−8y+16

8x+4y=36

2x+y=9 (1)

Again, PB

2

=PC

2

(x−0)

2

+(y−4)

2

=(x−6)

2

+(y−2)

2

x

2

+y

2

−8y+16=x

2

+y

2

−12x−4y+40

12x−4y=24

3x−y=6 (2)

Solving equation 1 and 2, we get,

x=3,y=3

Therefore, the coordinates of circumcentre of △ABC are P(3,3).

Circumradius = PA=

(4−3)

2

+(6−3)

2

=

10

units

Hope it is helpful to you

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft