A triangle has corners at #(5 ,6 )#, #(4 ,3 )#, and #(2 ,5 )#. What is the area of the triangle's circumscribed circle?

Question

A triangle has corners at #(5  ,6   )#, #(4   ,3   )#, and #(2   ,5  )#.  What is the area of the triangle's circumscribed circle?

Aurora Ayala · Accepted Answer

#(25pi)/8#

There are no square roots needed.

Archimedes' Theorem says for a triangle with sides #a,b,c:#

# 16 \ text{area}^2 = 4a^2b^2-(c^2-a^2-b^2)^2#

The radius of the circumcircle equals the product of the sides of the triangle divided by four times the area of the triangle. This is more useful squared:

# r^2 = {a^2 b^2 c^2}/{16 \ text{area}^2} = {a^2 b^2 c^2}/{ 4a^2b^2-(c^2-a^2-b^2)^2} #

We get #a^2,b^2 and c^2# from #(5,6),(4,3),(2,5)#

# a^2 = (5-4)^2+(6-3)^2=10#

#b^2 = (4-2)^2+(3-5)^2=8#

#c^2=(5-2)^2+(6-5)^2=10#

Isosceles triangle, not that it matters.

# text{circumcircle area} = pi r^2 = pi {10(8)(10) }/{ 4(10)(8) - (10-8-10)^2 } =(25pi)/8#

The other answer is correct, after much more work.

Allison Allen · Answer

#color(blue)((25pi)/8" units squared")#

The vertices of the triangle all lie on the circumference of the circumscribed circle. In order to find the area of the circle we need to find its radius.

The general equation of a circle is given as:

#(x-h)^2+(y-k)^2=r^2#

Where #bbh# and #bbk# are the x and y coordinates of the centre respectively.

We have three point so we can construct three different equations:

#(5-h)^2+(6-k)^2=r^2 \ \ \ \[1]#

#(4-h)^2+(3-k)^2=r^2 \ \ \ \[2]#

#(2-h)^2+(5-k)^2=r^2 \ \ \ \[3]#

Solving simultaneously:

Subtract #[2]# form #[1]#

#9-2h+27-6k=0#

#36-2h-6k=0 \ \ \ \[4]#

Subtract #[3]# from #[2]#

#12-4h-16+4k=0#

#-4-4h+4k=0 \ \ \ \[5]#

Using #[5]#

#h=k-1#

Substituting in #[4]#

#36-2(k-1)-6k=0=>k=19/4#

Substituting in #h=k-1#

#h=19/4-1=>h=15/4#

We now have the coordinates of the centre of the circle.

#(15/4,19/4)#

The distance from the centre to any point on the circumference is the radius. Using the distance formula and point #(4,3)#

#|r|=sqrt((15/4-4)^2+(19/4-3)^2)=(5sqrt(2))/4#

Area of circumscribed circle is:

#pi((5sqrt(2))/4)^2=(25pi)/8# units squared

PLOT:

Sadie Adams · Answer

undefined To find the area of the circumscribed circle of a triangle, you need to find the circumradius, which is the radius of the circle that passes through all three vertices of the triangle. Once you have the circumradius, you can use the formula for the area of a circle to find the area.

1. Find the lengths of the sides of the triangle using the distance formula.
2. Use Heron's formula to find the area of the triangle.
3. Use the formula for the circumradius of a triangle, $ R = \frac{abc}{4A} $, where $ a $, $ b $, and $ c $ are the side lengths of the triangle and $ A $ is the area of the triangle.
4. Once you have the circumradius, use the formula for the area of a circle, $ A = \pi R^2 $, where $ R $ is the radius of the circle.

Following these steps:

1. Calculate the lengths of the sides of the triangle:
   - Side $ a $ from (5, 6) to (4, 3)
   - Side $ b $ from (4, 3) to (2, 5)
   - Side $ c $ from (2, 5) to (5, 6)
2. Use Heron's formula to find the area of the triangle.
3. Use the formula $ R = \frac{abc}{4A} $ to find the circumradius, where $ A $ is the area of the triangle.
4. Finally, use the formula $ A = \pi R^2 $ to find the area of the circumscribed circle.