A triangle has corners at #(3 ,1 )#, #(4 ,9 )#, and #(7 ,4 )#. What is the area of the triangle's circumscribed circle?

Question

A triangle has corners at #(3  ,1   )#, #(4   ,9   )#, and #(7   ,4  )#.  What is the area of the triangle's circumscribed circle?

Noah Atkinson · Accepted Answer

Area of circumscribed circle is #51.6# sq.unit.

The three corners are #A (3,1) B (4,9) and C (7,4)# Distance between two points #(x_1,y_1) and (x_2,y_2)# is #D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2# Side #AB= sqrt ((3-4)^2+(1-9)^2) ~~ 8.06#unit Side #BC= sqrt ((4-7)^2+(9-4)^2) ~~5.83#unit Side #CA= sqrt ((7-3)^2+(4-1)^2) = 5.0# unit The semi perimeter of triangle is #s=(AB+BC+CA)/2# or #s= (8.06+5.83+5.0)/2~~ 9.45# unit Area of Triangle is #A_t = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|# #A_t = |1/2(3(9−4)+4(4−1)+7(1−9))|# or #A_t = |1/2(15+12-56)| = | -29/2| =14.5# sq.unit Radius of circumscribed circle is #R=(AB*BC*CA)/(4*A_t)# or #R=(sqrt(65)*sqrt(34)*sqrt(25))/(4*14.5) ~~ 4.05# Area of circumscribed circle is #A_c=pi*R^2=pi*4.05^2~~51.6# sq.unit [Ans]

Allison Allen · Answer

#pi r^2 = {pi (65)(34)(25) }/{ 4(25)(34) - (65-34-25)^2} = (27625 pi)/1682 #

The other answer gives an approximation for this question that may be answered exactly. I don't blame the other answer; we've all been taught to do this. I prefer the exact answer, which generally means avoiding square roots.

The circumcircle is just the circle through the three vertices; the triangle almost doesn't matter. Except, miraculously, the circumradius #r# equals the product of the triangle sides #a,b,c# divided by four times the triangle's area #A#.

#r = {abc}/{4A}#

It's much more useful squared, and we're looking for #pi r^2# anyway.

#pi r^2 = {pi a^2 b^2 c^2}/{16A^2}#

The coordinates give the squared distances easily. Archimedes' Theorem relates the squared distances to the triangle area:

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2#

So,

#pi r^2 = {pi a^2 b^2 c^2}/{ 4a^2b^2 - (c^2-a^2-b^2)^2}#

We form the squared distances from pairs of points #(3,1),(4,9),(7,4)#

#c^2=(4-3)^2+(9-1)^2=65#

#a^2=(7-4)^2+(4-9)^2=34#

#b^2=(7-3)^2+(4-1)^2=25#

We're free to play with the assignment to #a,b# and #c.# I prefer making #c# the biggest one which gives me the smallest thing to square. (I try to do these by hand sometimes to keep in shape.)

#pi r^2 = {pi (65)(34)(25) }/{ 4(25)(34) - (65-34-25)^2} = (27625 pi)/1682

The other answer said #51.6,# this is #51.597203956847822956320... quad sqrt#

Allison Allen · Answer

undefined To find the area of the circumscribed circle of a triangle, we need to first find the circumradius of the triangle, which is the radius of the circle passing through all three vertices of the triangle. Then, we can use the formula for the area of a circle, which is π times the square of the radius.

To find the circumradius (R), we can use the formula:

$$ R = \frac{abc}{4A} $$

Where a, b, and c are the lengths of the sides of the triangle, and A is the area of the triangle.

We can find the lengths of the sides of the triangle using the distance formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Once we have the lengths of the sides, we can use Heron's formula to find the area of the triangle:

$$ A = \sqrt{s(s - a)(s - b)(s - c)} $$

Where s is the semiperimeter of the triangle, calculated as (a + b + c) / 2.

After finding the circumradius (R), we can use the formula for the area of the circle:

$$ Area = \pi R^2 $$

Let's calculate:

$$ d_{1} = \sqrt{(4 - 3)^2 + (9 - 1)^2} = \sqrt{1 + 64} = \sqrt{65} $$

$$ d_{2} = \sqrt{(7 - 4)^2 + (4 - 9)^2} = \sqrt{9 + 25} = \sqrt{34} $$

$$ d_{3} = \sqrt{(7 - 3)^2 + (4 - 1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $$

$$ s = \frac{d_{1} + d_{2} + d_{3}}{2} = \frac{\sqrt{65} + \sqrt{34} + 5}{2} $$

$$ A = \sqrt{s(s - d_{1})(s - d_{2})(s - d_{3})} $$

$$ A = \sqrt{\left(\frac{\sqrt{65} + \sqrt{34} + 5}{2}\right)\left(\frac{\sqrt{65} + \sqrt{34} + 5}{2} - \sqrt{65}\right)\left(\frac{\sqrt{65} + \sqrt{34} + 5}{2} - \sqrt{34}\right)\left(\frac{\sqrt{65} + \sqrt{34} + 5}{2} - 5\right)} $$

$$ R = \frac{d_{1}d_{2}d_{3}}{4A} $$

$$ Area = \pi R^2 $$

You can plug in these values into the respective formulas to find the area of the circumscribed circle.

Allison Allen · Answer

undefined To find the area of the circumscribed circle of a triangle, we first need to find the lengths of the triangle's sides and then use those lengths to calculate the radius of the circumscribed circle. Once we have the radius, we can use the formula for the area of a circle to find the area of the circumscribed circle.

Using the distance formula, we find the lengths of the sides of the triangle:

Side 1: From (3, 1) to (4, 9)
        Length = √((4 - 3)^2 + (9 - 1)^2) = √(1 + 64) = √65.

Side 2: From (4, 9) to (7, 4)
        Length = √((7 - 4)^2 + (4 - 9)^2) = √(9 + 25) = √34.

Side 3: From (7, 4) to (3, 1)
        Length = √((3 - 7)^2 + (1 - 4)^2) = √(16 + 9) = √25 = 5.

Now that we have the lengths of the sides, we can use Heron's formula to find the area of the triangle:

S = (1/2) * √(s * (s - a) * (s - b) * (s - c)),
where a, b, and c are the lengths of the sides, and s is the semi-perimeter of the triangle given by s = (a + b + c) / 2.

For our triangle:
s = (65 + 34 + 5) / 2 = 52 / 2 = 26.

Using Heron's formula:
Area = (1/2) * √(26 * (26 - √65) * (26 - √34) * (26 - 5)).

Now, we find the radius of the circumscribed circle (R) using the formula:

R = (abc) / (4 * Area),
where a, b, and c are the lengths of the sides of the triangle.

Substituting the values:
R = (5 * √34 * √65) / (4 * Area).

Now, we can substitute the value of the area we found earlier and calculate the radius (R).

Finally, once we have the radius (R), we can use the formula for the area of a circle, A = πR^2, to find the area of the circumscribed circle.