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

Area of circumscribed circle is

sq.unit [Ans]

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

So,

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

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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.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(pi)/2 #, and the triangle's area is #12 #. What is the area of the triangle's incircle?
- What is the equation of the circle with a center at #(7 ,1 )# and a radius of #2 #?
- A circle's center is at #(2 ,1 )# and it passes through #(3 ,2 )#. What is the length of an arc covering #(5pi ) /12 # radians on the circle?
- A triangle has sides with lengths of 2, 9, and 8. What is the radius of the triangles inscribed circle?
- Two circles have the following equations: #(x +3 )^2+(y -5 )^2= 64 # and #(x -2 )^2+(y +4 )^2= 81 #. Does one circle contain the other? If not, what is the greatest possible distance between a point on one circle and another point on the other?

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