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

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

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Answer 1 · 2018-01-18T08:36:06

$Details are shown below. Please check my math.$ $"Details are shown below. Please check my math."$

Explanation:

enter image source here

The corner coordinates of the ABC triangle are on the circumference circle.
the first step is to find the edge lengths of triangle a, b, c.
We can find the distance between two known coordinates by using the following formula.

$P_{1} (x_{1}, y_{1}), P_{2} (x_{2}, y_{2})$ $P_1(x_1,y_1)" , "P_2(x_2,y_2)$

$l = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $l=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

The length of a side:
$a = \sqrt{{(5 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{3^{2} + 1^{2}} = \sqrt{9 + 1} = \sqrt{10} units$ $a=sqrt((5-2)^2+(2-1)^2)=sqrt(3^2+1^2)=sqrt(9+1)=sqrt(10)" units"$
The length of b side:
$b = \sqrt{{(9 - 5)}^{2} + {(7 - 2)}^{2}} = \sqrt{4^{2} + 5^{2}} = \sqrt{16 + 25} = \sqrt{41} units$ $b=sqrt((9-5)^2+(7-2)^2)=sqrt(4^2+5^2)=sqrt(16+25)=sqrt(41)" units"$
The length of c side:
$c = \sqrt{{(9 - 2)}^{2} + {(7 - 1)}^{2}} = \sqrt{7^{2} + 6^{2}} = \sqrt{49 + 36} = \sqrt{85} units$ $c=sqrt((9-2)^2+(7-1)^2)=sqrt(7^2+6^2)=sqrt(49+36)=sqrt(85)" units"$

-In the second step, we can calculate the area of the triangle known as corner coordinates.

$A (x_{1}, y_{1}), B (x_{2}, y_{2}), C = (x_{3}, y_{3})$ $A(x_1,y_1)" , "B(x_2,y_2)" , "C=(x_3,y_3)$
$A (9, 7), B (2, 1), C (5, 2)$ $A(9,7)" , "B(2,1)" , "C(5,2)$

enter image source here

$triangle's area= \frac{1}{2} \cdot | 9 \cdot 1 + 2 \cdot 2 + 5 \cdot 7 - 2 \cdot 7 - 5 \cdot 1 - 9 \cdot 2 |$ $"triangle's area="1/2*|9*1+2*2+5*7-2*7-5*1-9*2 |$

$triangle's area= \frac{1}{2} \cdot | 9 + 4 + 35 - 14 - 5 - 18 |$ $"triangle's area="1/2*|9+4+35-14-5-18 |$

$triangle's area= \frac{1}{2} \cdot | 9 + 4 + 35 - 14 - 5 - 18 | = 5.5 {units}^{2}$ $"triangle's area="1/2*|9+4+35-14-5-18 |=5.5" units"^2$

now we can use the formula given below.

$a r e a (A B C) = \frac{a \cdot b \cdot c}{4 \cdot r}$ $area(ABC)=(a*b*c)/(4*r)$

$5.5 = \frac{\sqrt{10} \cdot \sqrt{41} \cdot \sqrt{85}}{4 \cdot r}$ $5.5=(sqrt (10)*sqrt(41)*sqrt(85))/(4*r)$

$22 r = \sqrt{10 \cdot 41 \cdot 85}$ $22r=sqrt(10*41*85)$

$r = \frac{\sqrt{10 \cdot 41 \cdot 85}}{22}$ $r=(sqrt(10*41*85))/22$

$r = \frac{\sqrt{34850}}{22}$ $r=(sqrt(34850))/22$

$r = 8.49 units$ $r=8.49" units"$

the area of the triangle's circumscribed circle:

$a r e a = π \cdot r^{2}$ $area=pi*r^2$

$a r e a = 3.14 \cdot {(8.49)}^{2}$ $area=3.14*(8.49)^2$

$a r e a = 226.33 {units}^{2}$ $area=226.33" units"^2$

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

1 Answer

Explanation:

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