A triangle has corners at $(3, 7)$ $(3 ,7 )$ , $(2, 5)$ $(2 ,5 )$ , and $(8, 4)$ $(8 ,4 )$ . What is the area of the triangle's circumscribed circle?

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A triangle has corners at $(3, 7)$ $(3 ,7 )$ , $(2, 5)$ $(2 ,5 )$ , and $(8, 4)$ $(8 ,4 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2016-11-07T19:54:46

Area of the triangle's circumscribed circle: $2.65 sq.units$ $color(green)(2.65 " sq.units")$ (approx.)

Explanation:

Perhaps there is a simpler way, but here is how I would do it:

Part 1: Find the Lengths of the Three Sides
Let $a$ $a$ be the side $(3, 7) : (2, 5)$ $(3,7):(2,5)$
$XXX | a | = \sqrt{{(2 - 3)}^{2} + {(5 - 7)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$ $color(white)("XXX")abs(a)=sqrt((2-3)^2+(5-7)^2)=sqrt(1+4)=sqrt(5)$

Let $b$ $b$ be the side $(2, 5) : (8, 4)$ $(2,5):(8,4)$
$XXX | b | = \sqrt{{(8 - 2)}^{2} + {(4 - 5)}^{2}} = \sqrt{36 + 1} = \sqrt{37}$ $color(white)("XXX")abs(b)=sqrt((8-2)^2+(4-5)^2)=sqrt(36+1)=sqrt(37)$

Let $c$ $c$ be the side $(8, 4) : (3, 7)$ $(8,4):(3,7)$
$XXX | c | = \sqrt{{(3 - 8)}^{2} + {(7 - 4)}^{2}} = \sqrt{25 + 9} = \sqrt{34}$ $color(white)("XXX")abs(c)=sqrt((3-8)^2+(7-4)^2)=sqrt(25+9)=sqrt(34)$

Part 2: Find the Perimeter, Semi-Perimeter, and Area of the Triangle
(I will need the help of a calculator here)
Perimeter: $p = \sqrt{5} + \sqrt{37} + \sqrt{34} \approx 14.1497824026$ $p=sqrt(5)+sqrt(37)+sqrt(34)~~14.1497824026$

Semi-perimeter: $s = \frac{p}{2} \approx 7.0748912013$ $s=p/2~~7.0748912013$

Area of triangle (using Heron's formula):
$XXX A r e a_{△} = \sqrt{s (s - a) (s - b) (s - c)} \approx 6.5$ $color(white)("XXX")"Area_triangle=sqrt(s(s-a)(s-b)(s-c))~~6.5$

Part 3: Find the Radius and Area of the Circumscribed Circle
(more calculator usage)
Radius of Circumscribed Circle: $r = \frac{{Area}_{△}}{s} = 0.9187420435$ $r=("Area"_triangle)/s=0.9187420435$
(ask as a separate question if you are unfamiliar with this relation).

Area of Circle: ${Area}_{\circ} = π r^{2} = 2.6517773377$ $"Area"_circ = pir^2 =2.6517773377$

Answer 2 · 2016-11-07T20:10:22

$A = \frac{6290 π}{676}$ $A = (6290pi)/676$

Explanation:

The standard form for the equation of a circle is:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x - h)^2 + (y - k)^2 = r^2$

where $(h, k)$ $(h, k)$ is the center and $r$ $r$ is the radius

Because the triangles vertices are points on the circumscribed circle, we can use the 3 points and the standard form to write 3 equations:

${(3 - h)}^{2} + {(7 - k)}^{2} = r^{2}$ $(3 - h)^2 + (7 - k)^2 = r^2$ $[1]$ $" [1]"$
${(2 - h)}^{2} + {(5 - k)}^{2} = r^{2}$ $(2 - h)^2 + (5 - k)^2 = r^2$ $[2]$ $" [2]"$
${(8 - h)}^{2} + {(4 - k)}^{2} = r^{2}$ $(8 - h)^2 + (4 - k)^2 = r^2$ $[3]$ $" [3]"$

We have 3 equations and we can use them to find the values of, h, k, and $r^{2}$ $r^2$ .

Temporarily eliminate $r^{2}$ $r^2$ by setting the left side of equation [1] equal to the left side of equation [2] and the same for left side of equation [3]:

${(3 - h)}^{2} + {(7 - k)}^{2} = {(2 - h)}^{2} + {(5 - k)}^{2}$ $(3 - h)^2 + (7 - k)^2 = (2 - h)^2 + (5 - k)^2$ $[1 = 2]$ $" [1 = 2]"$
${(3 - h)}^{2} + {(7 - k)}^{2} = {(8 - h)}^{2} + {(4 - k)}^{2}$ $(3 - h)^2 + (7 - k)^2 = (8 - h)^2 + (4 - k)^2$ $[1 = 3]$ $" [1 = 3]"$

Expand the squares, using the pattern, ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a - b)^2 = a^2 - 2ab + b^2$ :

$9 - 6 h + h^{2} + 49 - 14 k + k^{2} = 4 - 4 h + h^{2} + 25 - 10 k + k^{2}$ $9 -6h + h^2 + 49 -14k + k^2 = 4 -4h + h^2 + 25 -10k + k^2$
$9 - 6 h + h^{2} + 49 - 14 k + k^{2} = 64 - 16 h + h^{2} + 16 - 8 k + k^{2}$ $9 -6h + h^2 + 49 -14k + k^2 = 64 -16h + h^2 + 16 -8k + k^2$

Please notice that for every $h^{2} and k^{2}$ $h^2 and k^2$ on the left there is a corresponding one on the right so they cancel:

$9 - 6 h + 49 - 14 k = 4 - 4 h + 25 - 10 k$ $9 -6h + 49 -14k = 4 -4h + 25 -10k$
$9 - 6 h + 49 - 14 k = 64 - 16 h + 16 - 8 k$ $9 -6h + 49 -14k = 64 -16h + 16 -8k$

Combine all of the constant terms into a single term on the right:

$- 6 h - 14 k = - 4 h - 10 k - 29$ $- 6h - 14k = - 4h- 10k - 29$
$- 6 h - 14 k = - 16 h - 8 k + 22$ $- 6h- 14k =- 16h- 8k + 22$

Combine all of the h terms into a single term on the right:

$- 14 k = 2 h - 10 k - 29$ $- 14k = 2h- 10k - 29$
$- 14 k = - 10 h - 8 k + 22$ $- 14k =- 10h- 8k + 22$

Combine all of the k terms into a single term on the left:

$- 4 k = 2 h - 29$ $-4k = 2h - 29$ $[4]$ $" [4]"$
$- 6 k = - 10 h + 22$ $-6k = -10h + 22$ $[5]$ $" [5]"$

Multiply equation [4] by -3 and equation [5] by 2 and add them:

$12 k - 12 k = - 6 h - 20 h + 87 + 44$ $12k - 12k = -6h - 20h + 87 + 44$

$0 = - 26 h + 131$ $0 = -26h + 131$

$h = \frac{131}{26}$ $h = 131/26$

Substitute $\frac{131}{26}$ $131/26$ for h in equation [4]:

$- 4 k = 2 (\frac{131}{26}) - 29$ $-4k = 2(131/26) - 29$

$- 4 k = - \frac{492}{26}$ $-4k = -492/26$

$k = \frac{123}{26}$ $k = 123/26$

Substitute the values for h and k into equation [3]

${(8 - \frac{131}{26})}^{2} + {(4 - \frac{123}{26})}^{2} = r^{2}$ $(8 - 131/26)^2 + (4 - 123/26)^2 = r^2$

${(\frac{77}{26})}^{2} + {(- \frac{19}{26})}^{2} = r^{2}$ $(77/26)^2 + (-19/26)^2 = r^2$

$r^{2} = \frac{6290}{676}$ $r^2 = 6290/676$

The area of the circle is $A = π r^{2}$ $A = pir^2$

$A = \frac{6290 π}{676}$ $A = (6290pi)/676$

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A triangle has corners at $(3, 7)$ $(3 ,7 )$ , $(2, 5)$ $(2 ,5 )$ , and $(8, 4)$ $(8 ,4 )$ . What is the area of the triangle's circumscribed circle?

2 Answers

Explanation:

Explanation:

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A triangle has corners at (3 ,7 )(3,7)(3 ,7 ), (2 ,5 )(2,5)(2 ,5 ), and (8 ,4 )(8,4)(8 ,4 ). What is the area of the triangle's circumscribed circle?

2 Answers

Explanation:

Explanation:

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A triangle has corners at $(3, 7)$ $(3 ,7 )$ , $(2, 5)$ $(2 ,5 )$ , and $(8, 4)$ $(8 ,4 )$ . What is the area of the triangle's circumscribed circle?