A triangle has corners at $(2, 4)$ $(2 ,4 )$ , $(3, 6)$ $(3 ,6 )$ , and $(4, 7)$ $(4 ,7 )$ . What is the area of the triangle's circumscribed circle?

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A triangle has corners at $(2, 4)$ $(2 ,4 )$ , $(3, 6)$ $(3 ,6 )$ , and $(4, 7)$ $(4 ,7 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-04-21T13:11:41

$\frac{65}{2} π$ $65/2 pi$

Explanation:

Just did one just like this.

The circumscribed circle is just the circle with the three vertices on it. The general equation is

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2 + (y-b)^2 = r^2$

Substituting three points,

${(2 - a)}^{2} + {(4 - b)}^{2} = r^{2}$ $(2-a)^2 + (4-b)^2=r^2$
${(3 - a)}^{2} + {(6 - b)}^{2} = r^{2}$ $(3-a)^2+(6-b)^2=r^2$
${(4 - a)}^{2} + {(7 - b)}^{2} = r^{2}$ $(4-a)^2+(7-b)^2=r^2$

Expanding,

$20 = 4 a + 8 b + r^{2} - a^{2} - b^{2}$ $20 = 4a + 8b + r^2-a^2-b^2$
$45 = 6 a + 12 b + r^{2} - a^{2} - b^{2}$ $45 = 6a + 12 b + r^2-a^2-b^2$
$65 = 8 a + 14 b + r^{2} - a^{2} - b^{2}$ $65 = 8a + 14b + r^2-a^2-b^2$

Subtracting pairs,

$25 = 2 a + 4 b$ $25 = 2a + 4b$
$20 = 2 a + 2 b$ $20 = 2a + 2b$

Subtracting,

$5 = 2 b$ $5 = 2b$
$b = \frac{5}{2}$ $b=5/2$
$a = 10 - b = \frac{15}{2}$ $a = 10 - b = 15/2$

Substituting,

$r^{2} = {(2 - \frac{15}{2})}^{2} + {(4 - \frac{5}{2})}^{2} = \frac{65}{2}$ $r^2 = (2-15/2)^2 + (4-5/2)^2= 65 /2$

So the circle's area is

$A = π r^{2} = \frac{65}{2} π$ $A = \pi r^2 = 65/2 pi$

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A triangle has corners at $(2, 4)$ $(2 ,4 )$ , $(3, 6)$ $(3 ,6 )$ , and $(4, 7)$ $(4 ,7 )$ . What is the area of the triangle's circumscribed circle?

1 Answer

Explanation:

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A triangle has corners at (2,4)(2 ,4 ), (3,6)(3 ,6 ), and (4,7)(4 ,7 ). What is the area of the triangle's circumscribed circle?

1 Answer

Explanation:

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A triangle has corners at $(2, 4)$ $(2 ,4 )$ , $(3, 6)$ $(3 ,6 )$ , and $(4, 7)$ $(4 ,7 )$ . What is the area of the triangle's circumscribed circle?