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

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

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Answer 1 · 2018-04-26T19:01:50

$\frac{5365 π}{578}$ $color(blue)((5365pi)/578)$ units squared.

Explanation:

The vertices of the triangle are all points on the circumference of the given circle. The equation for a circle is given by:

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

Where $h and k$ $bbh and bbk$ are the $x$ $bbx$ and $y$ $bby$ coordinates of the centre respectively anf $r$ $bbr$ is thr radius.

We can make 3 equations from this:

$(7-h)^2+(3-k)^2=r^2 \ \ \ \ \ [1]$

$(5-h)^2+(8-k)^2=r^2 \ \ \ \ \ [2]$

$(4-h)^2+(2-k)^2=r^2 \ \ \ \ \ [3]$

Subtracting $[3]$ from $[2]$

$69-2h-12k=0 \ \ \ \ [4]$

Subtract $[2]$ from $[1]$

$10k-4h-31=0 \ \ \ \ \ [5]$

From $[5]$ :

$k=(4h+31)/10$

Substituting in $[4]$

$69-2h-12((4h+31)/10)=0$

$69-2h-(6(4h+31))/5=0$

$345-10h-24h-186=0$

$h=159/34$

Substituting this in $[5]$

$10k-4(159/34)-31=0$

$k=(31+4(159/34))/10$

$k=169/34$

So we have the centre, we now find the radius:

Using vertex $(4,2)$

$(4-159/34)^2+(2-169/34)^2=r^2$

$r^2=5365/578$

Area:

$(5365pi)/578$

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

1 Answer

Explanation:

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

1 Answer

Explanation:

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