A triangle has corners at $(1, 5)$ $(1 , 5 )$ , $(4, 8)$ $(4 ,8 )$ , and $(9, 7)$ $(9 ,7 )$ . What is the radius of the triangle's inscribed circle?

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A triangle has corners at $(1, 5)$ $(1 , 5 )$ , $(4, 8)$ $(4 ,8 )$ , and $(9, 7)$ $(9 ,7 )$ . What is the radius of the triangle's inscribed circle?

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Answer 1 · 2017-06-08T14:31:38

The radius of the inscribed circle is $= 1.02$ $=1.02$

Explanation:

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Let the radius of the inscribed circle be $= r$ $=r$

The area of the triangle is

$A = \frac{1}{2} \cdot r \cdot (a) + \frac{1}{2} \cdot r \cdot (b) + \frac{1}{2} \cdot r \cdot (c)$ $A=1/2*r*(a)+1/2*r*(b)+1/2*r*(c)$

$A = \frac{1}{2} r (a + b + c)$ $A=1/2r(a+b+c)$

$r = \frac{2 A}{a + b + c}$ $r=(2A)/(a+b+c)$

Let,

$A = (4, 8)$ $A=(4,8)$

$B = (1, 5)$ $B=(1,5)$

$C = (9, 7)$ $C=(9,7)$

$a = \sqrt{{(9 - 1)}^{2} + {(7 - 5)}^{2}} = \sqrt{64 + 4} = \sqrt{68}$ $a=sqrt((9-1)^2+(7-5)^2)=sqrt(64+4)=sqrt68$

$b = \sqrt{{(9 - 4)}^{2} + {(7 - 8)}^{2}} = \sqrt{25 + 1} = \sqrt{26}$ $b=sqrt((9-4)^2+(7-8)^2)=sqrt(25+1)=sqrt26$

$c = \sqrt{{(4 - 1)}^{2} + {(8 - 5)}^{2}} = \sqrt{9 + 9} = \sqrt{18}$ $c=sqrt((4-1)^2+(8-5)^2)=sqrt(9+9)=sqrt18$

so,

$(a + b + c) = \sqrt{68} + \sqrt{26} + \sqrt{18}$ $(a+b+c)=sqrt68+sqrt26+sqrt18$

The area of the triangle is

$A=1/2|(4,8,1),(1,5,1),(9,7,1)|$

$=1/2(4*|(5,1),(7,1)|-8*|(1,1),(9,1)|+1*|(1,5),(9,7)|)$

$=1/2((4*-2)-(8*-8)+(1*-38))$

$=1/2(-8+64-38)$

$=9$

Therefore,

$r=(2*9)/(sqrt68+sqrt26+sqrt18)=1.02$

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A triangle has corners at $(1, 5)$ $(1 , 5 )$ , $(4, 8)$ $(4 ,8 )$ , and $(9, 7)$ $(9 ,7 )$ . What is the radius of the triangle's inscribed circle?

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Explanation:

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A triangle has corners at (1 , 5 )(1,5)(1 , 5 ), (4 ,8 )(4,8)(4 ,8 ), and (9 ,7 )(9,7)(9 ,7 ). What is the radius of the triangle's inscribed circle?

1 Answer

Explanation:

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A triangle has corners at $(1, 5)$ $(1 , 5 )$ , $(4, 8)$ $(4 ,8 )$ , and $(9, 7)$ $(9 ,7 )$ . What is the radius of the triangle's inscribed circle?