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Question

In $$\displaystyle \triangle ABC,$$ the incircle touches the sides $$BC, CA$$ and $$AB$$, respectively, at $$D, E$$ and $$F$$. 
If the radius of the incircle is $$4$$ units and $$BD, CE$$ and $$AF$$ are consecutive integers, 
then the value of $$\dfrac{s}{3}$$, where $$s$$ is a semi-perimeter of the triangle, is?


Solution


Given in radius $$r=4$$

$$BD,CE,AF$$ are in A.P of 

common difference $$1$$

$$s=\dfrac { \triangle  }{ r } =\dfrac { \sqrt { s(s-a)(s-b)(s-c) }  }{ r } \\$$

$$ =2y=x+z\\$$

$$ z-y=y-x=1\\$$

$$ a=x+z,b=y+z,c=x+y\\$$

$$ S=\left( \dfrac { a+b+c }{ 2 }  \right) =(3y)\\$$

$$ \triangle =\sqrt { (3y)\left[ 3y-(x+z) \right] \left[ 3y-(y+z) \right] \left[ 3y-(x+4) \right]  } \\$$

$$ =\sqrt { (3y)(y)+(2y-z)(2y-x) } \\ $$

$$=\sqrt { (3y)y(y-1)(y+1) } \\$$

$$ =\dfrac { \sqrt { 3{ y }^{ 2 }({ y }^{ 2 }-1) }  }{ r } \\$$

$$ s=\dfrac { \triangle  }{ r } ={ s }^{ 2 }=\dfrac { { \triangle  }^{ 2 } }{ { r }^{ 2 } } \\$$

$$ 9{ y }^{ 2 }=\dfrac { 3{ y }^{ 2 }({ y }^{ 2 }-1) }{ 16 } \Rightarrow 3\times 16=({ y }^{ 2 }-1)\\$$

$$ { y }^{ 2 }=49\Rightarrow y=7\\ $$

$$\dfrac { s }{ 3 } =\dfrac { 3y }{ 3 } =7$$

906548_142564_ans_902c604697064bb1bf019cbab3c6ad4f.png

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