How to find a radius of a circle inscribed inside a right triangle with sides 9.15 and 17.6 ?This problem was solved r=ab/(a+b+c) = 9.15(17.6)/[9.15 + 17.6+SQR(9.15^2 + 17.6^2)]= 3.46. However, I...

How to find a radius of a circle inscribed inside a right triangle with sides 9.15 and 17.6 ?

This problem was solved r=ab/(a+b+c) = 9.15(17.6)/[9.15 + 17.6+SQR(9.15^2 + 17.6^2)]= 3.46. However, I dont understand how to derive to this formular. Thanks

Asked on by thi-dallas

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lfryerda | High School Teacher | (Level 2) Educator

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The radius of an inscribed circle is usually called the inradius or the radius of the incircle, and is found from the area of the triangle and its perimeter (see attached link).  The inradius is found by the formula

`r={2K}/P`  

where K is the area of the triangle and P is the perimeter of the triangle.  In the case of a right-angled triangle, with legs of length `a` and `b` , the hypotenuse is `sqrt{a^2+b^2}` , and the area is `K=1/2 ab` , to get:

`r={2{ab}/2}/{a+b+sqrt{a^2+b^2}}`

`={ab}/{a+b+sqrt{a^2+b^2}}`

Now putting in the numbers that you have of `a=9.15` and `b=17.6` , we get the inradius of `r=3.46` .

The way to establish the formula `r={2K}/P` , is by noting that for the right-angled triangle with side lengths `a` , `b` and `c=sqrt{a^2+b^2}` , that the area of the three triangles with the angle bisectors of the triangle vertices that meet at the circle centre are given by 

`{ar}/2` , `{br}/2`  and `{cr}/2`

which means that the area of the triangle must be the sum of those areas, to get

`K={(a+b+c)r}/2`  

so by substituting in `P=a+b+c` , and rearranging, we get:

`r={2K}/P`

The formula has been derived.

Sources:

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