Question 1158199


<a href="https://www.imageupload.net/image/DdaZy"><img src="https://imagehost.imageupload.net/2020/05/05/radius-circle-inscribed-isosceles-triangle.th.png" alt="radius-circle-inscribed-isosceles-triangle.png" border="0" /></a>



If the base length of the isosceles triangle is {{{b }}}and the two legs are {{{a}}} then prove that the radius of the inscribed circle is given by formula:


{{{r=(b/2)sqrt((2a-b)/(2a+b))}}}


if {{{a=13}}} and {{{b=10}}}, we have


{{{r=(10/2)sqrt((2*13-10)/(2*13+10))}}}


{{{r=5sqrt(16/36)}}}


{{{r=5(sqrt(16)/sqrt(36))}}}


{{{r=5(4/6)}}}


{{{r=20/6}}}


{{{r=3.33in}}}