Question 192085
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Finding for "Radius" inscribed in a triangle:

Working Eqn:
Radius = {{{Area[T]/k}}}= {{{sqrt(k(k-a)(k-b)(k-c))/k}}}, * See properties of Triangle



But solving for "k":
We know an Isosceles Triangle has 2 equal sides, {{{a=b=12}}}, and the other side, {{{c=8}}}:


{{{k=(1/2)(a+b+c)=(1/2)(12+12+8)=32/2=red(16)}}}



Subst. in our Working Eqn:
{{{Radius=sqrt(16(16-12)(16-12)(16-8))/16=sqrt(16*4*4*8)/16=sqrt(16*16*8)/16}}}
{{{R=(sqrt(16*16)*sqrt(8))/16=sqrt(256)*sqrt(8)/16=cross(16)*sqrt(8)/cross(16)}}}
{{{R=sqrt(8)=sqrt(4*2)=sqrt(4)*sqrt(2)}}}
{{{red(R=2*sqrt(2))}}}, Answer


Thank you,
Jojo</font>