Question 1175877
What is the area of the smaller of the two concentric circles if the side length of the equilateral triangle is 8cm?


let radius of greater circle be {{{R}}}, radius of smaller circle be {{{r}}},  the height of the equilateral triangle{{{h}}}, and  the side of the equilateral triangle{{{a}}}

from the given image you can see that radius of greater circle is equal to height {{{R=h}}} of  the equilateral triangle, and {{{r=R/2=h/2}}} 
 
if the side {{{a=8cm}}}, then

{{{h=sqrt(a^2-(a/2)^2)}}}
{{{h=sqrt(8^2-(8/2)^2)}}}
{{{h=sqrt(64-16)}}}
{{{h=sqrt(48)cm}}}


the area of the smaller of the two concentric circle is {{{r=h/2}}}

{{{r=sqrt(48)/2}}}


the area of the smaller circle is:

{{{A=r^2*pi}}}
{{{A=(sqrt(48)/2)^2*pi}}}
{{{A=(48/4)*pi}}}
{{{A=12*pi*cm^2}}}