Question 313783
if the area of a circle is equal to the area of an equilateral triangle,
 then the ratio of the side of the triangle to the radius of the circle is closest to which number?
:
Choose a value for the area; A = 60
{{{pi*r^2}}} = 60
r^2 = {{{60/pi}}}
r = {{{sqrt(19.1)}}}
r = 4.37 is the radius
:
The triangle (s=side)
{{{1/2}}}*s*h = 60
Find the height of the triangle in terms of s
h = {{{sqrt(s^2 - (.5s)^2)}}}
h = {{{sqrt(s^2 - .25s^2)}}}
h = {{{sqrt(.75s^2)}}}
Area of the triangle
{{{1/2}}}*s*h = 60
s * h = 120; mult both sides by 2
Replace h with {{{sqrt(.75s^2)}}}
s * {{{sqrt(.75s^2)}}} = 120
Square both sides
s^2 * .75s^2 = 14400
.75s^4 = 14400
s^4 = {{{14400/.75}}}
s^4 = 19200
Find the 4th root of both sides
s = 19200^(1/4)
s = 11.77 is the side of triangle
:
{{{11.77/4.37}}} ~ 2.7, closest to 3
:
:
seems like there should be a more elegant way to do this, but I can't come up with it.