Question 1090034
what is the area of ellipse with an aspect ratio of two and total perimeter of {{{25}}} inches? 


For an ellipse, the aspect ratio denotes the ratio of the major axis {{{a}}}  to the minor axis {{{b}}}.
 
{{{a}}} and {{{b}}} are measured from the center, so they are like "radius" measures

given:  {{{a/b=2}}} => {{{a=2b}}}

and perimeter {{{P=25}}} inches

an approximation within about 5% of the true value, so long as a is not more than 3 times longer than b (in other words, the ellipse is not too "squashed"):


{{{P=2pi*sqrt((a^2+b^2)/2)}}}

{{{25=2pi*sqrt(((2b)^2+b^2)/2)}}}.............solve for {{{b}}}

{{{b = -(5 sqrt(5/2))/pi}}}

{{{b }}}≈  {{{-2.51773699057992}}}


{{{a=2*(-5sqrt(5/2))/pi = (-10sqrt(5/2))/pi}}}

{{{a}}} ≈ {{{ -5.032921210448}}}

Ellipse area is:

{{{Area =ab*pi}}}

{{{Area =(-10 sqrt(5/2))/pi*(-5 sqrt(5/2))/pi)*pi}}}

{{{Area =(-10 sqrt(5/2))/pi*(-5sqrt(5/2))}}}

{{{Area =125/pi}}}

{{{Area=39.8}}} in^2

check equation:

{{{x^2/((-10 sqrt(5/2))/pi)^2 + y^2/( -(5 sqrt(5/2))/pi)^2=1}}}

or
{{{x^2/(-5.032921210448)^2 + y^2/(-2.51773699057992)^2=1}}}

it will give you:

area | {{{39.808917197}}}
perimeter | {{{24.383816643}}}  which is an approximation  within about 5% of the true value