Question 1184839


Here the length of the major axis is given as {{{2a=48}}} -> {{{a=24}}} and the height of the semi-ellipse is given as {{{b=20}}}.

Now the standard equation of an ellipse with center at origin is:

{{{x^2/a^2+y^2/b^2=1}}}

Putting the values of a and b in the above equation of an ellipse, we have

{{{x^2/24^2 + y^2/20^2 = 1}}}

For the height of {{{10}}} feet, the equation becomes

{{{x^2/576 + 10^2/400 = 1}}}

{{{x^2/576 + 100/400 = 1}}}

{{{x^2/576+1/4 = 1 }}}

{{{x^2/576 = 1 -1/4}}}

{{{x^2/576 = 3/4}}}

{{{x^2 = (3/4)576}}}

{{{x^2 = 3*144}}}

{{{x = sqrt(3*144)}}}

{{{x = 12sqrt(3) }}} ft 

at a height of {{{10 }}}ft above the base,  the arch is {{{ 12sqrt(3) }}} ft  wide