Question 1187105
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A rectangle is inscribed in a circle of radius  {{{sqrt(10)}}}. If the area of the rectangle is 16, find its dimensions.
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<pre>
Since the radius of the circle is  {{{sqrt(10)}}},  its diameter is  {{{2*sqrt(10)}}}.


The diagonal of the rectangle is the diameter of the circle.


So, if x and y are the rectangle dimension, we have these two equations


    x^2 + y^2 = {{{(2*sqrt(10))^2}}} = 40,     (1)

    xy = 16.                        (2)


It implies


    x^2 + 2xy + y^2 = 40 + 2*16 = 72

    x^2 - 2xy + y^2 = 40 - 2*16 =  8,

or

    {{{(x+y)^2}}} = 72,

    {{{(x-y)^2}}} =  8.


Taking square roots from both sides of the equations, we get


    x + y = {{{6*sqrt(2)}}}    (3)

    x - y = {{{2*sqrt(2)}}}.   (4)


By adding      equations  (3) and (4), you get  2x = {{{8*sqrt(2)))}}};  hence,  x = {{{4*sqrt(2)}}}.

By subtracting equations  (3) and (4), you get  2y = {{{4*sqrt(2)))}}};  hence,  y = {{{2*sqrt(2)}}}.


<U>ANSWER</U>.  The dimensions of the rectangle are  {{{4*sqrt(2)}}}  and  {{{2*sqrt(2)}}}.
</pre>

Solved.