Question 1201846
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Answer: <font color=red size=4>8.23 inches by 18.23 inches</font>
Each value is approximate.



Explanation:


Here's what the diagram probably looks like
{{{
drawing(400,400,-5,5,-5,5,
circle(0,0,4),
line(-3.4551,-2.0155,3.458,-2.0105),line(3.458,-2.0105,3.4551,2.0155),line(3.4551,2.0155,-3.458,2.0105),line(-3.458,2.0105,-3.4551,-2.0155),
line(-3.458,2.0105,3.458,-2.0105),

circle(-3.458,2.0105,0.05),circle(-3.458,2.0105,0.1),circle(-3.458,2.0105,0.15),circle(3.4551,2.0155,0.05),circle(3.4551,2.0155,0.1),circle(3.4551,2.0155,0.15),circle(3.458,-2.0105,0.05),circle(3.458,-2.0105,0.1),circle(3.458,-2.0105,0.15),circle(-3.4551,-2.0155,0.05),circle(-3.4551,-2.0155,0.1),circle(-3.4551,-2.0155,0.15),

locate(-3.458,2.5105,"A"),locate(3.4551,2.5155,"B"),locate(3.458,-2.1105,"C"),locate(-3.7551,-2.1155,"D"),

locate(-3.2,0,"y"),locate(-0.2,-1.5,"x"),locate(0.4,0.4,"20"),

locate(-4,-4.5,matrix(1,4,Diagram,not,to,scale))
)
}}}
x = horizontal width of the rectangle
y = vertical height of the rectangle
These are positive real numbers
x > 0 and y > 0


area of the rectangle = x*y = 150
Solve for y to get
y = 150/x


Let's focus on triangle ADC.
This is a right triangle.
To see a proof why angle ADC = 90 degrees, search out "Thale's Theorem". 


Because ADC is a right triangle, we can use the pythagorean theorem to say:
{{{a^2 + b^2 = c^2}}}


{{{(DC)^2 + (AD)^2 = (AC)^2}}}


{{{x^2 + y^2 = 20^2}}}


{{{x^2 + (150/x)^2 = 400}}}


Let's solve for x.
{{{x^2 + (150/x)^2 = 400}}}


{{{x^2 + 22500/(x^2) = 400}}}


{{{x^4 + 22500 = 400x^2}}}


{{{x^4 - 400x^2 + 22500 = 0}}}


As a detour, let {{{w = x^2}}}
So 
{{{w^2 = (x^2)^2 = x^4}}}
and
{{{x^4 - 400x^2 + 22500 = 0}}}
{{{w^2 - 400w + 22500 = 0}}}


Use the quadratic formula to solve for w.
Plugging in a = 1, b = -400, c = 22500
{{{w = (-b +- sqrt(b^2 - 4ac))/(2a)}}}


{{{w = (-(-400) +- sqrt((-400)^2 - 4(1)(22500)))/(2(1))}}}


{{{w = (400 +- sqrt(160000-90000))/(2(1))}}}


{{{w = (400 +- sqrt(70000))/(2)}}}


{{{w = (400 +- 100*sqrt(7))/(2)}}}


{{{w = (2(200 +- 50*sqrt(7)))/(2)}}}


{{{w = 200 +- 50*sqrt(7)}}}


{{{w = 200 + 50*sqrt(7) }}} or {{{ w = 200 - 50*sqrt(7)}}}


{{{w = 332.287566 }}} (approximate) or {{{ w = 67.712434}}} (approximate)



If w = 332.287566, then,
{{{w = x^2}}}
{{{x^2 = 332.287566}}}
{{{x = sqrt(332.287566)}}} or {{{x = -sqrt(332.287566)}}}
{{{x = 18.228757}}} or {{{x = -18.228757}}}
We'll ignore the negative x value because a negative width makes no sense.


Now use that x value to find y
{{{y = 150/x}}}
{{{y = 150/18.228757}}}
{{{y = 8.228756}}}


The rectangle has dimensions of:
x = 18.228757
y = 8.228756



If w = 67.712434, then,
{{{w = x^2}}}
{{{x^2 = w}}}
{{{x^2 = 67.712434}}}
{{{x = sqrt(67.712434)}}}
{{{x = 8.228757}}}
Then plugging that into y = 150/x will lead to y = 18.228757
We arrive at the same dimensions as before, but now the x and y values swapped places.
Turns out the order doesn't matter when listing the length and width of a rectangle. 


Therefore, the only possible dimensions of the rectangle are roughly <font color=red size=4>8.23 inches by 18.23 inches</font> when rounding those previous figures to two decimal places.


As a check
x*y = 18.228757 * 8.228756 = 149.999994
which is really close to 150
There's likely some rounding error from a previous step.
And also, we can use the pythagorean theorem to confirm the answer as well
a^2+b^2 = c^2
x^2 + y^2 = 20^2
(18.228757)^2 + (8.228756)^2 = 20^2
400.000007072584 = 400
which isn't too far off either
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