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# Help

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Let ABCD and FBED be two 2 x 3 rectangles, as shown. The part where the two rectangles overlap is shaded. Find the area of the shaded region. Dec 24, 2019

#1
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The shaded area forms a rhombus

The long diagonal  =   √[2^2 + 3^2 ] = √13   units

The   center of the  rhombus  is  ( 1, 3/2)

The equation of the line  containing the long diagonal is

y = (3/2)x

The line perpendicular to this one has the equation

y = (-2/3) (x - 1)  +  3/2

y = (-2/3)x + 2/3 +3/2

y = (-2/3)x +  13/6

So    point F is     (0, 13/6)

And the distance  between F  and the center of the rhombus  =  sqrt  [ (0- 1)^2  +  ( 13/6 - 3/2)^2  ]  =

√ [1 +  (2/3)^2 ]  =  √  [ 9 +4 ] / 3  =  √13 / 3

So   the length of the other diagonal  =   (2/3)√13

So......the shaded area  =   (1/2) product of the diagonal lengths =  (1/2) *√13 * (2/3)√13  =   (13 / 3 )  units^2   Dec 24, 2019
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Is it possible to inscribe the ellipse in that shaded rhombus, and what would be the area of it?

Dec 24, 2019