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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
avatar+106516 
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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

 

 

cool cool cool

 Dec 24, 2019
 #2
avatar+141 
0

Is it possible to inscribe the ellipse in that shaded rhombus, and what would be the area of it?

 Dec 24, 2019

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