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Trapezoid ABCD has base AB = 20 units and base CD = 30 units. Diagonals AC and BD intersect at X. If the area of trapezoid ABCD is 300 square units, what is the area of triangle BXC?

 

thank

Guest Mar 23, 2018
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We can find the height of the ttrapezoid thusly

 

300  =  h ( 30 + 20) /2

300 = h * 25

!2  = h

 

Let A  = (5,12)   B  = (25,12)  C  = (30,0)  and D  = (0,0)

 

 

The area  of triangle DBC  = (1/2) CD * 12  =  (1/2) (30) * 12  =  180  units^2     (1)

 

The  equation  of line segment BD  is

y  =(12/25)x

 

The y coordinate  of X   is the height of triangle CXD and   can be found as

 

y  =  (12/25) (15)   = 3/5 * 12  = 7.2

 

So....the area of triangle  CXD  = (1/2)(CD)(7.2)  = (1/2)(30)(7.2)  = 108 units^2     (2)

 

So...the area  of triangle BXC  =  (1)  - (2)  =  180  - 108  =  72 units^2

 

Here's a pic :

 

 

cool cool cool

CPhill  Mar 23, 2018

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