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In a trapezoid ABCD with AB parallel to CD, the diagonals AC and BD intersect at E. If the area of triangle ABE is 50 units, and the area of triangle ADE is 20 square units, what is the area of trapezoid ABCD?

Mathgenius  Aug 26, 2018

Best Answer 

 #2
avatar+91027 
+2

This one was a little difficult...!!!

Note : angle AEB  = angle DEC

and angle CDB  = angle DBA

So...by AA congruency....ΔBEA ~ ΔDEC  →  DC /AB  = DE / BE

 

Area of  ΔDBA  = area of  ΔADE  + area of ΔABE  = 20 + 50  = 70

So  area of Δ DBA  = (1/2)sin DBA(AB * DB)  = 70

So...sin DBA = 140 / ( AB * DB)

 

And sin (DBA) = sin CDB

So area of  ΔCDB  = (1/2) sin (DBA)(DB * DC)  = (1/2) (140 / [(AB * DB] ) (DB * DC) =

70 (DC/AB)  = 70 ( DE /BE)

 

Now

Area ΔAEB  = (1/2)sin AEB ( AE * BE) = 50   (1)

Area Δ ADE  = (1/2) sin DEA (AE * DE)  = 20  (2)

But  angle AEB  is  supplemental to angle DEA  so thier sines are equal

So  sin AEB  = sinDEA

 

Rearranging (1), (2)  we have that

 

sin AEB  = 100 / [ AE * BE]      =     sin DEA = 40 / [ AE * DE ]

 So

100 / BE   = 40 / DE  →   DE / BE  = 40/100  = 2/5

 

But area of ΔCDB  was found to be  70 (DE / BE) = 70 (2/5)  =  28

 

So...area of trapezoid ABCD  = area of ΔDBA  + area of ΔCDB  =

                                                               70        +        28             =      98

 

 

cool cool cool

CPhill  Aug 26, 2018
edited by CPhill  Aug 26, 2018
edited by CPhill  Aug 26, 2018
 #1
avatar
+1

Should be 98, I think...

Guest Aug 26, 2018
 #2
avatar+91027 
+2
Best Answer

This one was a little difficult...!!!

Note : angle AEB  = angle DEC

and angle CDB  = angle DBA

So...by AA congruency....ΔBEA ~ ΔDEC  →  DC /AB  = DE / BE

 

Area of  ΔDBA  = area of  ΔADE  + area of ΔABE  = 20 + 50  = 70

So  area of Δ DBA  = (1/2)sin DBA(AB * DB)  = 70

So...sin DBA = 140 / ( AB * DB)

 

And sin (DBA) = sin CDB

So area of  ΔCDB  = (1/2) sin (DBA)(DB * DC)  = (1/2) (140 / [(AB * DB] ) (DB * DC) =

70 (DC/AB)  = 70 ( DE /BE)

 

Now

Area ΔAEB  = (1/2)sin AEB ( AE * BE) = 50   (1)

Area Δ ADE  = (1/2) sin DEA (AE * DE)  = 20  (2)

But  angle AEB  is  supplemental to angle DEA  so thier sines are equal

So  sin AEB  = sinDEA

 

Rearranging (1), (2)  we have that

 

sin AEB  = 100 / [ AE * BE]      =     sin DEA = 40 / [ AE * DE ]

 So

100 / BE   = 40 / DE  →   DE / BE  = 40/100  = 2/5

 

But area of ΔCDB  was found to be  70 (DE / BE) = 70 (2/5)  =  28

 

So...area of trapezoid ABCD  = area of ΔDBA  + area of ΔCDB  =

                                                               70        +        28             =      98

 

 

cool cool cool

CPhill  Aug 26, 2018
edited by CPhill  Aug 26, 2018
edited by CPhill  Aug 26, 2018
 #3
avatar+91027 
+2

Here's a pic  (not to scale) :

 

 

 

cool cool cool

CPhill  Aug 26, 2018

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