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?
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
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