Area of the trapeziod

Since BE is ll to CD that makes triangle ABE and triangle ACD similar triangles.

Hence AB/AC = AE / AD = BE / CD

now we can get AB=3, AE = 4, BE=5
AC= 6, CD = 10, AD = 8.

Area of triangle ABE = sqrt(s*(s-a)*(s-b)*(s-c)) Where s=(a+b+c)/2.

Solving this we will get

Area of ABE= sqrt 36 = 6
Area of ACD = sqrt 672 = 24

Hence area of Trapezoid BEDC= 18

Ans: B

Hi,

Mid Point Theorem can be used to solve the above problem:

The straight line joining the mid-points of two sides of a triangle is parallel to and equal to half the third side.

Converse of Mid-Point Theorem
The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side


Hope it is clear...

I have solved it another way...if there is any problem with my approach then please let me know.

AB = BC = 3
Thus AC = 6.

On the other hand, since BE II CD and B is the mid point of AC, E should be the mid point of AD.

Since AE = ED = 4, AD = 8.

Thus the triangle ACD has arms of 6:8:10 and must be a right triangle. LCAD = 90 deg.

Its area should be: 1/2*AC*AD = 24.

On the other hand, area of triangle ABE is : 1/2*AB*AE = 6.

So the difference (answer) = 18 (B)

Hi Want to beat...

I think there is no flaw in the way you solved the problem and I would like to appreciate the solution... it is really simple and good.