Question

ABCD is a parallelogram. E and F are points on side AB such that AE=EF=FB. show that ar (∆DAE) =1/6 ar (ABCD).

Answer 1

solution:-

Here, Height of parallelogram And triangles DAE, EDF , ECF and FCB remain same = h , As all these triangles and parallelogram lies between same parallel lines PQ | | RS .

Also given AE = EF = FB ,

SO,

AB = AE + EF + FB

AB = AE + AE + AE

AB = 3 AE .

We know area of parallelogram = Base Height
So,
Area of parallelogram ABCD = AB × h ---------- ( 1 )

And
We know Area of triangle = 1/2×Base×Height

So,

Area of triangle DAE = 1/2×AE× h , Substitute value from AB = 3 AE , we get

Area of triangle DAE = 1/2×AB/3×h

Area of triangle DAE = 1/6×AB×h , Substitute value from equation 1 , we get

Area of triangle DAE = 1/6AreaofparallelogramABCD
( Hence proved )

Answer 2

ABCD is a parallelogram.

The side AB of parallelogram is divided into three equal parts. According to Area Theorem the area of parallelogram and the triangle drawn on the same base and between the same parallel line have a ratio of 2:1 that area of the triangle is half of that of the parallelogram.

Ok so triangle DAE is drawn on AE as base therefore its area is half of the area of parallelogram drawn on AE. The whole parallelogram is made of three such parallelograms. Therefore the area of triangle
DAE = 1/3 X 1/2 X ar(ABCD) = 1/6 X ar(ABCD).