what does it mean the red part Sir.?

it is unclear

Let AB=DE=x and
AE=BD=y
Area of triangle ABE=xy/2.

since, BC || DF and BF||CD and BC=CD
so BCFD either square or rhombus
so CF is perpendicular to BD( or AE)
CF||DE and CD||FE
so CF=DE=x
so Area of BCDF=(diagonal1 x diagonal2)/2=xy/2
So Quantity A=Quantiy B
Answer: C

This is quite interesting:

Since AB=ED and they form right angles with the base AE, then have a rectangle and BD=AE ---------- (1)

Also, one must recognize that parallel lines will form alternate anterior angles and since we have BC=CD, then similarly BF=DF

Traiangles BCD and BFD are equal ---------- (2)

Note that F is the center of the rectangle. Therefore the length CF is equal to AB and ED ---------- (3)

Now the area of triangle ABE = 1/2*AE*AB

The area of the quadrilateral BCDF = BCD+BFD = 2*1/2*CF*BD using findings of (1) and (3), we can replace CF with AB and BD with AE

Answer is (C)

Visualization work best here!

There is no enough info to prove ABDE is a square or 30-60-90 for any triangles.

It is easy to prove that 1) BCDF and CDEF are parallelogram, 2) BF = FE (using midsegment).
So within trapezoid BCDE, it is easy to see that area BCDF equal to the area of triangle BDE.
And it is clear and area BDE equal to area ABE
Therefore BCDF = ABE.

Sir, can you explain this part?

No disrespect but I found all the solutions unnecessarily complicated or just invalid hence thought of posting my approach

From the given statements AB=DE, AB||DE, AB perpendicular AE
we can conclude that quad ABDE is a rectangle, and hence AE=BD

From the given statements BE||CD, BC||DF, BC=CD
we can conclude that quad BCDF is a rhombus

Lets say the diagonals of the rhombus BCDF bisect each other at point M on line BD

From above statement and property - the diagonals of a rhombus are perpendicular bisectors of one another,
we can conclude that CM=FM

From above statement if we extend line CF to line AE such that it intersects at point N
we conclude that FN is perpendicular bisector of AE

Quad ABDE is a rectangle and the diagonals bisect at a point equidistant from the opposite sides of the rectangle
from this we conclude that MF=NF

Thus CM=FM=FN,
CF = CM + MF
MN = MF + FN
MN = AB
AB=CF

Area of rhombus is half the product of its diagonals = 1/2*CF*BD

Area of triangle is half the product of its base and height = 1/2*AB*AE = 1/2*CF*BD

Hence, the answer is option C

Official solution for this question please

OE

Because of the various parallel and equal length constraints, you can see that quadrilateral BCD F is a rhombus with side lengths equal to half diagonal BE. In fact, the whole shape can be split into 10 identical right triangles, as shown:



A is The area of triangle ABE - 4 small triangles

B is The area of quadrilateral BCDF - 4 small triangles

C is the answer

Note on strategy: If you were running out of time and wanted to make a guess, you could make a reasonable guess here.

The amount of information they provide suggests that the figure is actually drawn to scale. If you see this, you can visualize the two areas, by seeing how many triangles of the same size they have in common. It looks basically equal to my eye, so I guessed C.

Most of the time you can't make visual guesses like this, but the amount of info made me think, there is really no way for me to change how this looks without violating one of the conditions they gave me.

I think it's a good quick reasoning, I did something similar (not exactly same) to get same answer.

Draw a line from B to D. Now ABDE is a rectangle as AB = DE (given) and angle BAE and DEA are right angles (given).
BC = CD and BC || DF and BF || CD. This means BCDF is not only a parallelogram but also a rhombus (all sides have to be equal and opposite sides parallel).
The diagonal BD of rhombus BCDF is equal to BD
Another diagonal CF of BCDF is equal to AB as half of CF is equal to half of AB (diagonals of a rhombus bisect each other -> CF is cut into two halves. one half inside the rectangle starting from midpoint of BD to center of ABDE).

Hence area of rhombus BCDF =1/2 * diagonal 1 * diagonal 2 = 1/2 * BD * AB
Area of triangle ABE = 1/2* base * height = 1/2 * AB * BD

They are both equal, hence C.

BF and FD will be the same in length. Draw an imaginary line connecting B and D. make the midpoint as S. On connecting F and S, we get a triangle BFS. Now, BFS will be similar to triangle BAE.

Area of the quad will be 1/2 * AE * AB/2 * 2 = 1/2 * AB* AE which is also equal to the area of triangle BAE.