pclawong wrote:

explain please

This is a tough ques, let me try

consider AB=DE =x

now join Band D

in the quadrilateral ABDE we have angle A =90 degree

angle E = 90 degree

and AB=DE = x

So quadrilateral ABDE is a square.

and Area of square ABDE = x^2

Now we need the area of triangle ABE = Area of square ABDE - area of triangle BDE - area of triangle DEF

Area of triangle BDF = 1/2 * base * altitude =

\frac{1}{2} * x *\frac{1}{2}*x(since side BD =x and 1/2 * total distance = altitude of triangle BDF, as BDEA is square)

and similarly Area of Triangle DEF =

\frac{1}{2} *x * \frac{1}{2}*xTherefore Area of Triangle ABE =

x^2 - (

\frac{1}{4}*x^2 +

\frac{1}{4}*x^2)

=

\frac{1}{2} *x^2Now in quadrilaterl BCDF we have

BC=CD , BC parallel to FD and BE parallel to CD and angle F =90degree (since the diagonal cuts at 90 degree )

Therefore we can consider BCDF is a square

Area of square BCDF=

\frac{diagonal ^2}{2} =

\frac{1}{2} *x^2Hence option C.

All queries are welcome!!!

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