Question

ABCD is a rectangle and N is the midpoint of AB and F is the midpoint of DA. If DA is produced which meets CN produced at M, then the ratio of the region FNM to the are of the rectangle ABCD is:

Answer 1

Answer:

Let the side of the square ABCD be 2x.

P being the mid-point of AB, BP = x.

Q being the mid-point of BC, BQ = x.

Thus PQ = x*2^0.5.

BD is the diagonal of the square ABCD and its length = 2x*2^2.

Let PQ intersect BD at R. RB = PQ/2 = x*2^0.5/2

Length DR = BD - RB = 2x*2^2 - x*2^0.5/2 = 3x/2^0.5

Area of triangle PQD = PQ *DR/2 = [x*2^0.5/2]*[3x/2^0.5]/2= 1.5*x^2.

Area of square ABCD = 4x^2.

So the Area of triangle PQD : Area of square ABCD =1.5*x^2:4x^2=0.375 or 37.5 %

Step-by-step explanation:

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