Farmer Pythagoras has a field in the shape of a right triangle.

From the image attached, we have found diagonal of square=(2/5)
2(side^2)=(diagonal^2) (In a square, relation between diagonal and side)
side^2= area of square S =2/25
total area of the field=(1/2)*4*3=6 ((1/2*base*height))
area planted=6-(2/25)=148/25
fraction of the field planed=(148/25)/6=74/75
IMO:OPTION E

Let the side of square S be x. Hypotenuse is 5 from Pythagoras theorem. Draw lines from top right upper vertex of S to two end points of the hypotenuse of the triangle. So, now we have divided this into square S + 1 thin triangle with base x and height (3-x) + one thin triangle with base x and height (4-x) + one triangle with base 5 and height 2. All this must add up to total area of 0.5*3*4 = 6

Hence, \(x^2 + 0.5*x*(3-x) + 0.5*x*(4-x) + 0.5*2*5 = 6\), which gives

\(x^2 + 1.5*x - 0.5*x^2 + 2*x - 0.5*x^2 + 5 = 6\)

\(3.5*x = 1\) or \((7/2)*x = 1\) or x = 2/7

Area of S = \((2/7)^2\) = 4/49

Area of planted region = 6 - (4/49)

Fraction needed = [6 - (4/49)]/6 = 1 - \(\frac{4}{(49*6)}\) = 1 - \(\frac{2}{147}\) = 145/147

Hence Option D

Notice that the incorrect responses assume that the diagonal of the square is the continuation of the perpendicular from the hypotenuse.

The correct responses avoid this assumption by terminating the perpendicular at the vertex of box S and then dealing with the resulting triangles.

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