AT=(1/2)(a/2)((3a/2)+x) and AQ=(1/2)(a/2)(3a/2)+(a/2)(3a/2)-(1/2)(x)(a/2) where :

AR=area of the rectangular paper , AU=area of the triangle , AQ=area of the quadrilateral

a=shorter side of the rectangle and x=the difference between one side of the triangle and the half of the longer side of the rectangle

Hence we get .

a^2=(8AT-2ax)/3=(8AQ+2ax)/9 as well as AQ=3AT-ax and taking a=16,17,18… and x=1,2,3,… we can get to the solution by trial and error.

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# If the paper is x by 3.x, the areas of the triangles and the quadrilaterals # are then 5.x^2 / 12 and 13.x2 / 12 respectively; these are integer when x # is 6.k for integer k so they become 15.k^2 and 39.k^2 and for each of these # to be 3 digit values reqires k to be either 3, 4 or 5 for k in (3, 4, 5): a, b = 15 * k ** 2, 39 * k ** 2 if set(str(a)) == set(str(b)): print('The area of each triangle is {} square centimetres.'.format(a)) |