Two triangles have integral side lengths, with all sides being less than 40 units. All the angles of first triangle are same as that of second one. But their areas are not same. Further the smaller triangle has two side lengths identical with the larger triangle. Find the difference in perimeter of the two triangles.

A. 27

B. 19

C. 12

D. 8

Let the corresponding sides of the two similar triangles are: {a, b, c} & {b, c, d} in that order so that two side lengths of the smaller triangle are identical with the larger triangle.

As triangles are similar, we have., a/b = b/c = c/d i.e. a, b, c, d are in GP.

Also common ratio of this GP should be less than 2 i.e. < 2 so that a, b, c forms the three sides of a triangle.

Now smallest case possible is that b/a = 3/2 , so that four numbers are: a, 3a/2 , 9a/4 , 27a/8 .

If we take a = 8, then a, b, c, d are 8, 12, 18, 27. It can be easily seen that this is the only case such that all the sides are smaller than 40.

Hence required difference in perimeter is = (12+18+27) – (8+12+18) = 19.

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