Question

Prove that the perimeter of a square is less than that of an equivalent parallelogram on the same Base.

Answer 1

★ GEOMETRIC RESOLUTION ★

Let's assume a geometric system of a square and a Parallelogram having the same base ,

Having the equivalent side of the square be " a " Then the respective two parallel side of the Parallelogram will also be equal to " a "

Lets rotate the one of the parallel lines with angle Φ clockwise , then then the side lengths which aren't equal and parallel to the first set will get increased and it's incremental rate will be responsible for the increase in perimeter of the figures aslike , perimeter is the sum of the sides of the figure ,

Now lets rotate alternate side length at maximum at angle θ , aslike it should touch the square at diagonally opposite angles ,

calculation of perimeters of both the geometric figures -

Square of side length " a " = a + a + a + a = 4a

Parallelogram of two base side length " a " and two parallel set lines of length √2a = a + a + √2a + √2a = 2a + 2√2a = 2a [ 1 + √2 ]

Clearly , the acquired perimeter of Parallelogram is greater than the that of square

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