There is a circle of radius 1 cm. Each member of a sequence of regular polygons S1(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is circumscribing the circle: and each member of the sequence of regular polygons S2(n), n = 4, 5, 6, … here n is the number of sides of

Question

There is a circle of radius 1 cm. Each member of a sequence of regular polygons S1(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is circumscribing the circle: and each member of the sequence of regular polygons S2(n), n = 4, 5, 6, … here n is the number of sides of the polygon, is inscribed in the circle. Let L1(n) and L2(n) denote the perimeters of the corresponding
polygons of S1(n) and S2(n), then $\frac{{L 1 (13) + 2 π}}{L 2 (17)}$ is

Accepted Answer

greater than 2 — Following rule should be used in this case: The perimeter of any polygon circumscribed about a circle is always greater than the circumference of the circle and the perimeter of any polygon inscribed in a circle is always less than the circumference of the circle. Since, the circles is of radius 1, its circumference will be 2 &pi; . Hence, L1(13) > 2&pi; and L2(17) < 2&pi;. So {L1(13) + 2 &pi; } > 4 &pi; . Hence, { L 1 ( 13 ) + 2 &pi; } L 2 ( 17 ) will be greater than 2. Hence, option (c).

CAT 1999 — QA Question 17

Answer & solution