The dimensions of a rectangle of maximum area inscribed in a semi circle of radius 8 cms is
- $8\sqrt{2},4\sqrt{2}$
- $8\sqrt{2},8\sqrt{2}$
- $4\sqrt{2},4\sqrt{2}$
- $4\sqrt{2},\sqrt{2}$
For a rectangle inscribed in a semicircle of radius R, the area is maximized when the width is R * sqrt(2) and height is R / sqrt(2). With R = 8, width = 8 * sqrt(2) and height = 8 / sqrt(2) = 4 * sqrt(2).
Place the semicircle of radius 8 on a coordinate plane with its diameter on the x-axis and center at the origin. Let the corners of the inscribed rectangle on the diameter be at x and -x, meaning the width is 2x and the height is y = sqrt(64 - x^2). The area is A = 2x * sqrt(64 - x^2), and maximizing it using calculus gives x = 4*sqrt(2). The dimensions are a width of 2 * 4*sqrt(2) = 8*sqrt(2) cm and a height of sqrt(64 - 32) = sqrt(32) = 4*sqrt(2) cm.