Plz solve the given para

Dear Student,

T h e q u a d r i l a t e r a l O A C B i s t h e k i t e . s o t h e d i a g o n a l s a r e a t 90 d e g r e e a n d o n e o f t h e d i a g o n a l b i s e c t . A M = B M = \sin \frac{x}{2} . ∠ B C A = π - x . now from △ ABC, Tan (π / 2 - x / 2) = \frac{sinx / 2}{CM} CM = \frac{sinx / 2}{Tan (π / 2 - x / 2)} = \frac{sinx / 2}{cotx / 2} = \frac{\sin^{2} \frac{x}{2}}{\cos \frac{x}{2}} . AB = 2 AM = 2 sinx / 2 A (△ ABC) = \frac{1}{2} AB \times CM = \frac{1}{2} 2 . \sin \frac{x}{2} . \frac{\sin^{2} \frac{x}{2}}{\cos \frac{x}{2}} . = \frac{\sin^{3} \frac{x}{2}}{\cos \frac{x}{2}} . . . . . . . i T (X) = \frac{\sin^{3} \frac{x}{2}}{\cos \frac{x}{2}} = \sin^{2} \frac{x}{2} . \tan \frac{x}{2} A (segment AB - shaded region) = \frac{R^{2}}{2} [x - sinx] = \frac{1}{2} [x - sinx] . s (X) = \frac{1}{2} [x - sinx] . 1) \lim_{x \to 0} \frac{T (x)}{x^{3}} = \lim_{x \to 0} \frac{\sin^{2} \frac{x}{2} . \tan \frac{x}{2}}{x^{3}} = \frac{1}{8} \lim_{x \to 0} \frac{\sin^{2} \frac{x}{2} . \tan \frac{x}{2}}{{\frac{x}{8}}^{3}} = \frac{1}{8} \lim_{x \to 0} \frac{\sin^{2} \frac{x}{2}}{x^{2} / 4} . \frac{\tan \frac{x}{2}}{x / 2} = \frac{1}{8} (1) (1) = \frac{1}{8} . 2) \lim_{x \to 0} \frac{s (x)}{x} = \lim_{x \to 0} \frac{\frac{1}{2} [x - sinx] .}{x} = \frac{1}{2} [1 - \frac{sinx}{x}] = \frac{1}{2} (1 - 1) = 0,