a. Area of sector MON = \(\frac{MON}{360} \times\pi r^2\) = \(\frac{60}{360}\times\frac{22}{7}\times7^2\) = \(\frac{77}{3}cm^2\) \(\frac{NT}{7}\) = sin60^\circ NT = 7sin60^\circ = \(7\times\frac{\sqrt{3}}{2}\) = \(7\times\frac{\sqrt{3}}{2}\)cm. \(\frac{OT}{9} = cos60^0\) OT = 7cos60^\circ = \(7\frac{1}{2}\)cm Area of △ONT = \(\frac{1}{2}\times OT\times NT\) = \(\frac{1}{2}\times\frac{7}{2}\times\frac{7\sqrt{3}}{2} = 49\sqrt{3}{8}cm^2\) Area of shaded portion = Area of sector MON - Area of △ONT = \(\frac{77}{3} - \frac{49\sqrt{3}}{8}\) = 25.67 - 10.61 = \(15.06cm^2\) = 15.1\(cm^2\) b. Given : y = mx + c; hence ; x = \(\frac{ y - c }{ m } = \frac{ y }{ m } - \frac{ c }{ m }\) x-intercept = \(\frac{ - c }{ m } = \frac{ - 3 }{ 4 }\) Therefore, m = \(\frac{4}{3}\)c ....................... (1) But y-intercept, c = \(\frac{2}{7}\) Therefore, m = \(\frac{4}{3}\times\frac{2}{7} = \frac{8}{21}\) Equation of straight: y = mx + c y = \(\frac{8}{21}x + \frac{2}{7}\) OR 21y = 8x + 6 = 8x - 21y = - 6