text: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:(i) Minor segment(ii) Major sector (Use π = 3.14)
  
  Let AB be the chord of the circle subtending 90° angle at centre O of the circle.
  
  Area of major sector OADB = () * r2
  
  = () r2
  
  = * 3.14 * 10 * 10
  
  = 235.5 cm2
  
  Area of minor sector OACB = * r2
  
  = * 3.14 * 10 * 10
  
  = 78.5 cm2
  
  Area of ΔOAB = * OA * OB
  
  = * 10 * 10
  
  = 50 cm2
  
  Area of minor segment ACB = Area of minor sector OACB -Area of ΔOAB
  
  = 78.5 - 50
  
  = 28.5 cm2
  
  9
metadata: {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}