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picture1_Ssc Cgl Notes 120941 | Circle


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File: Ssc Cgl Notes 120941 | Circle
circle notes for ssc cgl and chsl circle sector l length of arc ab r radius always in radian 180 ic c 180 c i 180 length of arc 2 ...

icon picture PDF Filetype PDF | Posted on 08 Oct 2022 | 3 years ago
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                                                                            Circle Notes for SSC CGL and CHSL  
               
              ➢          Circle :→ 
                         Sector :- 
                                      l                                                                                             
                         ⇒θ= →length of arc AB 
                                      r→radius
                              ↓
                              always in Radian
                                  180°
                         IC =                
                                      π
                           C
                         π =180° 
                                      C
                         I° = π  
                                  180
                         ⇒Length of arc = 2πr θ   
                                                                     360°
                                                                                   θ
                         ⇒Area of sector OAB = πr2                                        
                                                                                360°
                         ⇒Perimeter of sector = πr θ +2r 
                                                                              180°
               
              Segment :- 
                                                                                                                             
              → Area of segment = area of sector OACB – area of ∆OAB  
              =πr2 θ −1r2sinθ  
                           360°        2
              → Perimeter = length of arc ACB + Chord length AB 
                                 θ                       θ
                   (        )
              = 2πr 360°+2rsin(2)  
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         Q1. Find the area of a segment of a circle with a central angle of 120 degrees and a radius of 8 cm.  
         Sol. Area of segment = πr2 θ −1r2sinθ 
                                           360°    2
         = π(8)2120° −1(8)2120° 
                   360°    2
         = 83.047 
          
         Q2. Find the area of a sector with an arc length of 30 cm and a radius of 10 cm. 
         Sol. Length of arc = 2πr θ  =30 
                                        360°
         πr θ  =15 
             360°
         Area of sector OAB = πr2 θ  = (πr θ ) r = 15 × 10 = 150 cm 
                                         360°        360°
         Q3. In a circle of radius 21 cm and arc subtends an angle of 72 at centre. The length of arc is? 
         Sol. Length of arc = 2πr  θ   
                                         360°
         = 2 π  × 21  ×  72°  = 26.4 cm 
                           360°
         Important Properties Of Circle : - 
          
         ➢      Perpendicular from the centre of a circle to a chord bisects the chord. 
                AM = MB                                                                 
         Q1. AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the center of the circle. 
         The distance between them is 1 cm. Find the length of the radius? 
         Sol.  
         Let ON = x , AO = r             
         In triangle AOE 
          2                2 
         r = 16 + (x-1)  
         In triangle OCN  
          2         2
         r  = 9 +x   
                      2        2
         16 + (x-1)  = 9 +x  
         x=4  
          2
         r  = 9 +16, r = 5 cm 
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      ➢   Chords corresponding to equal arcs are equal. 
            ̂ ̂                                           
          If AB = CD,  then chord , AB = CD       
      ➢   Equal Chords of Circle Subtends equal angles at the centre. 
          If AB = CD                                      
          then ∠1 = ∠2                            
      ➢   Equal chords of a circle are equidistance from the centre. 
          If AB = CD, Then OX = OY                         
      ➢   The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any 
          point on  the remaining part of the circle. 
          x = 2y                                          
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        Q1.   The length of chord of a circle is equal to the radius of the circle .The angle which this chord 
              subtends in the major segment of the circle is equal to? 
        Sol.  
              OA = OB = r                
              AB is equal to radius 
              Therefore triangle OAB is an equilateral triangle  
              Angle OAB = 60°                                             60°
              Angle ACB, angle which chord subtends at major angle =  2  = 30° 
               
        ➢     Angle in same segment of a circle are equal. 
                                                            ∠1 = ∠2         
        ➢     Angle in a semicircle is always a right angle.      
                                                                             
        Q1.  AC is the diameter of a circumcircle of triangle ABC. Chord 
              ED is parallel to the diameter AC. If Angle CBE = 50°, then 
              the measure of angle DEC is? 
        Sol. 
                                         
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