Question

Plz keep the detailed solution of 1,2,3,4,5,6,7,8,9th problem (Suggested to keep pics) keep it as fast as possible??????

Answer 1

1) ∠ACB = ∠AOB / 2 = 50°   (as the arc ADB makes twice the angle at the center compared to the angle it makes at the circle).
    ACBD is a cyclic quadrilateral.  Hence, ∠ACB + ∠ADB = 180°  (sum of opposite angles).
    ∠ADB = 180° - 50° = 130°

2) 
   This exercise can be solved in two ways.
   Arc BD makes equal angles in the major segment BACD.
   ∠BAD = ∠BCD = 40°
   ANother way:  ΔABO, ΔODC are congruent. (SAS)  OD= OC=OB=OA.
    ∠AOB = ∠COD.  ∠ODC = ∠OAB = 40°

3)
   ∠PQR = ∠POR /2 = 60°  (angles at the center and at the circle by PSR.
   ∠PSR = 180 - ∠PQR = 120°  : cyclic quadrilateral.

4) In a parallelogram ABCD, ∠A = ∠C,   ∠B = ∠D.
    In a cyclic quadrilateral,  sum of opposite angles = 180°
     So  ∠A + ∠C = 180°  = ∠B + ∠D
          =>   ∠A = ∠C = ∠B = ∠D = 90°

5) It seems that OM ⊥ MB.  Then OM bisects AB.  So OM = 8/2 = 4 cm
In the ΔOMB, (by Pythagoras theorem)
         OB² = OM² + MB² = 3² + 4² = 25 
         radius = 5 cm

6)   The distances OM & ON from the center O to the chords RS and PQ are equal.  Hence, the lengths of the chords are also equal.
              RS = PQ = 6 cm.

   Another way: OQ² = OM² + MQ² = ON² + (PQ/2)² 
                         OR² = ON² + NR² = ON² + (RS/2)²
       OQ = OR = radius
          so PQ = RS.

7.  ABCD is a square.  Hence the diagonals are equal.
       radius = AC = BD = 4 cm

8.   Draw a circle with radius of 4 cm and center O.   Draw a diameter AOB. Then draw perpendicular bisectors of AO and OB.  These perpendicular bisectors are equal chords at equal distances from O.

9. 
    In ΔAOB, ΔCOD :  radius = OA = OB = OC = OD.     AB = CD.
    By  SSS method, ΔAOB and ΔCOD are congruent.
    ∠AOB = ∠COD = 70°
    ΔCOD is isosceles.   ∠OCD = ∠ODC 
    ∠OCD + ∠ ODC + 70° = 180°
    ∠OCD = ∠ODC = 65°