NCERT Class 9 Maths Solutions Chapter - 10 Circles, Ex - 10.4

Ex - 10.4

Question 1.  Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution

Let radius of circle centered at O and O' be 5 cm and 3 cm respectively.
    OA = OB = 5 cm
    O'A = O'B = 3 cm
    OO' will be the perpendicular bisector of chord AB.
     AC = CB

    Given that OO' = 4 cm
    Let OC be x. so, O'C will be 4 - x
    In OAC
    OA² = AC² + OC² 
     5² = AC² + x²
     25 - x² = AC²            ... (1)
    In O'AC
    O'A² = AC² + O'C²
     3² = AC² + (4 - x)²
     9 = AC² + 16 + x² - 8x
     AC² = - x² - 7 + 8x        ... (2)
From equations (1) and (2), we have
    25 - x² = - x² - 7 + 8x
           8x = 32
             x = 4
So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.

Now, AC² = 25 - x² = 25 - 4² = 25 - 16 = 9
 AC = 3 m

The length of the common chord AB = 2 AC = (2  3) m = 6 m

Question 2.  If two equal chords of a circle intersect within the circle, prove that the  segments of one chord are equal to corresponding segments of the other  chord.

Solution

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

 Draw perpendiculars OV and OU on these chords.
    In OVT and OUT
    OV = OU                                (Equal chords of a circle are equidistant from the centre)
   OVT = OUT                     (Each 90^o)
   OT = OT                                (common)

OVT  OUT              (RHS congruence rule)

 VT = UT                             (by CPCT)        ... (1)

    It is given that 
    PQ = RS                                           ... ... ... ... (2)

 PV = RU                                    ... ... ...  ... (3)
    On adding equations (1) and (3), we have
    PV + VT = RU + UT
 PT = RT                                    ... ... ...   ... (4)
    On subtracting equation (4) from equation (2), we have
    PQ - PT = RS - RT
 QT = ST                                      ... ... ... ... (5)
    Equations (4) and (5) shows that the corresponding segments of
    chords PQ and RS are congruent to each other.

Question 3.  If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution

 Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.
     Draw perpendiculars OV and OU on these chords.
     In OVT and OUT
     OV = OU                           (Equal chords of a circle are equidistant from the centre)
    OVT = OUT                 (Each 90^o)
    OT = OT                            (common)
OVT OUT            (RHS congruence rule)
OTV = OTU                 (by CPCT)    

    Hence, the line joining the point of intersection to the centre makes equal angles with the chords.

Question 4.  If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.

Solution

Let us draw a perpendicular OM on line AD.

 Here, BC is chord of smaller circle and AD is chord of bigger circle.
    We know that the perpendicular drawn from centre of circle bisects the chord.
 BM = MC             ... (1)

     And AM = MD      ... (2)
     Subtracting equations (2) from (1), we have
     AM - BM = MD - MC
 AB = CD

Question 5.  Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Solution

Draw perpendiculars OA and OB on RS and SM respectively.
Let R, S and M be the position of Reshma, Salma and Mandip respectively.

 AR = AS =  = 3cm

   OR = OS = OM = 5 m     (radii of circle)
    In OAR
    OA² + AR² = OR²            
    OA² + (3 m)² = (5 m)²
    OA² = (25 - 9) m² = 16 m²
    OA = 4 m                             
    We know that in an isosceles triangle altitude divides the base, so in RSM
    RCS will be of 90^o and RC = CM           
    Area of ORS =   OARS    

    
                      RC = 4.8

    RM = 2RC = 2(4.8)= 9.6   

    So, distance between Reshma and Mandip is 9.6 m.

Question 6.  A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Solution

Given that AS = SD = DA
             So, ASD is a equilateral triangle 
             OA (radius) = 20 m.
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC. 
We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

  AB = OA + OB = (20 + 10) m = 30 m.

    In ABD

    AD² = AB² + BD²
    AD² = (30)² +   

       
    So, length of string of each phone will be  m.

Ex-10.3

Ex-10.5