NCERT Solutions Class 9 Maths Chapter 10 Circles Ex 10.4

NCERT Solutions Class 9 Maths 
Chapter – 10 (Circles) 

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 10 Circles Exercise 10.4 has been provided here to help the students in solving the questions from this exercise.

Chapter 10: Circles

• NCERT Solution Class 9 Maths Ex – 10.1
• NCERT Solution Class 9 Maths Ex – 10.2
• NCERT Solution Class 9 Maths Ex – 10.3
• NCERT Solution Class 9 Maths Ex – 10.5
• NCERT Solution Class 9 Maths Ex – 10.6

Exercise – 10.4

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.
Answer – The perpendicular bisector of the common chord passes through the centers of both circles.

Given that the circles intersect at two points, so we can draw the above figure. Let AB be the common chord. Let O and O’ be the centers of the circles, respectively.
O’A = 5 cm, OA = 3 cm
OO’ = 4 cm       (Given distance between the centres is 4cm)
Since the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of the smaller circle lies inside, the bigger circle itself.
OO’ is the perpendicular bisector of AB.
So, OA = OB = 3 cm
AB = 3 cm + 3 cm = 6 cm (Since, O is the mid point of AB)
The length of the common chord is 6 cm.
It is also evident that the common chord is the diameter of the smaller circle.

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.
Answer – Let AB and CD be the 2 equal chords. AB = CD. Let the chords intersect at point E. Join OE.

To prove AE = CE and BE = DE.
Draw perpendiculars from the center O to the chords. This Perpendicular bisects the chord AB at M and CD at N.
Thus, AM = MB = CN = DN           —————- (i)
In ∆OME and ∆ONE
∠M = ∠N = 90°
OE = OE
OM = ON (Equal chords are equidistant from the center.)
By RHS criteria, ∆OME and ∆ONE are congruent.
So by CPCT, ME = NE           —————- (ii)
We know that: CE = CN + NE and AE = AM + ME
From (i) and (ii), it is evident CE = AE
DE = CD – CE and BE = AB – AE
AB and CD are equal, CE and AE are equal. So, DE and BE are also equal. It is proved corresponding segments of equal chords are equal.

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.
Answer – Let AB and CD be the two equal chords. AB = CD.

Let the chords intersect at point E. Join OE.
Draw perpendiculars from the center O to the chords. The Perpendicular bisects the chord AB at M and CD at N.
To prove: ∠OEN = ∠OEM.
In ∆OME and ∆ONE,
∠M = ∠N = 90°
OE = OE
OM = ON (Equal chords are equidistant from the center.)
By RHS criteria, ∆OME and ∆ONE are congruent. So, by CPCT, ∠OEN = ∠OEM
Hence proved that the line joining the point of intersection of two equal chords to the center makes equal angles with the chords.

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 (see Fig. 10.25).

Answer – Draw a perpendicular from the center of the circle OM to the line AD.

We can see that BC is the chord of the smaller circle, and AD is the chord of the bigger circle.
We know that perpendicular drawn from the center of the circle bisects the chord.
∴ BM = MC                   ————- (i)
and,
AM = MD                    ————- (ii)
Subtracting (ii) from (i), we obtain
AM − BM = DM − CM
∴ AB = CD

5. Three girls, Reshma, Salma and Mandip, are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Answer – Let O be the center of the circle, and R, M and S denote Reshma, Mandip, and Salma respectively.
Draw a perpendicular OA to RS from O. Then RA = AS = 3 m.

Using the Pythagoras theorem, we get OA = 4 m.
We can see that quadrilateral ORSM takes the shape of a kite. (Because OR = OM and RS = SM).
We know that the diagonals of a kite are perpendicular, and the main diagonal bisects the other diagonal.
∠RNS will be 90° and RN = NM
Area of ∆ORS = × RS × OA
= × 6 × 4
= 12                ——————- (1)
Also
Area of ∆ORS = × OS × RN
= × 5 × RN  ———————- (2)
From equation (1) and (2)
× 5 × RN = 12
RN = = 4.8m
RM = 2 × RN = 2 × 4.8 = 9.6 m
Distance between Reshma and Salma is 9.6 m.

6. A circular park of radius 20m is situated in a colony. Three boys, Ankur, Syed and David, are sitting at equal distances on its boundary, each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Answer – Centre and Centroid are the same for an equilateral triangle, and it divides the median in the ratio 2 : 1.
Let A, D, S denote the positions of Ankur, David, and Syed, respectively.

∆ADS is an equilateral triangle since all the 3 boys are equidistant from one another.
Let B denote the mid-point of DS, and hence AB is the median and perpendicular bisector of DS.
Hence ∆ABS is a right-angled triangle with ∠ABS = 90º.
O (centroid) divides the line AB in the ratio 2 : 1. So OA : OB = 2 : 1.
OA/OB = 2/1
Since OA = 20 
thus, OB = 10m
AB = OA + OB = 20 + 10 = 30m         —————(1)
Let the side of the equilateral triangle ∆ADS be 2x.
AD = DS = SA = 2x             —————(2)
Since B is the mid-point of DS, we get BS = BD = x        —————(3)
Applying Pythagoras theorem to ∆ABD, we get:
AD² = AB² + BD²
(2x)² = 30² + x²
4x² = 900 + x²
3x² = 900
x² = 300
x = 10√3
x = 17.32
AD = DS = SA = 2x = 2 × 10√3 = 20√3
AD = DS = SA = 2x = 2 × 17.32 = 34.64 m
Length of the string = Distance between them = AD or DS or SA = 34.64 m. 0r 20√3m.