# NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 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:
Two circles with centre O, and A having radius 5 cm and 3 cm respectively.
The distance between the centres i.e. OA = 4 cm CD is the common chord. OA is perpendicular to CD and bisects CD. OA ⊥ CD and AC = AD
In ∆OAC, OC = 5 cm, OA = 4 cm.
By Pythagoras theorem
AC2 = OC2 – OA2
⇒ AC2 = 9
⇒ AC = 3 cm
Hence CD passes through the centre of a smaller circle and equal to the diameter of the circle.
CD = 2AC = 2 × 3 = 6 cm Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are corresponding segments of the other chord.
Solution:
Given: AB and CD are two equal chords of the circle which intersect each other at P. To prove that: Segment ACB ≅ Segment CBD.
Proof: We have given,
chord AB = chord CD
∴ $$\widetilde{\mathrm{AB}} \cong \widetilde{\mathrm{CD}}$$
We know that, the segment made between congruent arcs and equal chords are congruent.
∴ Segment ACB ≅ Segment CBD.

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:
Given: AB and CD are two equal chords of a circle whose centre is O. Chord AB and CD intersect each other at E. To prove that: ∠1 = ∠2
Construction: Draw OM ⊥ AB and ON ⊥ CD, and join OE.
Proof: In ∆OME and ∆ONE
OM = ON (Equal chords are equidistance from the centre)
∠OME = ∠ONE (Each 90°)
OE = OE (Common)
So, by R-H-S congruency condition
∆OME ≅ ∆ONE
∠1 = ∠2 (By CPCT) 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 (see Fig. 10.25) Solution:
Let OM be perpendicular from O on line l.
We know that the perpendicular from the centre of a circle to a chord; bisect the chord. Since BC is a chord of the smaller circle and OM ⊥ BC.
∴ BM = CM ……(i)
Again, AD is a chord of the larger circle and OM ⊥ AD.
∴ AM = DM ……(ii)
Subtracting equation (i) from (ii), we get
AM – BM = DM – CM
or, AB = CD.

Question 5.
Three girls Reshma, Salma, and Mandeep are playing a game by standing on & circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Resma. 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:
Let Reshma, Salma and Mandip are standing on the point R, S, and M respectively where O is the centre of the circle. RS = SM = 6m
OR = 5m
∆RSM is Issosclus Triangle
∴ SP ⊥ RM and also P is mid-point of RM
RP = PM
ar (∆ORS) = $$\frac {1}{2}$$ × OS × PR
= $$\frac {1}{2}$$ × 5 × x …….(i)
Draw OT ⊥ RS here OR = OS (Radii of circle)
So, ∆ORS is Issosceles Triangle
∴ RT = TS = 3 cm
In right ∆OTS,
OS = 5 cm, TS = 3 cm
Using Pythagoras theorem,
OT2 = OS2 – TS2
⇒ OT2 = (5)2 – (3)2
⇒ OT2 = 16
⇒ OT = √16 = 4 cm
Again, ar (∆ORS) = $$\frac {1}{2}$$ × RS × OT
= $$\frac {1}{2}$$ × 6 × 4 ……(ii)
From equation (i) and (ii)
$$\frac {1}{2}$$ × 5 × x = $$\frac {1}{2}$$ × 6 × 4
or $$\frac {5x}{2}$$ = 12
x = $$\frac{12 \times 2}{5}=\frac{24}{5}$$
We have to calculate
RM = 2RP = 2 × $$\frac{24}{5}$$ = $$\frac{48}{5}$$
∴ RM = 9.6 m
Distance between Reshma and Mandip is equal to 9.6 m. Question 6.
A circular park of a radius of 20 m is situated in a colony. Three boys Ankur, Syed, and David are sitting at an equal distance 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.
Solution:
Let three boys Ankur, Syed and David are sitting on the point A, B, and C respectively.
O is the centre of the circle.
According to question AB = BC = CA = x (Let) So, ∆ABC is an equilateral triangle
OA = OB = OC = Radius of die circle = 20 meter
Join A, O and extand to D.
As ∆ABC is an equilateral triangle
$$\frac{\mathrm{OA}}{\mathrm{OD}}=\frac{2}{1}$$
or $$\frac{\mathrm{20}}{\mathrm{OD}}=\frac{2}{1}$$ (∵ OA = 20 m)
OD = 10 meter
In right angle triangle ACD
⇒ (x)2 = $$\left(\frac{x}{2}\right)^{2}$$ + (30)2 (∵ AD ⊥ BC and D is the mid point)
⇒ x2 = $$\frac{x^{2}}{4}$$ + 900
⇒ x2 – $$\frac{x^{2}}{4}$$ = 900
⇒ $$\frac{4 x^{2}-x^{2}}{4}$$ = 900
⇒ $$\frac{3 x^{2}}{4}$$ = 900
⇒ x2 = $$\frac{900 \times 4}{3}$$