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

AC^{2} = OC^{2} – OA^{2}

⇒ AC^{2} = 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,

OT^{2} = OS^{2} – TS^{2}

⇒ OT^{2} = (5)^{2} – (3)^{2}

⇒ OT^{2} = 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

(AD)^{2} = (CD)^{2} + (AD)^{2}

⇒ (x)2 = \(\left(\frac{x}{2}\right)^{2}\) + (30)^{2} (∵ AD ⊥ BC and D is the mid point)

⇒ x^{2} = \(\frac{x^{2}}{4}\) + 900

⇒ x^{2} – \(\frac{x^{2}}{4}\) = 900

⇒ \(\frac{4 x^{2}-x^{2}}{4}\) = 900

⇒ \(\frac{3 x^{2}}{4}\) = 900

⇒ x^{2} = \(\frac{900 \times 4}{3}\)

⇒ x^{2} = 1200

⇒ x = √1200

⇒ x = 20√3 meteres

Therefore, the length of the string of each phone is 20√3 meters.