In this chapter, we provide KSEEB SSLC Class 9 Maths Chapter 12 Circles Ex 12.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 9 Maths Chapter 12 Circles Ex 12.4 pdf, free KSEEB SSLC Class 9 Maths Chapter 12 Circles Ex 12.4 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 9 Maths Chapter 12 Circles Ex 12.4**

## Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.4

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:

R = 5 cm, r = 3 cm. d = 4 cm.

length of the common chord = ?

When measured Chord CD = 6 cm.

Circles with centre A and B, intersect at points C and D.

CD is common chord.

Perpendicular bisector of CD is AB.

AC = 5cm. BC = 3cm.

∴ Chord, CD= 2 × OC = 2 × 3

∴ Chord, CD = 6 cm.

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:

Data : Circle with centre ‘O’, AB and CD chords are equal and intersect at E.

To Prove: Segment EAD = Segment ACB.

Construction: Draw OP⊥CD and OQ⊥AB.

Join OE.

Proof: AQ = QB (∵ Chord drawn perpendicular CP = PD ,to centre bisects chord.).

In ∆OPE and ∆OQE,

∠OPE = ∠OQE = 90° (Construction)

OP = OQ

OE is common.

∴ ∆OPE ≅ ∆OQE (RHS Theorem)

QE = PE ……….. (i)

But QB = PD

∴ QB – QE = PD – PE

EB = ED ……….. (ii)

From (i) and (ii),

AE = CE.

CBD is minor segment formed by Chord CD.

ADB is minor segment formed by Chord AB.

∴ Minor segment CBD = Minor segment ADB

Major segment CAD = Major segment ACB.

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:

Data: O is the centre of Circle. Chord AB = Chord CD. These chords intersects at E. OE is joined.

To Prove: ∠OEP = ∠OEQ Construction:

Draw OP⊥CD and OQ⊥AB.

Proof: In ∆OPE and ∆OQE,

∠OPE = ∠OQE = 90° (construction)

OE is common.

OP = OQ (Equal chords are equidistant)

∴ ∆OPE ≅ ∆OEQ (RHS postulate)

∴ ∠OEP = ∠OEQ.

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:

Data: A line intersects two concentric circles with centre O at A, B, C, and D.

To Prove: AB = CD.

Construction: Draw OM⊥AD.

Proof: AD is the chord of a bigger circle.

OM⊥AD

∴ AM = MD ………(i) (∵ Perpendicular bisects the chord from centre)

Similarly, BC is the chord of the smaller circle.

OM⊥BC.

∴ BM = MC ………. (ii)

Subtracting (ii) from (i),

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 5m 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:

Data: Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 cm drawn in a park. Distance between Reshma and Salma and between Salma and Mandip is 6 cm. each.

To Prove: Distance between Reshma and Mandip = ?, RM = ?

Construction: Draw OX⊥RS and OY⊥SM. OR, OS and OM are joined.

Proof: OR = OS = OM = 5 m.

RX = XS = 3 m.

Now, in ⊥∆ORX,

OR^{2} = OX^{2} + RX^{2}

(5)^{2} = (OX)^{2} + (3)^{2}

25 = OX^{2} + 9

25 – 9 = OX^{2}

16 = OX^{2}

OX^{2} = 16

∴ OX = 4 WI …………… (i)

In quadrilateral ORSM,

OR = OM and SR = SM

∴ ORSM is a kite.

Diagonals of a kite bisect at the right angle.

∠RAS = 90°

∴ RA = AM.

Now, RM = 2RC = 2 × 4.8 = 9.8 m

∴ RM = 9.8 m.

∴ Distance between Reshma and Mandip is 9.8 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:

Data: 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 to each other.

To Prove: Length of wire of telephone =?

Proof: In this figure, AS = SD = DA

∴ ASD is an equilateral triangle.

∴ OA = 20 m. (∵ radius)

Median passed through centre of circle.

∴ AB is the median.

AD ⊥ SD

Median of an equilateral triangle divides in the ratio 2 : 1 in the centre?

∴ AO : OB = 2 : 1

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