NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.4 – Part 1

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 PQR and ABC is triangle ,∠B=∠Q=90,AB=PQ or BC=QR

RHS Congruence Rule

PQR and ABC is triangle ,∠B=∠Q=90,AB=PQ or BC=QR

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

Solution:

Circle c Circle c: Circle through B_1 with center O Circle d Circle d: Circle through D_1 with center C Segment f Segment f: Segment [E, F] Segment j Segment j: Segment [O, C] Segment k Segment k: Segment [A, C] Segment l Segment l: Segment [A, O] Point O O = (-1.42, 2.72) Point O O = (-1.42, 2.72) Point O O = (-1.42, 2.72) Point C C = (1.22, 2.7) Point C C = (1.22, 2.7) Point C C = (1.22, 2.7) Point A Point A: Intersection point of c, d Point A Point A: Intersection point of c, d Point A Point A: Intersection point of c, d Point D Point D: Point on f Point D Point D: Point on f Point D Point D: Point on f Point B Point B: Intersection point of c, d Point B Point B: Intersection point of c, d Point B Point B: Intersection point of c, d

Two Circle With Radius 5cm and 3cm

Two circle with radii 5cm and 3cm respectively, OP = 5cm, PS =3 cm and OS=4cm

Given, two circles with radius 5cm and 3cm .

  • OA=5cm , AC=3cm and OC=4cm .

  • Also ,AB=2AD (as we proved above)

  • Let DC be x .

In ΔAOD ,

  • OA2=OD2+AD2

  • OA2=(OCDC)2+AD2 (OC=OD+DC)

  • 52=(4x)2+AD2

  • 25=168x+x2+AD2

  • AD2=9+8xx2 ………..equation (1)

In ΔADC ,

  • AC2=AD2+DC2

  • 32=AD2+x2

  • AD2=9x2 ………….equation (2)

Equating (1) and (2),

  • AD2=AD2

  • 9+8xx2=9x2

  • 8x=0

  • x=0

Putting the value of x in (1) we get,

  • AD2=902

  • AD2=9

  • AD=3cm

Therefore, length of the chord AB=2AD=2×3=6cm

Q-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:

  • Given, PQ and SR are chords intersecting at T and PQ=SR

  • To prove, PT=TR And ST=TQ

Circle c Circle c: Circle through Q with center O Angle ? Angle ?: Angle between T, M, T' Angle ? Angle ?: Angle between T, M, T' Angle ? Angle ?: Angle between O, N, O' Angle ? Angle ?: Angle between O, N, O' Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [R, S] Segment h Segment h: Segment [O, T] Segment i Segment i: Segment [O, M] Segment j Segment j: Segment [O, N] Point O O = (-1.26, 3.56) Point O O = (-1.26, 3.56) Point O O = (-1.26, 3.56) Point Q Q = (0.76, 1.22) Point Q Q = (0.76, 1.22) Point Q Q = (0.76, 1.22) Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c Point T Point T: Intersection point of f, g Point T Point T: Intersection point of f, g Point T Point T: Intersection point of f, g Point M Point M: Point on f Point M Point M: Point on f Point M Point M: Point on f Point N Point N: Point on g Point N Point N: Point on g Point N Point N: Point on g

in Circle PQ and SR Are Two Chords

In circle PQ and SR are two Chords, intersecting at T

Construction, draw perpendicular bisectors of PQ and SR. Line from the center which bisects a chord is perpendicular to the chord.

  • OM bisects PQ(OMPQ)

  • ON bisects SR(ONSR)

As PQ=SR

  • PM=NR ……equation (1)

  • Because M and N are midpoints of PQ and SR, MQ=SN ……equation (2)

In OMT and ONT

  • OMT=ONT (perpendiculars)

  • OT=OT (common line)

  • OM=ON ( PQ=RS and thus equidistant from the centre)

  • ΔOMTΔONT By Right Angle Hypotenuse congruence condition.

  • MT=TN by Corresponding Parts of Congruent Triangles…….equation (3)

From (1) and (2) we get,

  • MQ=SN

  • PM+MT=NR+TN (since we are adding equal parts (MT and TN) to equal quantities what we get according to Euclid is also equal)

  • Therefore, PT=TR

Again,

  • MQ=SN

  • MQMT=SNTN (since we are subtracting equal parts (MT and TN) from equal quantities what is left according to Euclid is also equal)

  • TQ=ST

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