## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 15 Circle Chapter Test

Question 1.

In the given figure, a chord PQ of a circle with centre O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :

(i) PQ

(ii) AP

(iii) BP

Solution:

Given, radius = 15 cm

⇒ OA = OB = OP = OQ = 15 cm

Also, OM = 9 cm

∴ MB = OB – OM = 15 – 9 = 6 cm

AM = OA + OM = 15 + 9 cm = 24 cm

In ∆OMP, by using Pythagoras Theoream,

OP^{2} = OM^{2} + PM^{2}

15^{2} = 9^{2} + PM^{2}

= PM^{2} = 225 – 81

PM = \( \sqrt{{144}} \) = 12 cm

Also, In ∆OMQ,

by using Pythagoras Theorem,

OQ^{2} = OM^{2} + QM^{2}

15^{2} = OM^{2} + QM^{2}

15^{2} = 9^{2 }+ QM^{2}

QM^{2} = 225 – 81

QM = \( \sqrt{{144}} \) = 12 cm

∴ PQ = PM + QM (As radius is bisected at M)

⇒ PQ = 12 + 12 cm = 24 cm

(ii) Now in ∆APM

AP^{2} = AM^{2} + OM^{2}

AP^{2} = 24^{2} + 12^{2}

AP^{2} = 576 + 144

AP = \( \sqrt{{720}} \) = 12\( \sqrt{{5}} \) cm

(iii) Now in ∆BMP

BP^{2} = BM^{2} + PM^{2}

BP^{2} = 6^{2} + 12^{2}

BP^{2} = 36 + 144

BP = \( \sqrt{{180}} \) = 6\( \sqrt{{5}} \) cm

Question 2.

The radii of two concentric circles are 17 cm and 10 cm ; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.

Solution:

A line PQRS intersects the outer circle at P and S and inner circle at Q and R. Radius of outer circle OP = 17 cm and radius of inner circle OQ = 10 cm.

QR = 12 cm

From O, draw OM ⊥ PS

∴ QM = \(\frac{1}{2}\)QR = \(\frac{1}{2}\) × 12 = 6cm

In right ∆OQM,

OQ^{2} = OM^{2} + QMv

⇒ (10)^{2} = OM^{2} + (6)^{2}

⇒ OM^{2} = 10^{2} – 6^{2}

= 100 – 36 = 64 = (8)^{2}

OM = 8 cm

Now in right ∆OPM,

OP^{2} = OM^{2} + PM^{2}

⇒ (17)^{2} = (8)^{2} + PM^{2}

⇒ PM^{2} = (17)^{2} – (8)^{2}

= 289 – 64 = 225 = (15)^{2}

PM = 15 cm

PQ = PM – QM = 15 – 6 = 9 cm

Question 3.

A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the centre of the circle.

Solution:

O is the centre of the circle

Length of chord AB = 48 cm

and chord CD = 20 cm

OL ⊥ AB and OM ⊥ CD are drawn

AL = LB = \(\frac{48}{2}\) = 24 cm

and CM = MD = \(\frac{20}{2}\) = 10 cm

OL = 10 cm

Now in right ∆AOL

OA^{2} = AL^{2} + OL^{2} (Pythagoras Theorem)

⇒ OA^{2} = (24)^{2} + (10)^{2 }= 576 + 100

= 676 = (26)^{2}

∴ OA = 26 cm

But OC = OA (radii of the same circle)

∴ OC = 26 cm

Now in right ∆OCM

OC^{2} = OM^{2} + CM^{2}

(26)^{2} = OM^{2} + (10)^{2}

676 = OM^{2} + 100 ⇒ OM^{2} = 676 – 100

⇒ OM^{2} = 576 = (24)^{2}

∴ OM = 24 cm

Question 4.

(a) In the figure (i) given below, two circles with centres C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.

(b) In the figure (ii) given below, P is a point of intersection of two circles with centres C and D. If the st. line APB is parallel to CD, Prove that AB = 2 CD.

Solution:

(a) Two circles with centre C and D intersect each other at P and Q. PQ is the common chord = 6 cm. The line joining the centres C and D bisects the chord PQ at M.

∴ PM = MQ = \(\frac{6}{2}\) = 3 cm

Now in right ∆CPM,

CP^{2} = CM^{2} + PM^{2}

⇒ (5)^{2} = CM^{2} + (3)^{2} ⇒ 25 = CM^{2} + 9

⇒ CM^{2} = 25 – 9 = 16 = (4)^{2}

∴ CM = 4 cm

and in right ∆PDM,

PD^{2} = PM^{2} + MD^{2}

⇒ (4)^{2} = (3)^{2} + MD^{2} ⇒ 16 = 9 + MD^{2}

⇒ MD^{2} = 16 – 9 = 7

∴ MD = \( \sqrt{{7}} \) = 2·65 cm

∴ CD = CM + MD = 4 + 2·65

= 6·65 cm

(b) Given : Two circles with centre C and D intersect each other at P and Q. A straight line APB is drawn parallel to CD.

To Prove : AB = 2 CD.

Construction : Draw CM and DN perpendicular to AB from C and D.

Proof: ∵ CM ⊥AP

∴ AM = MP or AP = 2 MP

and DN ⊥ PB

∴ BN = PN or PB = 2 PN

Adding

AP + PB = 2 MP + 2 PN

⇒ AB = 2 (MP + PN) = 2 MN

⇒ AB = 2 CD. Q.E.D.

Question 5.

(a) In the figure (i) given below, C and D are centres of two intersecting circles. The line APQB is perpendicular to the line of centres CD. Prove that:

(i) AP = QB

(ii) AQ = BP.

(b) In the figure (ii) given below, two equal chords AB and CD of a circle with centre O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, Prove that NOMP is a square.

Solution:

(a) Given : Two circles with centres C and D intersect each other. A line APQB is drawn perpendicular to CD at M.

To Prove : (i) AP = QB (ii) AQ = BP.

Construction : Join AC and BC, DP and DQ.

Proof: (i) In right ∆ACM and ∆BCM

Hyp. AC = BC (radii of same circle)

Side CM = CM (common)

∴ ∆ACM ≅ ∆BCM (R.H.S. axiom of congruency)

∴ AM = BM …(i)

Again in right ∆PDM and ∆QDM

Hyp. PD = QD (radii of the same circle)

Side DM = DM (common)

∴ ∆PDM = ∆QDM (R.H.S. axiom of congruency)

∴ PM = QM …(ii)

Subtracting (ii) from (i),

AM – PM = BM – QM ⇒ AP = QB

(ii) Adding PQ both sides,

AP + PQ = PQ + QB

⇒ AQ = PB Q.E.D.

(b) Given : Two chords AB and CD intersect each other at P at right angle in the circle. M and N are mid-points of the chord AB and CD.

To Prove : NOMP is a square.

Proof: ∵ M and N are the mid-points of AB and CD respectively.

∴ OM ⊥ AB and ON ⊥ CD

and OM = ON (∵ Equal chords are at equal distance from the centre)

∵ AB ⊥ CD

∴ OM ⊥ ON

Hence NOMP is a square.

Question 6.

In the given figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its centre O, prove that AD bisects ∠BAC and ∠BDC.

Solution:

Given : AB and AC are equidistant from its centre O

So, AB = AC

In ∆ABD and ∆ACD

AB = AC (given)

∠B = ∠C (∵ Angle in a semicircle is 90°)

AD = AD (common)

∴ ∆ABD ≅ ∆ACD (SSS rule of congruency)

∴ AD bisects ∠BAC and ∠BDC