ML Aggarwal Circle Chapter Test Class 9 ICSE Maths Solutions Ch-15. Step by Step Solutions of Ch-Test Questions on Circle of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

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

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-15 | Circle |

Topics | Solution of Ch-Test Questions |

Academic Session | 2024-2025 |

### Solution of Ch-Test Questions on Circle

ML Aggarwal Class 9 ICSE Maths Solutions Ch-15

**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

**Answer : **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,

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

⇒ 15^{2} = 9^{2 }+ PM^{2}

⇒ PM^{2} = 255 – 81

⇒ PM = √144 = 12 cm

Also, In ∆OMQ

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

⇒ 15^{2 }= OM^{2} + QM^{2}

⇒ 15^{2 }= 9^{2} + QM^{2 }**(QM ^{2 }= 225 – 81)**

⇒ QM = √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 = √720 = 12 √5 cm

**(iii) **Now in ∆BMP

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

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

⇒ BP^{2 }= 36 + 144

⇒ BP = √180 = 6√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.**

**Answer : **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 = ½ QR = ½ × 12 = 6 cm

In right ∆OQM

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

⇒ (10)^{2 }= OM^{2 }+ (6)^{2}

⇒ OM^{2 }= 10^{2} – 6^{2}

⇒ OM^{2 }= 100 – 36 = 64 = (8)^{2}

⇒ OM = 8 cm

Now in right ∆OPM

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

⇒ (17)^{2 }= OM^{2 }+ PM^{2}

⇒ PM^{2 }= (17)^{2} – (8)^{2}

⇒ PM^{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.**

**Answer : **Perpendicular from Center on Chord bisect it

chord of length 48cm

48/2 = 24 cm

Distance from Center = 10 cm

Radius² = 24² + 10²

Radius² = 24² + 10²

Radius² = 26²

Radius = 26 cm

Let say chord of length is at Distance d from center

Then Chord length = 20 cm

20/2 = 10 cm

d² = 26² – 10²

d² = 24²

d = 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.**

**Answer : **(a)

CD = CM + MD = 4 + 2.65

= 6.65

##### (b) Two circle with center C and D intersect each other at P and Q. A straight line APB is Drawn parallel to CD.

To prove : AB = 2 CD

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

**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.**

**Answer :**

**(a)** To circles with centers C and D intersect each other. A line APQB is drawn perpendicular to CD at M.

To prove : AP = QB

AQ = BP

Join AC and BC, DP and DQ

Proof : (i) In right triangle ACM and triangle BCM

AC = BC

CM = CM

ACM ≅ BCM

AM = BM

PD = QD

side DM = DM

PDM ≅ QDM

so, PM = QM

Subtracting

AM – PM = BM – QM

AP = QB

##### (ii) Adding PQ both sides,

AP + PQ = PQ + QB

AQ = PB

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

To prove : NOMP is a square

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

OM ⊥ AB and ON ⊥ CD

and OM = ON

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.**

**Answer : **AB and AC are equidistant from its centre O

So, AB = AC

In triangle ABD and ACD

AB = AC

angle B = angle C

AD = AD

so, triangle ABD ≅ triangle ACD

AD bisects angle BAC and angle BCD

— : end of ML Aggarwal Circle Chapter Test Class 9 ICSE Maths Solutions Ch-15 : —

Return to :- **ML Aggarawal Maths Solutions for ICSE Class-9**

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