NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Circles |

Exercise |
Ex 11.5 |

Number of Questions Solved |
12 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5

**Question 1.**

**In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.**

**Solution:**

∴ ∠AOC = ∠AOB + ∠BOC = 60P + 30° = 90°

∴ Arc ABC makes 90° at the centre of the circle.

∴ ∠ADC = \(\frac { 1 }{ 2 }\) ∠AOC

(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)

= \(\frac { 1 }{ 2 }\) x 90° = 45°

**Question 2.**

**A chord of a circle is equal to the radius of the circle, find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.**

**Solution:**

Let BC be chord, which is equal to the radius. Join OB and OC.

Given, BC=OB = OC

∴ ∆OBC is an equilateral triangle.

∠BOC =60°

∴ BAC = \(\frac { 1 }{ 2 }\) ∠BOC

= \(\frac { 1 }{ 2 }\) x 60° = 30°

(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)

Here, ABMC is a cyclic quadrilateral.

∴ ∠BAC + ∠BMC = 180°

(∵ In a cyclic quadrilateral the sum of opposite angles is 180°)

⇒ ∠BMC= 180° – 30° =150°

**Question 3.**

**In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.**

**Solution:**

∴ ∠POR = 2∠PQR = 2 x 100° = 200°

(Since, the angle subtended by the centre is double the angle subtended by circumference.)

Since, in ∆OPR, ∠POR = 360° – 200° = 160° .. (i)

Again, ∆ OPR, OP = OR (Radii of the circle)

∴ ∠OPR = ∠ORP (By property of isosceles triangle)

In ∆POR, ∠OPR + ∠ORP + ∠POR = 180° …(ii)

From Eqs. (i) and (ii), we get

∠OPR + ∠OPR + 160° = 180°

∴ 2 ∠OPR = 180° – 160° = 20°

∴ ∠OPR = \(\frac { { 120 }^{ circ } }{ 2 }\) = 10°

**Question 4.**

**In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.**

**Solution:**

∵ ∠BDC = ∠BAC …(i)

(Since, the angles in the same segment are equal)

Now , in ∆ABC

∴ ∠A + ∠B+ ∠C= 180°

⇒ ∠A+ 69°+ 31° = 180°

⇒ ∠A + 100° = 180°

∴ ∠A = 180° – 100° = 80°

⇒ ∠BAC=80°

∴ From Eq.(i)∠BDC = 80°

**Question 5.**

**In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.**

**Solution:**

∴ ∠AEB = 180° – 130° = 50° (Linear Pair) …(i)

⇒ ∠CED = ∠AEB = 50° (Vertically opposite)

Again ∠ABD = ∠ACD (Since, the angles in the same segment are equal)

∠ABE = ∠ECD

⇒ ∠ABE = 180° …(ii)

∴ In ∆ CDE

∠A+ 20° + 50° = 180° [From Eqs. (i) and (ii)]

∠A + 70° = 180°

∴ ∠A = 180°- 70° =110°

Hence ∠BAC = 110°

**Question 6.**

**ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.**

**Solution:**

Angles in the same segment are equal.

∴ ∠BDC = ∠BAC

∴ ∠BDC = 30°

In ∆ BCD, we have

∴ ∠BDC + ∠DBC + ∠BCD = 180° (Given, ∠DBC = 70° and ∠BDC = 30°)

∴ 30° + 70° + ∠BCD = 180°

∴ ∠BCD= 180°-30°-70° = 80°

If AB = BC, then ∠BCA = ∠BAC= 80° (Angles opposite to equal sides in a triangle are equal)

Now, ∠ECD = ∠BCD – ∠BCA = 80° – 30P = 50° (∵ ∠BCD = 80° and ∠BCA =30°)

Hence, ∠BCD = 80°

and ∠ECD = 50°

**Question 7.**

**If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.**

**Solution:**

**Given:** Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle through the vertices M, P, Q and N of the quadrilateral NQPM.

**To prove:** Quadrilateral NQPM is a rectangle.

**Proof:** ∵ ON = OP = OQ = OM (Radii of circle)

Now, ON = OP = \(\frac { 1 }{ 2 }\) NP

and OM = OQ = \(\frac { 1 }{ 2 }\) MQ

∴ NP = MQ

Hence, the diagonals of the quadrilateral MPQN are equal and bisect each other. So, quadrilateral NQPM is a rectangle.

**Question 8.**

**If the non-parallel sides of a trapezium are equal, prove that it is cyclic.**

**Solution:**

**Given:** Non-parallel sides PS and QR of a trapezium PQRS are equal.

**To prove:** ABCD is a cyclic trapezium.

**Construction:** Draw SM ⊥ PQ and RN ⊥ PQ.

Proof In ∆SMP and ∆RNQ, we get

SP = RQ (Given)

∠SMP = ∠RNQ (Each = 90°)

and SM = RN

(∵ Distance between two parallel lines is always equal)

∴ By RHS criterion, we get

∆ SMP ≅ ∆ RNQ

So, ∠P = ∠Q (By CPCT)

and ∠PSM = ∠QRN

Now, ∠PSM = ∠QRN

∴ 90° + ∠PSM = 90° + ∠QRN (Adding both sides 90°)

∴ ∠MSR + ∠PSM = ∠NRS + ∠QRN (∵∠MSR = ∠NRS = 90°)

So, ∠PSR = ∠QRS

i.e., ∠S = ∠R

Thus, ∠P = ∠Q and ∠R = ∠S …(i)

∴ ∠P+ ∠Q+ ∠R+ ∠S = 360° (∵ Sum of the angles of a quadrilateral is 360°)

∴ 2∠S + ∠Q = 360° [From Eq. (i)]

∠S+∠O = 180°

Hence, PQRS is a cyclic trape∠ium.

**Question 9.**

**Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A,D and P, Q respectively (see figure). Prove that ∠ ACP = ∠QCD.**

**Solution:**

**Given:** Two circles intersect at two points B and C. Through B two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q, respectively.

**To prove:** ∠ACP = ∠QCD

**Proof:** In circle I, ∠ACP = ∠ABP (Angles in the same segment) …(i)

In circle II, ∠QCD = ∠QBD{Angles in the same segment)…(ii)

∠ABP = ∠QBD (Vertically opposite angles)

From Eqs. (i) and (ii), we get ∠ACP = ∠QCD

**Question 10.**

**If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.**

**Solution:**

**Given:** Two circles are drawn with sides AC and AB of AABC as diameters . Both circles intersect each other at D.

**To prove:** D lies on BC.

**Construction:** Join AD.

**Proof:** Since, AC and AB are the diameters of the two circles.

∠ADB = 90° ( ∴ Angles in a semi-circle) …(i)

and ∠ADC = 90° (Angles in a semi-circle) …(ii)

On adding Eqs. (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90° = 180°

Hence, BCD is a straight line.

So, D lies on BC.

**Question 11.**

**ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.**

**Solution:**

Since, ∆ADC and ∆ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.

∵ Angles in the same segment are equal.

∴ ∠CBD = ∠CAD

**Question 12.**

**Prove that a cyclic parallelogram is a rectangle.**

**Solution:**

**Given:** PQRS is a parallelogram inscribed in a circle.

**To prove**: PQRS is a rectangle.

**Proof:** Since, PQRS is a cyclic quadrilateral.

∴ ∠P+∠R = 180°

(∵ Sum of opposite angles in a cyclic quadrilateral is 180°) …(i)

But ∠P = ∠R (∵ In a || gm opposite angles are equal) …(ii)

From Eqs. (i) and (ii), we get

∠P = ∠R = 90°

Similarly, ∠Q = ∠S = 90

∴ Each angle of PQRS is 90°.

Hence, PQRS is a rectangle.

We hope the NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5, drop a comment below and we will get back to you at the earliest.