NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5.

- Circles Class 9 Ex 10.1
- Circles Class 9 Ex 10.2
- Circles Class 9 Ex 10.3
- Circles Class 9 Ex 10.4
- Circles Class 9 Ex 10.5
- Circles Class 9 Ex 10.6

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 10.5 |

Number of Questions Solved |
11 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

Question 1.

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

Solution.

Here, ∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°

Since, arc ABC makes an angle of 90° at the center the circle.

∴ ∠ADC = ∠AOC

[since, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle]

= 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 PQ be chord. Join OP and OQ.

It is given that PQ = OP = OQ (∵ Chord = radius

Δ OPQ is equilateral.

⇒ ∠POQ = 60°

Since arc PBQ makes reflex ∠POQ = 360 °-60 ° = 300 ° at the center of the circle and ∠PBQ at a point in the minor arc of the circle.

∠PBC = (reflex ∠POQ)

Question 3.

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

Solution.

Question 4.

In figure, ∠ABC = 69° and ∠ACB = 31°. Find ∠BDC.

Solution.

Question 5.

In the figure, A, B, C, and D are four points on a circle. AC and BD intersect at a point E such that

∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Solution.

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 q ∠ECD.

Solution.

Question 7.

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

Solution.

Given: Diagonals NP and QM of a cyclic

quadrilateral NQPM are diameters of the circle passing through the vertices M, P, Q, and N.

To prove : Quadrilateral NQPM is a rectangle.

Here, ON = OP = OQ = OM [radii of same circle]

Then, ON = OP = NP

and OM = OQ = MQ

∴ NP = MQ

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

Hence proved

Question 8.

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

Solution.

Given: Non-parallel sides PS and QR of a trapezium PQRS are equal. To prove: PQRS is a cyclic trapezium.

Construction: Draw SM ⊥ PQ and RN ⊥ PQ.

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, P and 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 which intersect the circles at A, D, P and Q, respectively

Question 10.

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

Solution.

Given: Two circles are drawn with sides AC and

AB of Δ ABC 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° …(i) [angle in semi-circle]

and ∠ADC = 90° …(ii) [angle in semi-circle]

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

∠ADB + ∠ADC = 90°+ 90° = 180°

∠BDC = 180°

Hence, BDC is a straight line.

So, point of intersection D lies on the third side.

Hence proved.

Question 11.

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

Solution.

Since Δ ADC and Δ 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 = ∠CA

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° …(i)

[since, sum of pair of opposite angles in a cyclic quadrilateral is 180°]

∠P=∠R …(ii) [since, in a parallelogram, opposite angles are equal]

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