In this chapter, we provide KSEEB SSLC Class 9 Maths Chapter 12 Circles Ex 12.5 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 9 Maths Chapter 12 Circles Ex 12.5 pdf, free KSEEB SSLC Class 9 Maths Chapter 12 Circles Ex 12.5 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 9 Maths Chapter 12 Circles Ex 12.5**

## Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.5

Question 1.

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

Solution:

∠AOB + ∠BOC = 60° + 30° = 90°

∴ ∠AOC = 90° (Angle subtended at the centre)

∠ADC = ?

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ Angle subtended at the centre

∠AOC= 2 × ∠ADC

90° = 2 × ∠ADC 90

∴ ∠ADC = 902 = 45°

∴ ∠ADC = 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:

(i) Angle subtended in the circumference, ∠BAD =?

(ii) Angle subtended in the circumference, ∠BCD =?

In this figure BD is a chord, OB is the radius, it is equal to OD.

∴ OB = OD = BD

∴ ∆OBD is equilateral triangle.

∴ each angle is equal to 60°.

∴ angle subtended at the centre ∠BOD = 60°.

(i) Angle subtended in the circumference

∠BAD= 12 × angle subtended at centre ∠BOD

= 12 × 60°

∴ ∠BAD = 30°.

(ii) The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

∴ In cyclic quadrilateral ABCD,

∠BAD + ∠ACD = 180

30 + ∠ACD = 180 (∵ ∠BAD = 30°)

∴ ∠ADC = 180 – 30

∴ ∠ACD = 150°.

Question 3.

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

Solution:

If ∠PQR = 100°, then ∠OPR = ?

∠PQR = 100°.

∴ ∠POR = ?

∠POR = 2 × ∠PQR = 2 × 100°

∠POR = 200°

∴ ∠POR – Reflex angle ∠POR = 360°

∴ ∠POR = 360 – 200

∴ ∠POR = 160°

In ∆POR, OP = OR radii.

∴ ∠OPR = ∠ORB

∴ ∠OPR + ∠ORP + ∠POR = 180°

∠OPR + ∠OPR + 160 = 180

2∠OPR+ 160 = 180

2∠OPR = 180 – 160

2∠OPR = 20

∠OPR = 202

∴ ∠OPR= 10°.

Question 4.

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

Solution:

If ∠ABC = 69°, ∠ACB = 31°, then ∠BDC = ?

In ∆ABC,

∠ABC = ∠ACB + ∠BAC = 180°

69 + 31 + ∠BAC = 180

100 + ∠BAC = 180

∠BAC = 180 – 100

∠BAC = 80°

∠BAC and ∠BDC are angles in same segment. These are equal.

∠BDC = ∠BAC = 80°

∴ ∠BDC = 80°.

Question 5.

In Fig., 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:

∠BEC = 130°, ∠ECD = 20°, ∠BAC = ?

Angles formed by arc AD are ∠ABD, ∠ACD is equal.

∴ ∠ABD = ∠ACB = 20°

∠ABD = 20°

∠AEB + ∠BEC = 180 (adjacent angles)

∠AEB + 130° = 180

∠AEB = 180 – 130

∴ ∠AEB = 50°

Now, in ∆BAE,

∠BAE = ∠ABE + ∠AEB = 180°

∠BAE + 20 + 50 = 180

∠BAE + 70 = 180

∠BAE = 180 – 70

∴ ∠BAE = 110°

But ∠BAE and ∠BAC are the same.

∴ ∠BAC = 110°

Question 6.

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

Solution:

∠DBC = 70°, ∠BAC = 30°, then ∠BCD =?

AB = BC, then ∠ECD = ?

∠DAC and ∠DBC are angles in the same segment.

∴ ∠DAC = ∠DBC = 70°

∴ ∠DAC = 70°

ABCD is a cyclic quadrilateral.

∴ Sum of opposite angles is 180°.

∠DAB + ∠DCB = 180

100 + ∠DCB = 180

[∵ ∠DAC + ∠BAC = ∠DAB 70 + 30 = 100]

∠DCB = 180 – 100

∴ ∠DCB = 80

∠DCB = ∠BCD = 80

∴ ∠BCD = 80

In ∆ABC, AB = AC,

∴ ∠BAC = ∠BCA = 30°

∠BCA = 30°

∠ECD = ∠BCD – ∠BCA = 80 – 30

∴ ∠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:

Data: In cyclic quadrilateral ABCD, AC and BD are diameters of circle.

To Prove: ABCD is a rectangle.

Proof: AC is a diameter. ∠ABC is angle in semicircle. Angle in semicircle is a right angle.

∴ ∠ABC = 90° ∠ADC = 90°

Similarly, BD is a diamgers, ∠DAB, ∠DCB are angles in semicircle.

∠DAB = 90° ∠DCB = 90°

Now, four angles of quadrilateral ABCD are right angles.

∴ ∠A = ∠B = ∠C = ∠D = 90°

∴ ABCD is a rectangle.

Question 8.

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

Solution:

Data: ABCD is a trapezium, DC || AB and AD = BC which are non-parallel sides.

To Prove: ABCD is a cyclic quadrilateral.

Proof: ABCD is a trapezium.

AB || CD and AD = BC.

∴ ∠DAB + ∠CDA = 180° …………. (i)

(sum of interior angles)

Similarly, ∠DCB + ∠ABC = 180° …………. (ii)

As we know, AD = BC,

∴∠DAB = ∠CBA

Substituting Eqn. (i) in Eqn. (ii),

∠CBA + ∠CDA = 180°

∠DAB + ∠DCB = 180°

If sum of angles of a quadrilateral is 180°, then it is cyclic quadrilateral.

∴ ABCD is a cyclic quadrilateral.

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. Prove that ∠ACP = ∠QCD.

Solution:

Two circles are drawn taking PQ and PR of a triangle as diameter. Let these intersect at P and S.

To Prove: ∠ACP = ∠QCD

Proof: ∠ABP = ∠QBD ………….. (i) (vertically opposite angles)

∠ABP = ∠ACP ……….. (ii) (angles in the same segment)

Similarly, ∠QCD = ∠QBD …………. (iii)(angles in the same segment)

From (i), (ii), and (iii),

∠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:

Data: Two circles are drawn taking PQ and PR of a triangle as diameter. Let these intersect at P and S.

To Prove: The point of intersection ‘S’ is on the third side QR of ∆PQR.

Construction: Join PS.

Proof: QAP is a diameter.

∴ ∠QSP = 90° (angle in the semi-circle) Similarly, ABR is a diameter.

∠PSR – 90° (angle in the semicircle)

∠QSR = ∠QSP + ∠RSP = 90 + 90

∠QSR = 180°

∴ ∠QSR is a straight angle.

∴ QSR is a straight line.

∴ Point ‘S’ is on the third side QR of ∆PQR.

Question 11.

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

Solution:

Data: ∆ABC and ∆ADC are right-angled triangles having common hypotenuse AC.

To Prove: ∠CAD = ∠CBD

Proof: In ∆ABC, ∠ABC = 90°

∴ ∠BAC + ∠BCA = 90° …………. (i)

In ∆ADC, ∠ADC = 90°

∴ ∠DAC + ∠DCA = 90° …………… (ii)

Adding (i) and (ii),

∠BAC + ∠BCA + ∠DAC + ∠DCA = 90 + 90

(∠BAC + ∠DAC) + (∠BCA + ∠DCA) = 180°

∠BAD + ∠BCD = 180°

∴ ∠ABC + ∠ADC = 180°

If opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.

∴ ∠CAD = ∠CBD (∵ Angles in the same segment).

Question 12.

Prove that a cyclic parallelogram is a rectangle.

Solution:

ABCD is a cyclic parallelogram in the circle with ‘O’ centre.

To Prove: ABCD is a rectangle.

Proof: ABCD is a cyclic parallelogram.

∴ AB || DC and AD || BC

∠A = ∠C (Opposite angles of parallelogram)

But, ∠A + ∠C = 180° (Opposites angles of cyclic quadrilateral)

∠A + ∠A = 180°

2∠A = 180°

∠A = 1802

∴ ∠A = 90°

If each angle of the parallelogram is a right angle, it is a rectangle.

∴ ABCD is a rectangle.

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