**Download NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles**

**1. In given figure, 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****.**

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Solution :

Here, it can be clearly observed that

∠AOC = ∠AOB + ∠BOC

= 60° + 30°

= 90°

We know that,

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

=> ∠ADC = (1/2) ∠ AOC

=> ∠ADC = (1/2) x 90°

=> ∠ADC = 45°

**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 :

In ∆ AOB,

AB = OA = OB ( Given )

=> ∆ AOB is an equilateral triangle.

=> ∠ AOB = ∠ OAB = ∠ OBA = 60°

And that implies, ∠ ACB = (1/2) ∠ AOB = 30°

Here, ADBC is a cyclic quadrilateral.

=> ∠ ACB + ∠ ADB = 180°

=> ∠ ADB = 180° – 30°

=> ∠ ADB = 150°

Hence, the angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.

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

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Solution :

Given, ∠ PQR = 100°

Here, PR is the chord of the circle.

Let, S is any point on the major arc of the circle.

=> PQRS is a cyclic quadrilateral.

∠PQR + ∠ PSR = 180°

=> ∠ PSR = 180° – 100°

=> ∠ PSR = 80°

We know that,

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.

=> ∠ POR = 2 (∠ PSR )

= 2 x 80°

= 160°

In triangle POR,

OP = OR ( Both are Radii )

=> ∠ PRO = ∠ OPR

∠ POR + ∠ PRO + ∠ OPR = 180°

=> 2 ∠ OPR = 180° – 160°

=> 2 ∠ OPR = 20°

=> ∠ OPR = 10°

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

Solution :

We know that,

Angles subtend by same chord in same segment will be equal.

=> ∠ BAC = ∠ BDC ..(1)

In triangle ABC,

∠ BAC + ∠ ABC + ∠ ACB = 180°

=> ∠ BAC + 69° + 31° = 180°

=> ∠ BAC = 180° – 100°

=> ∠ BAC = 80°

From (1)

∠ BDC = 80°

**5. In given 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****.**

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Solution :

Since, BED is a straight line

∠ BEC + ∠ CED = 180°

=> ∠ CED = 180° – 130°

=> ∠ CED = 50°

From the triangle EDC,

∠ EDC + ∠ CED + ∠ ECD = 180°

=> ∠ EDC = 180° – 50° – 20°

=> ∠ EDC = 110°

We know that,

∠ EDC = ∠ BDC

=> ∠ BDC = 110°

We know that,

Angles in same segment make by same chord will be equal.

=> ∠ BDC = ∠ BAC

=> ∠ BAC = 110°

**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 :

We know that,

Angles in the same segment form by same chord will be equal.

For chord CD,

∠CBD = ∠CAD

=> ∠CAD = 70°

∠BAD = ∠CAD + ∠CAB

= 70° + 30°

= 100°

Since ABCD is a cyclic quadrilateral,

∠BAD + ∠BCD = 180°

∠BCD = 180° – 100°

∠BCD = 80° ..(1)

In the triangle ABC,

AB = BC ( given )

=> ∠BCA = ∠BAC

= 30°

∠BCD = ∠BCA + ∠ACD

=> ∠ ACD = 80° – 30°

=> ∠ ACD = 50°

We know that,

∠ ACD = ∠ ECD

=> ∠ ECD = 50°

**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 :

Here, ABCD is a cyclic quadrilateral intersecting their diagonals AC & BD at center of the circle O.

Consider the chord BD,

=> ∠BAD = (1/2) ∠BOD

=> ∠BAD = (1/2) ( 180°)

=> ∠BAD = 90°

In a cyclic quadrilateral ABCD,

∠BAD + ∠BCD = 180°

=> ∠BCD = 180° – 90°

=> ∠BCD = 90°

Consider the chord AC,

=> ∠ABC = (1/2) ∠AOC

=> ∠ABC = (1/2) ( 180°)

=> ∠ABC = 90°

In a cyclic quadrilateral ABCD,

∠ABC + ∠ADC = 180°

=> ∠ADC = 180° – 90°

=> ∠ADC = 90°

Each interior angle of the cyclic quadrilateral is 90°.

=> Cyclic quadrilateral having diameters of the circle as its diagonals will be a rectangle.

**Hence Proved.**

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

Solution :

Here, ABCD is a trapezium with AB || CD.

AD = BC ( Given )

Construction :

Draw perpendiculars from A & B on CD.

=> AM & BN are perpendicular to CD.

From the triangles AMD & BNC,

∠AMD = ∠BNC (Each 90˚)

AM = BN ( Perpendicular distance between parallel lines will be equal )

AD = BC ( given )

=> ∆ AMD & ∆ BNC are congruent under R.H.S. congruency rule.

=> ∠ADC = ∠BCD [By CPCT] ..(1)

( Corresponding angles are equal in congruent triangles )

AB || CD

AD is the transversal.

∠BAD + ∠ADC = 180°

From (1),

∠BAD + ∠BCD = 180° ..(2)

∠ABC + ∠ADC + ∠BAD + ∠BCD = 360°

=> ∠ABC + ∠ADC = 180° ..(3)

Equation (2) & Equation (3) shows that the opposite angles are supplementary.

We can say that,

If the non-parallel sides of a trapezium are equal, then it will be cyclic.**Hence Proved.**

**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 given figure). Prove that ****∠****ACP = ****∠**** QCD****.**

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Solution :

Construction :

Join chords AP & DQ

For the chord AP,

∠ACP = ∠ABP ( Angles by same chord in same segment will be equal ) ..(1)

For the chord DQ,

∠DBQ = ∠QCD ( Angles by same chord in same segment will be equal ) ..(2)

∠ABP = ∠DBQ ( Vertically opposite angles ) ..(3)

From (1), (2) & (3),

We can say that,

∠ACP = ∠ QCD**Hence Proved.**

**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 :

Consider the triangle ABC.

Here AB & AC are diameters of the respective circles.

Those two circles intersect at A & D

D do not lie on BC.

=> ∠ADB = ∠ADC = 90° ( Angle in a semi-circle is 90°)

We know that,

∠BDC = ∠ADB + ∠ADC

= 90° + 90°

= 180°

=> Therefore, BDC is a straight line (D should lie on the line BC)

=> Hence, our assumption was wrong

=> The intersecting point of circles should lie on the third side of the triangle.**Hence Proved.**

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

Solution :

From the triangle ABC,

∠ABC + ∠BAC + ∠ACB = 180° ( sum of interior angle in a triangle is 180°)

90° + ∠BAC + ∠ACB = 180°

=> ∠BAC + ∠ACB = 90° ..(1)

From the triangle ADC,

∠ADC + ∠DAC + ∠ACD = 180° ( sum of interior angle in a triangle is 180°)

90° + ∠DAC + ∠ACD = 180°

=> ∠DAC + ∠ACD = 90° ..(2)

Adding the equations (1) & (2),

=> ∠BAC + ∠ACB + ∠DAC + ∠ACD = 90° + 90°

= 180°

=> (∠BAC + ∠DAC) + (∠ACB + ∠ACD) = 180°

=> ∠BAD + ∠BCD = 180° ..(3)

We have,

∠ABC + ∠ADC = 180° ..(4)

From (3) & (4),

We can say that, ABCD is a cyclic quadrilateral.

Here it can be clearly see that

∠ CAD = ∠ CBD ( Angles in same segment by same chord will be equal. )

**Hence Proved.**

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

Solution :

Here ABCD is a cyclic parallelogram.

∠ DAB + ∠ DCB = 180° .. (1)

And we know that,

In a parallelogram opposite angles are equal.

=>∠ DAB = ∠ DCB .. (2)

From (1) & (2)

∠ DAB = ∠ DCB = 90°

Parallelogram will be a rectangle if one of its interior angle is 90°.

Hence, cyclic parallelogram is a rectangle.

**Download NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.5 – Circles**

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