**Q. 1Â Â Â Â In figure A, B and C are three points on a circle with centre O such that = 30Âº and = 60Âº. If D is a point on the circle other than the arc ABC, find .**

Since are ABC makes = 60Âº + 30Âº = 90Âº at the centre of the circle and at a point on the remaining part of the circle.

Therefore

**Q. 2Â Â Â Â A chord of a circle is equal to the radius of the circle. Find th eangle subtended by the chord at a point on the minor arc and also at a point on the major are.**

**Sol. **

Let PQ be chord. Join OP and OQ.

It is given that PQ = OP = OQ (Since Chord = radius)

Therefore OPQ is equilateral.

= 60ÂºSince are PBQ makes reflex POQ = 360Âº â€“ 60Âº = 300Âº at centre of the circle and at a point in the minor arc of the circle.

Therefore

Similarly, (60Âº) = 30Âº

Hence, angle subtended by the chord on the minor are 150Âº and on the major chord = 30Âº.

**Q.3Â Â Â Â Â In figure = 100Âº, there P, Q and R are points on a circle with centre O. Find .**

**Sol. **

Since the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same are at a point on the circumference.

Therefore Reflex

Reflex

= 360Âº â€“ 200Âº = 160Âº

In OPR, OP = OR [Radii of the same circle]

[Angles opp. to equal sides are equal]

and 160Âº ......... (1) [Proved above]

In ,

[Angle sum property]

**Q.4Â Â Â Â Â In figure 69Âº, 31Âº, Find **

In ABC,

Since angles in the same segment are equal

Therefore

**Q.5Â Â Â Â Â In figure A, B, C and D are four points on a circle. AC and BD intersect at a point E such that = 130Âº and = 20Âº. Â ****Find .**

**Sol. **

= 180Âº [Linear pair]

130Âº = 180Âº

180Âº â€“ 130Âº

= 50Âº

In ECD, = 180Âº

50Âº + 20Âº = 180Âº

180Âº â€“ 50Âº â€“ 20Âº

= 110Âº

= 110Âº

Since angles in the same segment are equal

Therefore 110Âº.

**Q.6Â Â Â Â ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If 70Âº, is 30Âº, find .Â ****Further, if AB = BC, find **

**Sol. **

[Angles in the same segment]

30Âº [Since 30Âº (given]

In BCD, we have

= 180Âº [Sum of of a ]

30Âº + 70Âº + = 180Âº [Since = 70Âº, = 30Âº]

= 180Âº â€“ 30Âº â€“ 70Âº = 80Âº

If AB = BC, than = 30Âº [Angles opp. to equal sides in a are equal]

Now ,

= 80Âº â€“ 30Âº = 50Âº

[Since 80Âº (found above) and 30Âº]

Hence, 80Âº and 50Âº

**Q.7Â Â Â Â If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.**

**Sol. **

Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quad. ABCD.

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

**Solution : -Â **Since all the radii of the same circle are equal

Therefore OA = OB = OC = OD

and

AC = BD

Therefore the diagonals of the quadrilateral ABCD are equal and bisect each other.

Quadrilateral ABCD is a rectangle.

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

**Sol. **

**Given :** Non - parallel sides AD and BC of a trapezium are equal.

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

**Construction :** Draw DE AB and CF AB.

**Proof :** In order to prove that ABCD is a cyclic trapezium it is sufficient to prove that = 180Âº.

In s DEA and CFB, we have

AD = BC [Given]

[Each = 90Âº]

and DE = CF [Distance between two|| lines is always equal]

Therefore by RHS criterion of congruence, we have

(Corresponding parts of congruent triangles are equal)

Now,

90Âº + 90Âº +

[Since 90Âº and 90Âº]

Thus ,

Therefore , 360Âº

[Since sum of the angles of a quad. is 360Âº]

360Âº

= 180Âº

Hence, ABCD is a cyclic trapezium.

**Q.9Â Â Â Â 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 **

Since angles in the same segment are equal.

Therefore ... (1)

and ... (2)

Also ... (3) [Vertically opp. angles]

Therefore From (1), (2) and (3) , we have

**Q.10Â Â Â 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.**

**Sol. **

**Given** : Two circles are drawn with sides AB and AC of ABC as diameters. The circle intersect at D.

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

**Construction :** Join A and D.

**Proof :** Since AB and AC are the diameters of the two circles. [Given]

Therefore 90Âº [Angles in a semi- circle]

and, 90Âº [Angles in a semi-circle]

Adding we get = 90Âº + 90Âº = 180Âº

BDC is a straight line.

Hence, D lies on BC.

**Q.11Â Â Â ABC and ADC are two right triangles with common hypotenuse AC. Prove that .**

**Sol. **

ABC and ADC are right with common hypotenuse AC. Draw a circle with AC as diameter passing through B and D.Â Join BD.

Clearly, [Since Angles in the same segment are equal]

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

**Sol. **

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

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

**Proof :** Since ABCD is a cyclic parallelogram.

Therefore = 180Âº ... (1)

But ... (2)

From (1) and (2), we have

= 90Âº

Similarly, = 90Âº

Therefore Each angle of ABCD is of 90Âº

Hence, ABCD is a rectangle.

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