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**Question 1. In the figure, A, B and C are three points on a circle with centre O such that âˆ BOC = 30Âº and âˆ AOB = 60Â°. If D is a point on the circle other than the arc ABC, find âˆ ADC. Solution: **We have a circle with centre O, such that

âˆ AOB = 60Âº and âˆ BOC = 30Âº

âˆµ âˆ AOB + âˆ BOC = âˆ AOC

âˆ´ âˆ AOC = 60Âº + 30Âº = 90Âº

Now, the arc ABC subtends âˆ AOC = 90Âº

at the centre and âˆ ADC at a point D on the circle other than the arc ABC.

âˆ´ âˆ ADC = (1/2)[âˆ AOC]

â‡’ âˆ ADC = (1/2)(90Âº) = 45Âº

**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:** We have a circle having a chord AB equal to radius of the circle.

âˆ´ AO = BO = AB

â‡’ Î”AOB is an equilateral triangle.

Since, each angle of an equilateral = 60Âº.

â‡’ âˆ AOB = 60Âº

Since, the arc ACB makes reflex âˆ AOB = 360Âº âˆ 60Âº = 300Âº at the centre of the circle and âˆ ABC at a point on the minor arc of the circle.

âˆ´ âˆ ACB = (1/2) [reflex âˆ AOB]

= (1/2)[300Âº] = 150Âº

Similarly, âˆ ADB = (1/2)[âˆ AOB]

= (1/2)x [60Âº] = 30Â°

Thus, the angle subtended by the chord on the minor arc = 150Âº and on the major arc = 30Âº.

**Question 3. In the figure, âˆ PQR = 100Âº, where P, Q and R are points on a circle with centre O. Find âˆ OPR. Solution:** âˆµ The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

âˆ´ reflex âˆ POR = 2âˆ PQR

But âˆ PQR = 100Âº

âˆ´ reflex âˆ POR = 2 x 100Âº = 200Âº

Since, âˆ POR + reflex âˆ POR = 360Âº

âˆ´ âˆ POR + 200Âº = 360Âº

â‡’ âˆ POR = 360Âº âˆ 200Âº

â‡’ âˆ POR = 160Âº

Since, OP = OR [Radii of the same circle]

âˆ´ In Î”POR,âˆ OPR = âˆ ORP [Angles opposite to equal sides of a triangle are equal]

Also, âˆ OPR + âˆ ORP + âˆ POR = 180Âº [Sum of the angles of a triangle = 180Â°]

â‡’ âˆ OPR + âˆ OPR + 160Âº = 180Âº [âˆµ âˆ OPR = âˆ ORP]

â‡’ 2âˆ OPR = 180Âº âˆ 160Âº = 20Âº

â‡’ âˆ OPR = (20^{0}/2)= 10Âº

**Question 4. In the figure, âˆ ABC = 69Âº, âˆ ACB = 31Âº, find âˆ BDC. Solution: **We have, in Î”ABC, âˆ ABC = 69Âº and âˆ ACB = 31Âº

But âˆ ABC + âˆ ACB + âˆ BAC = 180Âº

âˆ´ 69Âº + 31Âº + âˆ BAC = 180Âº

â‡’ âˆ BAC = 180Âº âˆ 69Âº âˆ 31Âº = 80Âº

Since, angles in the same segment are equal.

âˆ´ âˆ BDC = âˆ BAC

â‡’ âˆ BDC = 80Âº

**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 âˆ BEC = 130Â° and âˆ ECD = 20Âº. Find âˆ BAC. Solution: **In D CDE, Exterior âˆ BEC = {Sum of interior opposite angles} [âˆµ BD is a straight line.]

130Âº = âˆ EDC + âˆ ECD 130Âº

= âˆ EDC + 20Âº

â‡’ âˆ EDC = 130Âº âˆ 20Âº = 110Âº

â‡’ âˆ BDC = 110Âº

Since, angles in the same segment are equal.

âˆ´ âˆ BAC = âˆ BDC

â‡’ âˆ BAC = 110Âº

**Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If âˆ DBC = 70Âº, âˆ BAC is 30Âº, find âˆ BCD. Further, if AB = BC, find âˆ ECD. Solution**: âˆµ Angles in the same segment of a circle are equal.

âˆ´ âˆ BAC = âˆ BDC

â‡’ 30Âº = âˆ BDC Also âˆ DBC = 70Âº [Given]

âˆ´ In Î”BCD, we have âˆ BCD + âˆ DBC + âˆ CDB = 180Âº [âˆµ Sum of angles of a triangle is 180Âº]

â‡’ âˆ BCD + 70Âº + 30Âº = 180Âº

â‡’ âˆ BCD = 180Âº âˆ 70Âº âˆ 30Âº = 80Âº

Now, in Î”ABC,

âˆµ AB = BC [Given]

âˆ´ âˆ BCA = âˆ BAC [Angles opposite to equal sides of a triangle are equal]

â‡’ âˆ BCA = 30Âº [âˆµ âˆ BAC = 30Âº]

Now, âˆ BCA + âˆ ECD = âˆ BCD

â‡’ 30Âº + âˆ ECD = 80Âº

â‡’ âˆ ECD = 80Âº âˆ 30Âº = 50Âº

**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:** âˆµ AC and BD are diameters.

âˆ´ AC = BD [âˆµ All diameters of a circle are equal] â€¦(1)

Since a diameter divides a circle into equal parts.

âˆ´ âˆ BAD = 90Âº [âˆµ Angle formed in a semicircle is 90Âº]

Similarly, âˆ ABC = 90Âº âˆ BCD = 90Âº and âˆ CDA = 90Âº

Now, in right Î”ABC and right Î”BAD, AC = BD [From (1)]

AB = AB [Common]

âˆ´ Î”ABC â‰Œ Î”BAD [RHS criterion]

â‡’ BC = AD [c.p.c.t.]

Similarly, AB = CD Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is right angle.

âˆ´ ABCD is a rectangle.

**Question 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Solution**: We have a trapezium ABCD such that AB || CD and AD = BC.

Let us draw BE || AD such that ABED is a parallelogram.

âˆµ The opposite angles of a parallelogram are equal.

âˆ´ âˆ BAD = âˆ BED â€¦(1)

and AD = BE [Opposite sides of a parallelogram] â€¦(2)

But AD = BC [Given] â€¦(3)

âˆ´ From (2) and (3), we have BE = BC

â‡’ âˆ BEC = âˆ BCE [Angles opposite to equal sides of a triangle D are equal] â€¦(4)

Now, âˆ BED + âˆ BEC = 180Âº [Linear pairs]

â‡’ âˆ BAD + âˆ BCE = 180Âº [Using (1) and (4)]

i.e. A pair of opposite angles of quadrilateral ABCD is 180Âº.

âˆ´ ABCD is cyclic.

â‡’ The trapezium ABCD is a cyclic.

**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 the figure). Prove that âˆ ACP = âˆ QCD. Solution: **Since angles in the same segment of a circle are equal.

âˆ´ âˆ ACP = âˆ ABP â€¦(1)

Similarly, âˆ QCD = âˆ QBD â€¦(2)

Since âˆ ABP = âˆ QBD [Vertically opposite angles are equal]

âˆ´ From (1) and (2), we have âˆ ACP = âˆ 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: **We have a DABC, and two circles described with diameters as AB and AC respectively. They intersect in a point D.

Let us join A and D.

âˆµ AB is a diameter.

âˆ´ âˆ ADB is an angle formed in a semicircle.

â‡’ âˆ ADB = 90Âº â€¦(1)

Similarly, âˆ ADC = 90Âº â€¦(2)

Adding (1) and (2),

we have âˆ ADB + âˆ ADC = 90Âº + 90Âº = 180Âº

i.e. B, D and C are collinear points.

â‡’ BC is a straight line.

Thus, D lies on BC.

**Question 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that âˆ CAD = âˆ CBD. Solution: **We have right Î”ABC and right Î”ADC such that they are having AC as their common hypotenuse.

âˆµ AC is a hypotenuse.

âˆµ âˆ ADC = 90Âº = âˆ ABC

âˆ´ Both the triangles are in the same semicircle.

â‡’ A, B, C and D are concyclic.

Let us join B and D.

âˆµ DC is a chord.

âˆ´ âˆ CAD and âˆ CBD are formed in the same segment.

â‡’ âˆ CAD = âˆ CBD.

**Question 12. Prove that a cyclic parallelogram is a rectangle. Solution:** We have a cyclic parallelogram ABCD.

Since, ABCD is a cyclic quadrilateral.

âˆ´ Sum of its opposite angles = 180Âº

â‡’ âˆ A + âˆ C = 180Âº â€¦(1)

But âˆ A = âˆ C â€¦(2)

[âˆµ Opposite angles of parallelogram are equal]

From (1) and (2), we have âˆ A= âˆ C = 90Âº

Similarly, âˆ B= âˆ D = 90Âº

â‡’ Each angle of the parallelogram ABCD is of 90Âº.

Thus, ABCD is a rectangle.

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