Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

**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.

Contact Us

Awesome answers of question

Thank u

Awesome answers of question

Thank u very much

Best app for CBSE students. Love it.

Really its excellent

Awesome

Unbelievable

Thanks very much

hero re bhai

Thnx

Super vooper Hooper too good to

Understand

Thanks for uploading these simple solution

mathematics so easy everyone good going.

its beauty ful

it is easy to under standing .nice explained

Thanks

Thanks it had been of great help

Thanks

Loooove It!!!!!

Very nice i love this app

Nice .............â˜ºâ˜ºâ˜ºâ˜ºâ˜ºâ˜ºâ˜º. I liked it

it is so good nice

it is good for students

Thankx I like the wab site

Very helpful Google

Helpful Google....I m so happy

bakwash

Very nice

Very helpful form me