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(B) The locus of the centre of all circles of given radius r, in the same planes, passing through a fixed point is a circle.

(i) An arc is a semi-circle when its ends are the ends of a diameter.

(ii) Segment of a circle is the region between an arc and chord of the circle.

(iii) A circle divides the plane, on which it lies, in three parts.

Correct option is (C) 3

Three non-collinear points are required to describe a circle.

Correct option is  C) 3

1. a) 7

2. b) 5

3. d) 3

4. c) 24

5. b) 60

If ‘r’ is the radius of the small circle,

radius of the large circle = r + 1

Perimeter of the small circle = 2 π r

Perimeter of the large circle = 2 π (r + 1) = 2

π r + 2 π

i.e., perimeter of the large circle is 2 π units more than the perimeter of the small circle.

Radius of the wheel = 20 cm

When the wheel rotates once it will travel the distance same as its area

Perimeter of the wheel = 2 × π × radius = 2 × 3.14 × 20 = 125.64 cm

The distance travelled forward when the wheel rotates once = 125.64 cm

The distance travelled forward when the wheel rotates 10 times

= 125.64 × 10 = 1256.4 cm

If diameter is 2 meters, perimeter is 6.28 meter.

If the diameter is 1 metre, perimeter is 6.28/2 meter.

If the diameter is 3 metres, perimeter = \(\frac{6.28}{2}\times3=9.42m\)

(i) We know that OB = OC which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBC = ∠OCB = 55^o

In △ BOC

Using the angle sum property

∠BOC + ∠OCB + ∠OBC = 180^o

By substituting the values

∠BOC + 55^o + 55^o = 180^o

On further calculation

∠BOC = 180^o – 55^o – 55^o

By subtraction

∠BOC = 180^o – 110^o

So we get

∠BOC = 70^o

(ii) We know that OA = OB which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBA= ∠OAB = 20^o

In △ AOB

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180^o

By substituting the values

∠AOB + 20^o + 20^o = 180^o

On further calculation

∠AOB = 180^o – 20^o – 20^o

By subtraction

∠AOB = 180^o – 40^o

So we get

∠AOB = 140^o

We know that

∠AOC = ∠AOB – ∠BOC

By substituting the values

∠AOC = 140^o – 70^o

So we get

∠AOC = 70^o

Given: ∠OAB = 30°, ∠OCB = 57°

In ΔOAB, AO = BO (Both are radii of the circle).

Thus ∠OAB = ∠OBA = 30° (angles opposite to equal sides are equal)

In ΔAOB, sum of all angles of a triangle is 180°.

∴ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ 30° + 30° + ∠AOB = 180°

⇒ ∠AOB = 180° - 30° - 30°

⇒ ∠AOB = 120° …..…… (1)

Now, in triangle OBC, OC and OB are radius of the circle and are thus equal.

∴ ∠OBC = ∠OCB = 57° (angles opposite to equal sides are equal)

In ΔAOB, sum of all angles of a triangle is 180°.

∴ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 57° + 57° + ∠BOC = 180°

⇒ ∠BOC = 180° - 57° - 57°

⇒ ∠BOC = 66° …………… (2)

Now, from equation (1), we have:

∠AOB = 120°

⇒ ∠AOC + ∠COB = 120°

⇒ ∠AOC + 66° = 120°

⇒ ∠AOC = 120° - 66°

⇒ ∠AOC = 54°

Therefore, ∠AOC = 54° and ∠BOC = 66°.

Consider two different circles intersecting at three point A, B and C

We know that these points are non collinear and a unique circle can be drawn using these points

This shows that our assumption is wrong

Therefore, it is proved that two different circles cannot intersect each other at more than two points.

Let there be two circles which intersect at three points say at A, B and C. Clearly, A, B and C are not collinear. We know that through three non-collinear points A, B and C one and only one circle can pass. Therefore, there cannot be two circles passing through A, B and C. In other words, the two circles cannot intersect at more than two points.

(c) 5 cm

The diameter is the largest chord of a circle, so radius = \(\frac{1}{2}\)x AB= \(\frac{1}{2}\) x 10 cm = 5 cm.

Answer is 1.5 cm. Diameter

Centres of the circle lie on the same straight line.

(i) True : Because it is a one dimensional figure 

(ii) True : Since, line segment joining the centre to any point on the circle is a radius of the circle 

(iii) True : Because each arc measures equal 

(iv) False : Since, a circle has only infinite number of equal chords 

(v) True : Because, radius equal to times of its diameter 

(vi) True : Yes, sector is the region between the chord and its corresponding arc 

(vii) False : The degree measure of an arc is half of the central angle containing the arc 

(viii) True : Yes, The degree measure of a semi-circle is 180°

(i) T 

(ii) T 

(iii) T 

(iv) F 

(v) T 

(vi) T 

(vii) F 

(viii) T

(i) Interior/Exterior 

(ii) Concentric 

(iii) The Exterior 

(iv) Arc 

(v) Diameter 

(vi) Semi-circle 

(vii) Center 

(viii) Three

(i) True 

(ii) True 

(iii) True 

(iv) False 

(v) True 

(vi) True 

(vii) False 

(viii) True

(i) Interior/exterior 

(ii) Concentric 

(iii) The exterior 

(iv) Arc 

(v) Diameter 

(vi) Semi-circle 

(vii) Centre 

(viii) Three

A tangent is a line touching a circle at one point.

          PQ = PR = 5 cm

and    PQ = QS

∴         PS = 2PQ

       = 2 x 5 = 10 cm

A circle can have infinite tangents.

\(OB^2=8^2+15^2\)

\(OB=\sqrt{64\,+225}\)

\(OB=17cm\)

Answer is D. 7 cm

Given: 

AB = 12 cm 

BC = 8 cm 

AC = 10 cm

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By the above property, 

AD = AF (tangent from A) 

BD = BE (tangent from B) 

CF = CE (tangent from C) 

Clearly,

AB = AD + DB = 12 cm

BC = BE + EC = 8 cm 

AC = AF + FC = 10 cm 

Now, 

AB – BC = 12 cm – 8 cm 

⇒ (AD + DB) – (BE + EC) = 12 cm – 8 cm 

⇒ AD + DB – BE – EC = 12 cm – 8 cm 

⇒ AD + BE – BE – CF = 12 cm – 8 cm [∵ DB = BE and CF = CE] 

⇒ AD – CF = 12 cm – 8 cm 

⇒ AD – (10 cm – AF) = 12 cm – 8 cm [∵AF + FC = 10 cm 

⇒ FC = 10 cm – AF] 

⇒ AD – (10 cm – AF) = 4 cm 

⇒ AD – 10 cm + AF = 4 cm 

⇒ AD + AD = 4 cm + 10 cm [∵ AD = AF] 

⇒ 2AD = 14 cm

⇒ AD = \(\frac{14cm}{2}\)

⇒ AD = 7 cm 

Hence, AD = 7 cm

Answer is B.  AC = BC

Given: 

AP = PB 

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By the above property, 

AP = AQ (tangent from A) 

BR = BP (tangent from B) 

CQ = CR (tangent from C) 

Clearly, 

AP = BP = BR 

AQ = AP = BR 

Now, 

AQ + QC = BR + RC 

⇒ AC = BC [∵AC = AQ + QC and BC = BR + RC] 

Hence, AC = BC

(1) 6 cm 

(2) 10 cm 

(3) 34 cm 

(4) 9 cm

Option : (D)

Given : 

AB and CD are two chords of the circle. 

∠APC = 90° 

∠ACP = ∠PBD = y 

(Angles on the same segment) 

In ΔACP, 

∠ACP + ∠APC + ∠PAC = 180° 

y + 90° + y = 180° 

x + y = 90°

i. AB = 8 cm, OA = 4 cm, OB = 4 cm

ii. OA = 5 cm, OB = 5 cm, OP = 5 cm

Answer is C. 5 cm

Given: 

AB = 8 cm 

PE = 3 cm 

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By above property, 

AB = AC = 8 cm (tangent from A) 

PE = CE = 3 cm (tangent from E)

Now, 

AE = AC – CE 

⇒ AE = 8 cm – 3 cm 

⇒ AE = 5 cm 

Hence, AE = 5 cm

Consider △ OEP and △ OFP

We know that

∠OEP = ∠OFP = 90^o

OP is common i.e. OP = OP

From the figure we know that OP bisects ∠BPD

It can be written as

∠OPE = ∠OPF

By ASA congruence criterion

△ OEP ≅ △ OFP

OE = OF (c. p. c. t)

We know that AB and CD are equidistant from the centre

So we get

AB = CD

Therefore, it is proved that AB = CD.

It changes into a rectangle or a square.

Let ABCD be a cyclic quadrilateral and let O be the centre of the corresponding circle 

Then, 

Each side of the equilateral ABCD is a chord of the circle and the perpendicular bisector of a chord always passes through the centre of the circle 

So, 

Right bisectors of the sides of the quadrilateral ABCD will pass through the centre O of the corresponding circle.

It is given that ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD

Construct DE ⊥ AB and CF ⊥ AB

Consider △ ADE and △ BCF

We know that

∠AED + ∠BFC = 90^o

From the figure it can be written as

∠ADE = ∠ADC – 90^o = ∠BCD – 90^o = ∠BCF

It is given that

AD = BC

By AAS congruence criterion

△ ADE ≅ △ BCF

∠A = ∠B (c. p. c. t)

We know that the sum of all the angles of a quadrilateral is 360^o

∠A + ∠B + ∠C + ∠D = 360^o

By substituting the values

2 ∠B + 2 ∠D = 360^o

By taking 2 as common

2 (∠B + ∠D) = 360^o

By division

∠B + ∠D = 360/2

So we get

∠B + ∠D = 180^o

So, ABCD is a cyclic quadrilateral.

Therefore, it is proved that the points A, B, C and D lie on a circle.

Given that, 

∠BOC = 100° 

By degree measure theorem, 

∠AOC = 2 ∠APC 100° = 2∠APC 

∠APC = 50° 

Therefore, 

∠APC + ∠ABC = 180° 

(Opposite angles of a cyclic quadrilateral) 

50° + ∠ABC = 180° 

∠ABC = 130° 

Therefore, 

∠ABC + ∠CBD = 180° 

(Linear pair) 

130° + ∠CBD = 180° 

∠CBD = 50°