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Correct option is: (B) 90°

\(\because\) Tangent BC intersect/ touches circle at point O and OA is radius drawn from point of contact of circle and tangent.

\(\therefore\) OA \(\perp\) BC (Tangent and radius drawn from point of contact is perpendicular to each other)

\(\therefore\) \(\angle\) AOB = 90°

Correct option is: (B) 90°

Answer is  B. \(\sqrt{127}\,cm\)

Given: 

AP = 10 cm

OA = 6 cm 

OB = 3 cm 

Property : The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

By above property, ∆PAO is right-angled at ∠PAO (i.e., ∠PAO = 90°) and ∆PBO is right-angled at ∠PBO (i.e., ∠PBO = 90°). 

Therefore by Pythagoras theorem in ∆PAO, 

OP² = OA² + AP² 

⇒ OP² = 6² + 10²

 ⇒ OP² = 36 + 100 

⇒ OP = √136

Now by Pythagoras theorem in ∆PBO, 

OP² = OB² + BP² 

BP² = OP² – OB² 

⇒ BP² = (√136) ² – 3² 

⇒ BP² = 136 – 9 

⇒ BP= √127 

Hence, BP= √127 cm

The circles with centres P and Q touch each other at R. 

∴ By theorem of touching circles, 

P – R – Q 

i. In ∆PAR, 

seg PA = seg PR [Radii of the same circle] 

∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem] 

Similarly, in ∆QBR, 

seg QR = seg QB [Radii of the same circle] 

∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem] 

But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles]

∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)] 

But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.

∴ seg AP || seg BQ [Alternate angles test] 

ii. In ∆APR and ∆RQB, 

∠PAR ≅ ∠QRB [From (i) and (iii)] 

∠APR ≅ ∠RQB [Alternate angles] 

∴ ∆APR – ∆RQB [AA test of similarity] 

iii. ∠PAR = 35° [Given] 

∴ ∠RBQ = ∠PAR= 35° [From (iv)] 

In ∆RQB, 

∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°] 

∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)] 

∴ ∠RQB + 2 ∠RBQ = 180° 

∴ ∠RQB + 2 × 35° = 180° 

∴ ∠RQB + 70° = 180° 

∴ ∠RQB = 110°

(C) 50°

According to the question,

A circle with centre O, diameter AC and ∠ACB = 50°

AT is a tangent to the circle at point A

Since, angle in a semicircle is a right angle

∠CBA = 90°

By angle sum property of a triangle,

∠ACB + ∠CAB + ∠CBA = 180°

50° + ∠CAB + 90° = 180°

∠CAB = 40° … (1)

Since tangent to at any point on the circle is perpendicular to the radius through point of contact,

We get,

OA ⏊ AT

∠OAT = 90°

∠OAT + ∠BAT = 90°

∠CAT + ∠BAT = 90°

40° + ∠BAT = 90° [from equation (1)]

∠BAT = 50°

Given: X and Y are the centres of circle. 

To prove: radius XA || radius YB 

Construction: Draw segments XZ and YZ.

Proof:

By theorem of touching circles, points X, Z, Y are collinear.

∴ ∠XZA ≅ ∠BZY [Vertically opposite angles]

Let ∠XZA = ∠BZY = a ……… (i)

Now, seg XA seg XZ [Radii of the same circIe]

∴ ∠XAZ ≅∠XZA = a …………….. (ii) [Isosceles triangle theorem]

Similarly, seg YB ≅ seg YZ [Radii of the same circie]

∴ ∠BZY = ∠ZBY = a …………….. (iii) [Isosceles triangle theorem]

∴ ∠XAZ = ∠ZBY [From (i), (ii) and (iii)]

∴ radius XA || radius YB [Alternate angles test]

From the figure we know that ∠AOD and ∠OEC form right angles

So we get

∠AOD = ∠OEC = 90^o

We know that OD || BC and OC is a transversal

From the figure we know that ∠AOD and ∠OEC are corresponding angles

∠AOD = ∠OEC

We know that ∠DOC and ∠OCE are alternate angles

∠DOC = ∠OCE = 30^o

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.

So we get

∠DOC = 2 ∠DBC

It can be written as

∠DBC = ½ ∠DOC

By substituting the values

∠DBC = 30/2

By division

y = ∠DBC = 15^o

In the same way

∠ABD = ½ ∠AOD

By substituting the values

∠ABD = 90/2

By division

∠ABD = 45^o

We know that

∠ABE = ∠ABC = ∠ABD + ∠DBC

So we get

∠ABE = ∠ABC = 45^o + 15^o

By addition

∠ABE = ∠ABC = 60^o

Consider △ ABE

Using the angle sum property

∠BAE + ∠AEB + ∠ABE = 180^o

By substituting the values

x + 90^o + 60^o = 180^o

On further calculation

x = 180^o – 90^o – 60^o

By subtraction

x = 180^o – 150^o

So we get

x = 30^o

Therefore, the value of x is 30^o and y is 15^o.

Answer : (d) 90º 

 ∠QSR = ∠QRP = y° (Angles in alternate segment are equal) 

Also, ∠QRS = 90° (Angle in a semi-circle) 

∠PRS = ∠PRQ + ∠QRS = y° + 90° 

In ∆ PRS, ∠SPR + ∠PRS + ∠PSR = 180° 

⇒ x° + y° + 90° + y° = 180° 

⇒ x + 2y = 90°.

Correct option is  C) 240°

1. c) 150

2. a) 75

3. b) 75

4. a) 90

(d) 25°

∠ABD = ∠AFD = 25° (Angles in the same segment) 

∠EBC = ∠ABD = 25° (Vert. opp. ∠s) 

∠EFC = ∠EBC = 25° (Angles in the same segment)

True. Each point in the circumference are equidistant from the centre of the circle. That distance is called radius of the circle.

True. A circle has two dimension.

False. The part which joins centre of the circle to the end points of arc, area coveted with two radii and arc s is called sector.

False. A circle has infinite number of equal chords because in a circle there are infinite points.

True. Diameter is a line segment which passes through centre of the circle.

Option : (C)

Arc AB = 260° (Given) 

Let a point C on the circle 

We Know that, 

An angle subtended by an arc at the centre of the circle is double the angle subtended at any point on the circle. 

∠ACB = \(\frac{1}{2}\)∠AOB

∠ACB = \(\frac{1}{2}\) x 100

= 50°

False. If equilateral triangle is constructed in interior of the circle, the three equal parts are called minor arcs.

Option : (A)

The centre O lies in the interior of S.

Option : (C)

Arc ACB = 3 arc AB (Given) 

Central angle = 270° 

Degree measures of the two arcs are 90°

Solution: a_{22 =149 =a+7 x21}

➪​149 -147=2=a

Now, S_{22 = 22/2(2+149) = 11 x151 =1661}

Solution: S_{n  = 400 = n/2 (5+45)}

➪​ n = 400/25 = 16

Now, a_{16 = 45 = 5 + d x 15}

➪ ​15d = 40

d = 8/3

(i) The centre of a circle lies in the interior of the circle, (exterior/interior) 

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in the exterior of the circle, (exterior/interior) 

(iii) The longest chord of a circle is a diameter of the circle. 

(iv) An arc is a semicircle when its ends are the ends of a diameter. 

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

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

(i) The centre of a circle lies in interior of the circle.

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.

(iii) The longest chord of a circle is a diameter of the circle.

Diameter

= 2 x radius

= 2 x 3.5

= 7 cm

Radius = diameter ÷ 2

= 6 ÷ 2

= 3 cm

Radius = Diameter ÷ 2

= 5 ÷ 2

= 2.5 cm

Solution: -8 = a + 8d
4 = a + 2d
Or, -8 – 4 = 6d
Or, -12 = 6d
Or, d = -2
Hence, a = -8 + 16 = 8
0 = 8 + -2(n-1)
Or, 8 = 2(n-1)
Or, n-1 = 4

Or, n = 5

Let digits of the number be (a – d), a and (a + d) respectively.
∴ The required number is 100 (a – d) + 10a + (a + d) .
Given : The sum of the digits = 15
➪ (a-d) + a + (a+d) = 15
3a=15➪  a = 5
Now, the number on reversing the digits is 100 (a + d) + 10a + (a – d) .
According to the question
100(a – d) + 10a + a + d = 100 (a + d) + 10a + (a – d) + 594
on solving we get, d = -3
The digits of the number are (5 – (–3)), 5, (5 + (–3) = 8, 5, 2
And the required number is 8 × 100 + 5 × 10 + 2 = 852

The correct option is: (C) Both (A) and (B)

Explanation:

In the given figure, PQ is the common chord of the two circles.

=> AB bisects the common chord PQ at M.

 .'. PM = MQ

Moreover, PQ is perpendicular to AB.

.'. Option (C) is correct.