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Since, sum of the angles between radii and between intersection point of tangent is 180° .Angle at the point of intersection of tangents 

= 180° - 130° = 50°.

Solution: S_{l = 144 =9/2(a + 28)}

➪ 144 x 2/9 = 32 =a + 28

➪a = 32 -28 = 4

Construct OL ⊥ AB and OM ⊥ CD

Consider △ OLB and △ OMC

We know that ∠OLB and ∠OMC are perpendicular bisector

∠OLB = ∠OMC = 90^o

We know that AB || CD and BC is a transversal

From the figure we know that ∠OBL and ∠OCD are alternate interior angles

∠OBL = ∠OCD

So we get OB = OC which is the radii

By AAS congruence criterion

△ OLB ≅ △ OMC

OL = CM (c. p. c. t)

We know that the chords equidistant from the centre are equal

So we get

AB = CD

Therefore, it is proved that AB = CD.

Let the common difference of the A.P. be x
Given: The smallest angle = 8°
⇒ a = 8
And the largest is 72°
⇒ a_n = 72
⇒ a + (n – 1)d = 72
⇒ 8 + (n – 1)d = 72
⇒ (n – 1) d = 72 – 8 = 64 ..........(1)
We know that sum of all the angles of a circle is 360° Sn = n/2[(2a + (n-1)d] 

Putting the value of n in equation (1) we get
(9 – 1) d = 64
d = 8
Now angle in fourth sector = a₄ = a + (4 – 1) d
= a + 3d = 8 + 3 × 8 = 8 + 24 = 32
∴ The value of n = 9 and angle in fourth sector is 32°

Let the nth term be 84 more than the 13th term.
Now a/q,
a=3, d=10-3=7
So, 13th term= a+12d =3+12x7=87
Then nth term=84+87=171
171=a + (n-1)d
171=3 + (n-1)x7
171-3/7+1=n
168/7+1=n
24+1=25=n

Therefore 25th term of the ap will be 84 more than 13th term

a. Diameter of the circle = 6 × 2 = 12 cm

Side of the square = 12 cm

b. Area of the circle = n r²

= n × 6 × 6 = 36n = 36 × 3.14 = 113.04 cm²

c. Area of the square = 12 × 12 = 144 cm²

Area of the shaded portion = 144 – 113.04 = 30.96 cm²

i. Radius of the small circle = 2 cm

Area = π × 2² = 3.14 × 4

= 12.56 cm²

Daigonal of the square = 4 cm

One side of the square = 4/√2 cm

Area of the square = \(\frac{4}{\sqrt{2}}\times\frac{4}{\sqrt{2}}=\frac{16}{2}=8 \,cm\)

Differences between the areas = 12.56 – 8 = 4.56 cm

ii. Radius of the circle = 2 cm

Area= π × 2² = 3.14 × 4 = 12.56 cm²

One side of the regular hexagon = 2 cm

Area of the regular hexagon = \(6\times\frac{\sqrt{3}\times2^2}{4}=6\sqrt{3}\)

= 6 × 1.73 = 10.38 cm²

Differences between the areas = 12.56 – 10.38 = 2.18 cm²

Option : (B)

AB is a chord of circle P and Q are two points on circle 

∠APB = ∠AQB 

(Angles on the same segment)

Option 'B' is correct Answer

∠APB = ∠AQB 

  (Angles on the same segment) 

Solution: n₇ = a + 6d
And, n₁₀ = a + 9d
Or, a + 9d – a – 6d = 7
Or, 3d = 7
Or, d = 7/3

Let O be the circle circumscribing the cyclic rectangle ABCD. 

Since, 

∠ABC = 90°and AC is the chord of the circle. 

Similarly, 

BD is a diameter 

Hence, 

Point of intersection of AC and BD is the centre of the circle.

The correct answer is  (B) 20⁰. 

We have, 

∠ACB = 40° 

∠DPB = 120° 

∠ADB = ∠ACB = 40° 

(Angle on same segment) 

In triangle PDB, 

By angle sum property 

∠PDB + ∠PBD + ∠BPD = 180° 

40° + ∠PBD + 120° = 180° 

∠PBD = 20° 

Therefore, 

∠CBD = 20°

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.

∴ AB + CD = AD + BC

\(\Rightarrow\) 6 + 8 = AD = 9

\(\Rightarrow\) AD = 5 cm

Proof: 

diameter CD ⊥ chord AB [Given] 

∴ seg OE ⊥ chord AB [C-O-E, O-E-D] 

∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord] 

In ∆CEA and ∆CEB,

 ∠CEA ≅ ∠CEB [Each is of 90°] 

seg AE ≅ seg BE [From (i)]

seg CE ≅ seg CE [Common side] 

∴ ∆CEA ≅ ∆CEB [SAS test] 

∴ seg AC ≅ seg BC [c. s. c. t.] 

∴ ∆ABC is an isosceles triangle.

CD is diameter, O is the centre. 

CD ⊥ AB; 

Let M be the point of inter-section. 

Now in ΔAMD and ΔBMD 

AM = BM [ ∵ radius perpendicular to a chord bisects it] 

∠AMD =∠BMD [given] 

DM = DM (common)

∴ ΔAMD ≅ ΔBMD 

⇒ AD = BD [C.P.C.T]

Data: ∆ABC and ∆ADC are right angled triangles having common hypotenuse AC. 

To Prove: ∠CAD = ∠CBD 

Proof: In ∆ABC, ∠ABC = 90° 

∴ ∠BAC + ∠BCA = 90° …………. (i) 

In ∆ADC, ∠ADC = 90° 

∴ ∠DAC + ∠DCA = 90° …………… (ii) 

Adding (i) and (ii), 

∠BAC + ∠BCA + ∠DAC + ∠DCA = 90 + 90 

(∠BAC + ∠DAC) + (∠BCA + ∠DCA) = 180° 

∠BAD + ∠BCD = 180° 

∴ ∠ABC + ∠ADC = 180° 

If opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral. 

∴ ∠CAD = ∠CBD (∵ Angles in the same segment).

Answer : (b) 65º 

∠ROQ = 180° – 50° = 130° (∵ ∠OQP + ∠ORP + ∠OPR + ∠ROQ = 360° and ∠OQP = ∠ORP = 90°) 

RT = TM, QS = SM (Tangents to a circle from the same external point are equal) 

Also, OQ = OM = OR (Radii of the given circle) 

∴  ∠ROT = ∠TOM and ∠MOS = ∠SOQ. (∵ Tangents from the an external point subtend equal angles at the centre)

⇒ ∠SOT = ∠SOM + ∠TOM = \(\frac{1}{2}\) ∠QOM + \(\frac{1}{2}\) ∠ROM 

∠SOT = \(\frac{1}{2}\) ∠ROQ 

= \(\frac{1}{2}\) × 130° 

= 65°.

‘O’ is the centre of the circle. 

OM = ON and 0M ⊥ PQ; ON ⊥ RS 

Thus the chords FQ and RS are equal. 

[ ∵ chords which are equidistant from the centre are equal in length] 

∴ RS = PQ = 6cm

∠APO = 30° …given

From P we have two tangents PA and PB

We know that if we join point P and centre of circle O then the line PO divides the angle between tangents

⇒ ∠APO = ∠OPB = 30° …(i)

∠OAP = ∠OBP = 90° …radius is perpendicular to tangent …(ii)

Consider quadrilateral OAPB

⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°…sum of angles of quadrilateral

From figure ∠APB = ∠APO + ∠OPB

⇒ ∠OAP + ∠APO + ∠OPB + ∠PBO + ∠AOB = 360°

Using (i) and (ii)

⇒ 90° + 30° + 30° + 90° + ∠AOB = 360°

⇒ 240° + ∠AOB = 360°

⇒ ∠AOB = 120°

Hence ∠AOB is 120°

We know that tangent segments to a circle from the same external point are congruent. So, we have

EA = EC for the circle having center O₁

and

ED = EB for the circle having center O₁

Now, Adding ED on both sides in EA = EC. we get

EA + ED = EC + ED

\(\Rightarrow\) EA + EB = EC + ED

\(\Rightarrow\) AB = CD

Correct option is: (C) 5√3

\(\because\) \(\angle\) APB = 60°

=  \(\angle\) APO = \(\frac {\angle APB}2 = \frac {60^\circ}{2}\) = 30°

Also OA \(\perp\) AP (Angle between radius and tangent at point of contact)

\(\therefore\) \(\angle\) OAP = 90°

Now, in right \(\triangle\) OAP

\(\frac {PA}{OP}\) = cos (\(\angle\)APO)

= \(\frac {PA}{10}\) = cos 30° (\(\because\) \(\angle\) OPA = 30° & OP = 10 cm).

= PA = 10 cos 30° = \(\frac {\sqrt3}2 \times 10 = 5\sqrt3\).

Correct option is: (C) 5√3

Correct answer is (b) 2 cm

Given, AB = 5 cm, BC = 7 cm and CA = 6 cm.

Let, AR = AP = x cm.

BQ = BP = y cm

CR = CQ = z cm

(Since the length of tangents drawn from an external point arc equal)

Then, AB = 5 cm

\(\Rightarrow\) AP + PB = 5 cm

\(\Rightarrow\) x + y = 5   ......(i)

Similarly, y + z = 7   ....(ii)

and z + x = 6   .........(iii)

Adding (i), (ii) and (iii), we get:

(x + y) + (y + z) + (z + x) = 18

\(\Rightarrow\) 2(x + y + z) = 18

\(\Rightarrow\) (x + y + z) = 9    ........(iv)

Now, (iv) – (ii):

\(\Rightarrow\) x = 2

∴ The radius of the circle with center A is 2 cm.

Here, OB is the bisector of ∠CBD.

(Two tangents are equally inclined to the line segment joining the center to that point)

\(\therefore \angle CBO=\angle DBO=\frac{1}{2}\angle CBD=60^\circ\)

∴ From △BOD, ∠BOD = 30°

Now, from right – angled △BOD,

\(\Rightarrow\) \(\frac{BD}{OB}=sin30^\circ\)

\(\Rightarrow\) OB = 2BD

\(\Rightarrow\) OB = 2BC (Since tangents from an external point are equal. i.e., BC = BD)

∴ OB = 2BC

It can be clearly show that OB bisects ∠DBC.

∴∠OBC = ∠OBD = 60

In ?OBC,

∠OBC = 60, ∠OCB = 90

∠COB + ∠OBC +∠OCB = 180 [Angle sum property of triangle]

∠COB + 60 + 90 = 180

∠COB = 180 – 150 = 30

sin(∠COB) = BC/BO

1/2 = BC/BO

Hence, BO = 2BC

P and M are the tangents to the given circles.

(A) The ∠ACB is 48⁰.

Correct option is: (B) 140°

Correct option is: (C) 90°

The angle between a tangent and the radius drawn at the point of contact is 90°.

Correct option is: (C) 90°

Correct option is: (C) 12, 12

\(\because\) PA & PB are tangents to a circle from outer point P.

\(\therefore\) PA = PB (\(\because\) Tangents from a fixed outer point to circle are of equal length)

Among all given options only option (C) gives two equal length for tangents PA & PB.

\(\therefore\) Option (C) is correct.

Correct option is: (C) 12, 12

Solution: a_n = 350=17+9(n-1)

➪​9(n-1) =350-17=333

➪n-1=333/9=37

➪n=38

Now, S₃₈=38/2(17+350)=19 x367=6973

Solution: a₂₂  = 149 = a + 7 x 21

➪ 149 -147 = 2 =a

Now, S₂₂  = 22/2(2+149) =11 x151 =1661

Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be 

x² + y² + 2gx + 2fy + c = 0 …….(i) 

For point (9, 1), 

Substituting x = 9 and y = 1 in (i), we get 

81 + 1 + 18g + 2f + c = 0 

⇒ 18g + 2f + c = -82 …..(ii) 

For point (7, 9), 

Substituting x = 7 and y = 9 in (i), we get 

49 + 81 + 14g + 18f + c = 0 

⇒ 14g + 18f + c = -130 ……(iii) 

For point (-2, 12), 

Substituting x = -2 and y = 12 in (i), we get 

4 + 144 – 4g + 24f + c = 0

⇒ -4g + 24f + c = -148 …..(iv) 

By (ii) – (iii), we get 4g – 16f = 48 

⇒ g – 4f = 12 …..(v) 

By (iii) – (iv), we get 18g – 6f = 18 

⇒ 3g – f = 3 ……(vi) 

By 3 × (v) – (vi), we get -11f = 33 

⇒ f = -3 

Substituting f = -3 in (vi), we get 3g – (-3) = 3 

⇒ 3g + 3 = 3 

⇒ g = 0 

Substituting g = 0 and f = -3 in (ii), we get 

18(0) + 2(-3) + c = – 82 

⇒ -6 + c = -82

 ⇒ c = -76 

Equation of the circle becomes 

x² + y² + 2(0)x + 2(-3)y + (-76) = 0 

⇒ x² + y² – 6y – 76 = 0 ……(vii) 

Now for the point (6, 10), 

Substituting x = 6 and y = 10 in L.H.S. of (vii),

we get L.H.S = 6² + 10² – 6(10) – 76 

= 36 + 100 – 60 – 76 

= 0 

= R.H.S. 

∴ Point (6,10) satisfies equation (vii). ∴ the given points are concyciic.

Intercepts a and b on axes that implies circle passes through the points (a, 0) and (0, b) passing through origin 

⇒ c = 0 

(a, 0) a² + 2ga = 0 ⇒ g = \(\frac{a}{2}\)

(0, -b) b² + 2fb = 0 ⇒ f = \(\frac{-b}{2}\)

∴ The required equation is x² + y² – ax – by = 0.