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Correct option is: (B) 0

There is only one circle can be draw through three non-collinear points in a plane.

But when three points are collinear then no circle can be drawn through them.

Hence, number of circles passing through three collinear points in a plane is O.

Correct option is: (B) 0

Given: B is the centre of circle. arc APC ≅ arc DQE 

To prove: chord AC ≅ chord DE

Proof: 

[m(arc APC) = ∠ABC (i) [Definition of measure of 

m(arc DQE) = ∠DBE] (ii) minor arc] 

arc APC ≅ arc ∠DQE (iii) [Given] 

∴ ∠ABC ≅ ∠DBE [From (i), (ii) and (iii)] 

In ∆ABC and ∆DBE, 

side AB ≅ side DB [Radil of the same circle] 

side [CB] side [EB] [Radii of the same circle] 

∠ABC ≅∠DBE [From (iii), Measures of congruent arcs]

∴ ∆ABC ≅ ∆DBE [SAS test of congruency] 

∴ chord AC ≅ chord DE [c.s.c.t]

(D) 8.8 or 2.2 cm

Two circles can touch each other internally or externally. 

∴ Distance between centres = 5.5 + 3.3 or

5.5 – 3.3 = 8.8 or 2.2

In ∆ABC, 

∠ABC = 90° 

∴ ∠BAC < 90° and ∠ACB < 90° [Given] 

∴ ∠ABC > ∠BAC and ∠ABC > ∠ACB 

∴ AC > BC and AC > AB [Side opposite to greater angle is greater]

∴ Hypotenuse is the longest side in right angled triangle.

We use theorem, If two angles of a triangle are not equal, then the side opposite to the greater angle is greater than the side opposite to the smaller angle.

Answer : (a) ∠CPA = ∠DPB 

In the bigger circle, ∠APX = ∠ABP 

In the smaller circle, ∠CPX = ∠PDC {Angles in alternate segment are equal.}

⇒ ∠APX + ∠CPA = ∠CPX = ∠PDC 

⇒ ∠ABP + ∠CPA = ∠PDC (∵ ∠APX = ∠ABP) 

⇒ ∠ABP + ∠CPA = ∠DBP + ∠DPB (ext. ∠ theorem in ∆ PDB)

∠ABP + ∠DPB 

⇒ ∠CPA = ∠DPB.

Proof: 

For smaller circle, 

seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B] 

∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]

 For bigger circle, 

seg OM ⊥ chord AB [Construction] 

∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]

∴ AP + PM = MQ + QB [A-P-M, M-Q-B] 

∴ AP + MQ = MQ + QB [From (i)] 

∴ AP = BQ

Let ABCD be a cyclic quadrilateral, and let O be the center 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 center of the circle 

So, right bisectors of the sides of quadrilaterals ABCD, will pass through the circle O of the corresponding circle

In the given figure,

TP is the tangent from an external point T and

∠PBT = 30°

Now,

∠APB = 90° (Angle in a semicircle)

∠PBT = 30° (given)

So, ∠PAB = 90° – 30° = 60°

But, ∠PAT + ∠PAB = 180° (Linear pair)

∠PAT + 60° = 180°

∠PAT = 180° – 60° = 120°

Also, ∠APT = ∠PBA = 30° (Angles in the alternate segment)

In ∆PAT,

∠PTA = 180° – (120° + 30°) = 180° – 150° = 30°

PA = AT

In right ∆APB,

sin 30° = AP/AB

1/2 = AP/AB

AB = 2 AP

Since AP = AT

AB = 2 AT

or AB/AT = 2/1

AB:AT = 2:1

or BA:AT = 2:1. Hence Proved.

In the given figure,

O be the centre of circle which is inscribed in a quadrilateral ABCD.

And OP = OQ = r = radius of circle

The circle touches the sides of quadrilateral at P, Q, R and S respectively.

AB = 29 cm, AD = 23 cm, ∠B = 90°

DS = 5 cm

Join OP and OQ.

Now,

OP = OQ = r and ∠B = 90°

So, PBQO is a square.

DR and DS are the tangents to the circle.

DR = DS = 5 cm

AQ and AR are tangents to the circle.

AR = AD – DR = 23 – 5 = 18 cm

AQ = AR = 18 cm

And BQ = AB – AQ = 29 – 18 = 11 cm

Since PBQO is a square.

OP = OQ = BQ = 11 cm

Hence, radius of the circle is 11 cm

Let ABC be an equilateral triangle of side 9 cm 

Let, AD be one of its medians and G be the centroids of the triangle ABC 

Then, 

AG : GD = 2: 1 

We know that, 

In an equilateral triangle centroid coincides with the circumcentre 

Therefore, 

G is the centre of the circumference with circum radius GA 

Also, 

G is the centre and GD is perpendicular to BC 

Therefore, 

In right triangle ADB, we have 

AB² = AD² + DB² 

9² = AD² + DB² 

AD = \(\frac{9\sqrt 3}{2}\) cm 

Therefore, 

Radius = AG = \(\frac{2}{3}\)AD 

= 3\(\sqrt{3}\) cm

Data : Two congruent circles intersect each other at points A and B. 

Through A any line segment PAQ is drawn so that P, Q lie on the two circles. 

To Prove: BP = BQ 

Construction: Join AB. 

Proof: Two congruent triangles with centres O and O’ intersects at A and B. Through A segment PAQ is drawn so that P, Q lie on the two circles. 

Similarly, ∠AQB= 70° in circle subtended by chord AB. Because Angles subtended by circumference by same chord. 

∴ ∠APB = ∠AQB = 70°. 

Now, in ∆PBQ, ∠QPB = ∠PQB. 

∴ Sides opposite to each other are equal. 

∴ BP = BQ.

(C) The sum of ∠DAB + ∠ABD is120⁰.

Given : ∠DAB = 50° 

By degree measure theorem: ∠BOD = 2 ∠BAD 

so, x = 2( 50°) = 100° 

Since, ABCD is a cyclic quadrilateral, we have 

∠A + ∠C = 180° 

50° + y = 180° 

y = 130°

We have, 

∠A = 3∠C 

Let, ∠C = x 

Therefore, 

∠A + ∠C = 180° 

(Opposite angles of cyclic quadrilateral) 

3x + x = 180° 

4x = 180° 

x = 45° 

∠A = 3x 

= 3 x 45° 

= 135° 

Therefore, 

∠A = 135°

Given that, 

PQ is a diameter of circle which bisects chord AB to C 

To prove : 

PQ bisects ∠AOB

Proof : 

In ΔAOC and ΔBOC,

OA = OB (Radius of circle) 

OC = OC (Common) 

AC = BC (Given) 

Then,

Δ ADC ≅ Δ BOC

(By SSS congruence rule)

∠AOC = ∠BOC 

(By c.p.c.t) 

Hence, 

PQ bisects ∠AOB.

∠A + ∠C = 180° …..(1) [Opposite angles of cyclic quadrilateral] 

Since m ∠A = 3(m∠C) (given) 

=> ∠A = 3∠C …(2) 

Equation (1) => 3∠C + ∠C = 180° 

or 4∠C = 180° 

or ∠C = 45° From equation (2) 

∠A = 3 x 45° = 135°

A cyclic quadrilateral ABCD with AB||CD and B = 70°. 

∠B + ∠C = 180° (Co-interior angle)

70° + ∠C = 180° 

∠C = 110° 

And, 

=> ∠B + ∠D = 180° (Opposite angles of Cyclic quadrilateral) 

70° + ∠D = 180° 

∠D = 110° 

Again, ∠A + ∠C = 180° (Opposite angles of cyclic quadrilateral) 

∠A + 110° = 180° 

∠A = 70° 

Answer: ∠A = 70° , ∠C = 110° and ∠D = 110°

Given that, 

∠ACB is an angle in major segment 

To prove : ∠ACB > 90° 

Proof : 

By degree measure theorem, 

∠AOB = 2∠ACB 

And, 

∠AOB < 180° 

Then, 

2∠ACB < 180° 

∠ACB < 90° 

Hence, proved

Given: ABCD is a cyclic quadrilateral with AD ‖ BC 

=> ∠A + ∠C = 180° ………(1) [Opposite angles of cyclic quadrilateral] 

and ∠A + ∠B = 180° ………(2) [Co-interior angles] 

Form (1) and (2), we have

∠B = ∠C

Hence proved.

Given that,

ABCD is a cyclic trapezium with AD ǁ BC and ∠B = 70°

Since, 

ABCD is a quadrilateral 

Then, 

∠B + ∠D = 180° 

70° + ∠D = 180° 

∠D = 110° 

Since, 

AD ǁ BC 

Then, 

∠A + ∠B = 180° 

(Co. interior angle) 

∠A + 70° = 180° 

∠A = 110° 

Since, 

ABCD is a cyclic quadrilateral 

Then, 

∠A + ∠C = 180° 

110° + ∠C = 180° 

∠C = 70°

Since, 

ABCD is a cyclic quadrilateral with AD||BC 

Then, 

∠A + ∠C = 180° ...(i) 

(Opposite angles of cyclic quadrilateral) 

And, 

∠A + ∠B = 180° ...(ii) 

(Co. interior angles) 

Comparing (i) and (ii), we get 

∠B = ∠C 

Hence, proved

Gardner is standing near the well initially and he did not return to the well after
watering the last tree.
Distance covered by Gardner to water 1^st tree and return to the initial position =
10 m + 10 m = 20 m
Distance covered by Gardner to water 2^nd tree and return to the initial position =
15 m + 15 m = 30 m

Distance covered by Gardner to water 3^rd tree and return to the initial position =
20 m + 20 m = 40 m
∴Distances covered by the Gardner to water the plants are in A.P.
Here a = 20, d = 10
Distance covered to water 25^th tree = 20 + (25 – 1) x 10 = 20 + 240 =260
Total distance covered by the Gardner=25/2[(2x20+(25-1)x10] -260 =2470
Thus, the total distance covered by the Gardner is 2740m.

27, a, b - 6 are in A.P.
d= t₂-t₁=t_3-t₂= a -27 =b-6-a
 ➪a+a=b-6+27
➪2a=b+21
➪2a-b=21

We have, an = 5-6n
⇒ a₁ = 5 - 6 x 1 = -1
So, the given sequence is an A.P with first term a = a₁ = -1 and last term l = a_n = 5 - 6n
Therefore the sum of n terms is given by: S_n= n/2 (a+l)= n/2 (-1+5-6n)=2n-3n²

Given, the nth terms of the sequences 3, 10, 17,... and 63, 65, 67,... are equal.
since, nth term of A.P.= a + (n – 1) d
➪ 3+(n-1)7=63+(n-1)2
➪ 3 + 7n-7=63 + 2n-2
7n-2n= 61+4
5n = 65
n = 13
Therefore, 13^th terms of both these A.P.s are equal to each other.

Solution: a = 200, d = 50 and n =30

s₃₀ = 30/2(400+50 x29) = 27750

Solution: a = 1, d = 2, and n =25

S₂₅ =25/2(2+2 x24) =625

Solution: Smallest three digit number divisible by 7 is 105 Greatest three digit number divisible by 7 is 994 Number of terms
= {(last term – first term )/common difference }+1
= {(994-105)/7}+1
= (889/7)+1=127+1=128

Solution: In the first AP a = 63 and d = 2
In the second AP a = 3 and d = 7
As per question,
63+2(n-1) = 2+ 7(n-1)
Or, 61 = 5 (n-1)
Or, n-1 = 61/5
As the result is not an integer so there wont be a term with equal values  for both APs.