Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 8 - Circles

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Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 8 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 10 students for Maths Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 489:

Answer:

Consider the figure.
We know that the tangent is perpendicular to the radius of a circle.
So, OPB is a right angled triangle, with $∠ OBP = 90 °$
By using pythagoras theorem in $△ OPB$ , we get
$\Rightarrow {OB}^{2} + {PB}^{2} = {OP}^{2} \Rightarrow (20)^{2} + {PB}^{2} = (29)^{2} \Rightarrow 400 + {PB}^{2} = 841 \Rightarrow {PB}^{2} = 841 - 400 = 441 \Rightarrow PB = \sqrt{441} = 21$
So, length of the tangent from point P is 21 cm.

Page No 490:

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras' theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Page No 490:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ (6.5)² = (2.5)² + PA²
⇒ PA² = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.

Page No 490:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          .....(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm                    .....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   .....(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm           .....(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

Page No 490:

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Page No 490:

Answer:

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

Page No 490:

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Page No 490:

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Page No 491:

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Page No 491:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

Page No 491:

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Page No 491:

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 54 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3 \Rightarrow 36 = 15 + 6 + x + 9 + x$

$\Rightarrow 36 = 30 + 2 x \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

Page No 491:

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 3² = OR² + (2.4)²
⇒ OR² = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (2.4)²
⇒ x² = y² + 5.76 .....(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 1.8)² = x² + 3²
⇒ y² + 3.6y + 3.24 = x² + 9
⇒ y² + 3.6y = x² + 5.76 .....(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm

Page No 491:

Answer:

Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90^∘
Now, ∠OQC + ∠POQ = 180^∘ (co-interior angles)
⇒ ∠POQ = 180^∘ − 90^∘ = 90^∘
Similarly, Now, ∠ORA + ∠POR = 180^∘ (co-interior angles)
⇒ ∠POR = 180^∘ − 90^∘ = 90^∘
Now, ∠POR + ∠POQ = 90^∘ + 90^∘ = 180^∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^∘
Hence, QR is a straight line passing through centre O.

Page No 492:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.

Page No 492:

Answer:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180^∘ (Angle sum property of triangle)
⇒ ∠BAP = 180^∘ − 90^∘ − 30^∘ = 60^∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 60^∘ − 30^∘ = 30^∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\sin ∠ ABP = \frac{AP}{BA} \Rightarrow \sin 30 ° = \frac{AT}{BA} \Rightarrow \frac{1}{2} = \frac{AT}{BA}$
∴ BA : AT = 2 : 1

Page No 494:

Answer:

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
⇒6 + 8 = AD + 9
⇒ AD = 5 cm

Page No 494:

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘+ 90^∘ = 360^∘
⇒ 230^∘+ ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Page No 495:

Answer:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90^∘ + 90^∘ + 70^∘ = 360^∘
⇒ 250^∘ + ∠QOT = 360^∘
⇒ ∠QOT = 110^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ TRQ = \frac{1}{2} (∠ QOT) = 55 °$

Page No 495:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O₁
and
ED = EB for the circle having centre O₁
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD

Page No 495:

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

Page No 495:

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 21 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 42 = 7 \times 2 + (4 + x) \times 2 + (3 + x) \times 2 \Rightarrow 21 = 7 + 4 + x + 3 + x$

$\Rightarrow 21 = 14 + 2 x \Rightarrow 2 x = 7 \Rightarrow x = 3.5 cm$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Page No 495:

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Page No 495:

Answer:

Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn't passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90^∘ (OP ⊥ AB)
⇒ ∠ORP = 90^∘
∠ROP + ∠OPA = 180^∘ (Co interior angles)
⇒∠ROP + 90^∘ = 180^∘
⇒∠ROP = 90^∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

Page No 495:

Answer:

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ   (Radii)
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$\cos 60^{°} = \frac{PR}{OR} \Rightarrow \frac{1}{2} = \frac{PR}{OR} \Rightarrow OR = 2 PR \Rightarrow OR = PR + PR \Rightarrow OR = PR + RQ$

Page No 496:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm          .....(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm                    .....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   .....(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm           .....(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm

Page No 496:

Answer:

We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90^∘ + 90^∘= 360^∘
⇒ ∠APB + ∠AOB = 180^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.

Page No 496:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{AB}{2}$ = 4 cm
In right triangle AOP
AO² = OP² + PA²
⇒ 5² = OP² + 4²
⇒ OP² = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.

Page No 496:

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘          (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘         (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ PRQ = \frac{1}{2} (reflex ∠ POQ) = 120 °$

Page No 496:

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120⁰
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘

Page No 496:

Answer:

Let PA and PB be the two tangents drawn to the circle with centre O and radius a such that $∠ APB = 60 ° .$

In ∆OPB and ∆OPA
OB = OA = a    (Radii of the circle)
$∠ OBP = ∠ OAP = 90 °$ (Tangents are perpendicular to radius at the point of contact)
BP = PA    (Lengths of tangents drawn from an external point to the circle are equal)
So, ∆OPB ≌ ∆OPA           (SAS Congruence Axiom)
$∴ ∠ OPB = ∠ OPA = 30 °$ (CPCT)
Now,
In ∆OPB,
$\sin 30 ° = \frac{OB}{OP} \Rightarrow \frac{1}{2} = \frac{a}{OP} \Rightarrow OP = 2 a$
Thus, the length of OP is 2a.
Disclaimer: The answer given in the book is incorrect.

Page No 499:

Answer:

We can draw only two tangents from an external point to a circle.

Hence, the correct answer is option (b)

Page No 499:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
$OQ = \frac{1}{2} QS = 3 cm$ [∵Radius is half of diameter]
Now, in right triangle OQR
By using Pythagoras theorem, we have
OR² = RQ² + OQ²
= 4² + 3²
= 16 + 9
= 25
∴OR² = 25
⇒OR = 5 cm
Hence, the correct answer is option (c).

Page No 499:

Answer:

(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
$∴ O T ⊥ P T From right - angled triangle P T O, ∴ O P^{2} = O T^{2} + P T^{2} [U s i n g P y t h a g o r a s' t h e o r e m] \Rightarrow O P^{2} = {(7)}^{2} + {(24)}^{2} \Rightarrow O P^{2} = 49 + 576 \Rightarrow O P^{2} = 625 \Rightarrow O P = \sqrt{625} \Rightarrow O P = 25 cm$

Page No 499:

Answer:

Two diameters cannot be parallel as they perpendicularly bisect each other.
Hence, the correct answer is option (d)

Page No 499:

Answer:

In right triangle AOB
By using Pythagoras theorem, we have
AB² = BO² + OA²
= 10² + 10²
= 100 + 100
= 200
∴OR² = 200
⇒OR = $10 \sqrt{2}$ cm
Hence, the correct answer is option (c).

Page No 499:

Answer:

In right triangle PTO
By using Pythagoras theorem, we have
PO² = OT² + TP²
⇒ 10² = 6² + TP²
⇒100 = 36 + TP²
⇒TP² = 64
⇒ TP = 8 cm
Hence, the correct answer is option (a).

Page No 500:

Answer:

Construction: Join OT

We know that the radius and tangent are perperpendular at their point of contact
In right triangle PTO
By using Pythagoras theorem, we have
PO² = OT² + TP²
⇒ 26² = OT² + 24²
⇒676 = OT² + 576
⇒TP² = 100
⇒ TP = 10 cm
Hence, the correct answer is option (a).

Page No 500:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ [Angle sum property of a triangle]
⇒ 2∠OQP + 90^∘ = 180^∘
⇒ ∠OQP = 45^∘
Hence, the correct answer is option (b).

Page No 500:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBA = ∠OCA = 90^∘
Now, In quadrilateral ABOC
∠BAC + ∠OCA + ∠OBA + ∠BOC = 360^∘ [Angle sum property of a quadrilateral]
⇒ 40^∘ + 90^∘ + 90^∘ + ∠BOC = 360^∘
⇒ 220^∘ + ∠BOC = 360^∘
⇒ ∠BOC = 140^∘
Hence, the correct answer is option (d).

Page No 500:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBC = ∠OAC = 90^∘
Now, In quadrilateral ABOC
∠ACB + ∠OAC + ∠OBC + ∠AOB = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠ACB + 90^∘ + 90^∘ + 60^∘ = 360^∘
⇒ ∠ACB + 240^∘ = 360^∘
⇒ ∠ACB = 120^∘
Hence, the correct answer is option (d).

Page No 500:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ 10² = 6² + PA²
⇒ PA² = 64
⇒ PA = 8 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 8 cm
Now, AB = AP + PB = 8 + 8 = 16 cm
Hence, the correct answer is option (c).

Page No 500:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOB
By using Pythagoras theorem, we have
OA² = AB² + OB²
⇒ 17² = AB² + 8²
⇒289 = AB² + 64
⇒AB² = 225
⇒ AB = 15 cm
The tangents drawn from the external point are equal.
Therefore, the length of AC is 15 cm
Hence, the correct answer is option (b).

Page No 500:

Answer:

(b) 50°
$∠ A B C = 90^{0} (A ngle in a semicircle) I n △ A B C, w e h a v e : ∠ A C B + ∠ C A B + ∠ A B C = 180^{0} \Rightarrow 50^{0} + ∠ C A B + 90^{0} = 180^{0} \Rightarrow ∠ C A B = (180^{0} - 140^{0}) \Rightarrow ∠ C A B = 40^{0} Now, ∠ C A T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ C A B + ∠ B A T = 90^{0} \Rightarrow 40^{0} + ∠ B A T = 90^{0} \Rightarrow ∠ B A T = (90^{0} - 40^{0}) \Rightarrow ∠ B A T = 50^{0}$

Page No 501:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, OP = OQ
∵POQ is a isosceles right triangle
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ [Angle sum property of a triangle]
⇒ 70^∘ + 2∠OPQ = 180^∘
⇒ ∠OPQ = 55^∘
Now, ∠TPQ + ∠OPQ = 90^∘
⇒ ∠TPQ = 35^∘
Hence, the correct answer is option (a).

Page No 501:

Answer:

(c) $2 \sqrt{3} cm$

$O A ⊥ A T So, \frac{A T}{O T} = \cos 30^{0} \Rightarrow \frac{A T}{4} = \frac{\sqrt{3}}{2} \Rightarrow A T = (\frac{\sqrt{3}}{2} \times 4) \Rightarrow A T = 2 \sqrt{3}$

Page No 501:

Answer:

(c) 70°
$Given, PA and PB are tangents to a circle with centre O, with ∠ AOB = 110^{0} . Now, we know that tangents drawn from an external point are perpendicular to the radius at the point of contact . So, ∠ OAP = 90^{0} and ∠ OBP = 90^{0} . \Rightarrow ∠ OAP + ∠ OBP = 90^{0} + 90^{0} = 180^{0}, which shows that OABP is a cyclic quadrilateral . ∴ ∠ AOB + ∠ APB = 180^{0} \Rightarrow 110^{0} + ∠ APB = 180^{0} \Rightarrow ∠ APB = 180^{0} - 110^{0} \Rightarrow ∠ APB = 70^{0}$

Page No 501:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
AF = AE = 4 cm
BF = BD = 3 cm
EC = AC − AE = 11 − 4 = 7 cm
CD = CE = 7 cm
∴ BC = BD + DC = 3 + 7 = 10 cm
Hence, the correct answer is option (b).

Page No 501:

Answer:

We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180 degrees.
∴∠AOD + ∠BOC = 180^∘
⇒∠BOC = 180^∘ − 135^∘ = 45^∘
Hence, the correct answer is option (b).

Page No 501:

Answer:

$(a) 100 ° Given, ∠ Q P T = 50^{0} and ∠ O P T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O P Q = (∠ O P T - ∠ Q P T) = (90^{0} - 50^{0}) = 40^{0} O P = O Q (R adius of the same circle) \Rightarrow ∠ O Q P = ∠ O P Q = 40^{0} In ∆ P O Q, ∠ P O Q + ∠ O Q P + ∠ O P Q = 180^{0} ∴ ∠ P O Q = 180^{0} - (40^{0} + 40^{0}) = 100^{0}$

Page No 501:

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘+ 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘           [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘
Hence, the correct answer is option (b).

Page No 502:

Answer:

(c) $3 \sqrt{3} cm$

$Given, PA and PB are tangents to circle with cent re O and radius 3 cm and ∠ APB = 60^{0} . T angents drawn from an external point are equal; so, PA = PB . A nd OP is the bisector of ∠ APB, which gives ∠ OPB = ∠ OPA = 30^{0} . OA ⊥ PA . S o, from right - angled ∆ OPA, we have : \frac{OA}{AP} = \tan 30^{0} \Rightarrow \frac{OA}{AP} = \frac{1}{\sqrt{3}} \Rightarrow \frac{3}{AP} = \frac{1}{\sqrt{3}} \Rightarrow AP = 3 \sqrt{3} cm$

Page No 502:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Now, In △PQA
∠PQA + ∠QAP + ∠APQ = 180^∘            [Angle sum property of a triangle]
⇒ 90^∘ + ∠QAP + 27^∘ = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠QAP = 63^∘
In △PQA and △PRA
PQ = PR              (Tangents draw from same external point are equal)
QA = RA   (Radii of the circle)
AP = AP              (common)
By SSS congruency
△PQA ≅ △PRA
∠QAP = ∠RAP = 63^∘
∴∠QAR = ∠QAP + ∠RAP = 63^∘ + 63^∘ = 126^∘
Hence, the correct answer is option (c).

Page No 502:

Answer:

Construction: Join CA and CB

We know that the radius and tangent are perperpendular at their point of contact
∵∠CAP = ∠CBP = 90^∘
Since, in quadrilateral ACBP all the angles are right angles
∴ ACPB is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ CB = AP and CA = BP
Therefore, CB = AP = 4 cm and CA = BP = 4 cm
Hence, the correct answer is option (b).

Page No 502:

Answer:

(b) 50°
$Given, P A and P B are two tangents to a circle with centre O and ∠ A P B = 80^{0} . ∴ ∠ A P O = \frac{1}{2} ∠ A P B = 40^{0} [S ince they are equally inclined to the line segment joining the centre to that point] and ∠ O A P = 90^{0} [S ince tangents drawn from an external point are perpendicular to the radius at the point of contact] Now, in triangle A O P : ∠ A O P + ∠ O A P + ∠ A P O = 180^{0} \Rightarrow ∠ A O P + 90^{0} + 40^{0} = 180^{0} \Rightarrow ∠ A O P = 180^{0} - 130^{0} \Rightarrow ∠ A O P = 50^{0}$

Page No 502:

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠QPR = 90^∘    (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠APQ = ∠PRQ = 58^∘
Now, In △PQR
∠PQR + ∠PRQ + ∠QPR = 180⁰            [Angle sum property of a triangle]
⇒ ∠PQR + 58^∘ + 90⁰ = 180^∘
⇒ ∠PQR= 32^∘
Hence, the correct answer is option (a).

Page No 502:

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠DPC = 90^∘    (Angle in a semi circle is 90^∘)
Now, In △CDP
∠CDP + ∠DCP + ∠DPC = 180^∘            [Angle sum property of a triangle]
⇒ ∠CDP + ∠DCP + 90^∘ = 180^∘
⇒ ∠CDP + ∠DCP = 90^∘
By using alternate segment theorem
We have ∠CDP = ∠CPB
∴∠CPB + ∠ACP = 90^∘
Hence, the correct answer is option (b).

Page No 502:

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠BAC = 90^∘    (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠PAB = ∠ACB = 67^∘
Now, In △ABC
∠ABC + ∠ACB + ∠BAC = 180^∘            [Angle sum property of a triangle]
⇒ ∠ABC + 67^∘+ 90^∘ = 180^∘
⇒ ∠ABC= 23^∘
Now, ∠BAQ = 180^∘ − ∠PAB          [Linear pair angles]
= 180^∘ − 67^∘
= 113^∘
Now, In △ABQ
∠ABQ + ∠AQB + ∠BAQ = 180^∘            [Angle sum property of a triangle]
⇒ 23^∘ + ∠AQB + 113^∘ = 180^∘
⇒ ∠AQB = 44^∘
Hence, the correct answer is option (d).

Page No 503:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
NA = NC and NC = NB
We also know that angle opposite to equal sides are equal
∴ ∠NAC = ∠NCA and ∠NBC = ∠NCB
Now, ∠ANC + ∠BNC = 180^∘           [Linear pair angles]
⇒ ∠NBC + ∠NCB + ∠NAC + ∠NCA= 180^∘      [Exterior angle property]
⇒ 2∠NCB + 2∠NCA= 180^∘
⇒ 2(∠NCB + ∠NCA) = 180^∘
⇒ ∠ACB = 90^∘
Hence, the correct answer is option (c).

Page No 503:

Answer:

(a) 60 cm²

$G i v e n, O Q = O R = 5 cm, O P = 13 cm . ∠ O Q P = ∠ O R P = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) From right - angled ∆ P O Q : P Q^{2} = (O P^{2} - O Q^{2}) \Rightarrow P Q^{2} = 13^{2} - 5^{2} \Rightarrow P Q^{2} = 169 - 25 \Rightarrow P Q^{2} = 144 \Rightarrow P Q = \sqrt{144} \Rightarrow P Q = 12 cm ∴ a r (△ O Q P) = \frac{1}{2} \times P Q \times O Q \Rightarrow a r (△ O Q P) = (\frac{1}{2} \times 12 \times 5) {cm}^{2} \Rightarrow a r (△ O Q P) = 30 {cm}^{2} S i m i l a r l y, a r (△ O R P) = 30 {cm}^{2} ∴ a r (q u a d . P Q O R) = (30 + 30) {cm}^{2} = 60 {cm}^{2}$

Page No 503:

Answer:

(c) 40°

$S i n c e, A B ∥ P R, B Q is transversal . ∠ B Q R = ∠ A B Q = 70^{0} [A lternate angles] O Q ⊥ P Q R (T angents drawn from an external point are perpendicular to the radius at the point of contact) a n d A B ∥ P Q R ∴ Q L ⊥ A B; so, O L ⊥ A B ∴ O L bisects chord A B [P erpendicular drawn from the centre bisects the chord] From ∆ Q L A a n d ∆ Q L B : ∠ Q L A = ∠ Q L B = 90^{0} L A = L B (O L bisects chord A B) Q L is the common side . ∴ ∆ Q L A ≅ ∆ Q L B [By SAS congruency] ∴ ∠ Q A L = ∠ Q B L \Rightarrow ∠ Q A B = ∠ Q B A ∴ ∆ A Q B is isosceles . ∴ ∠ L Q A = ∠ L Q R ∠ L Q P = ∠ L Q R = 90^{0} ∠ L Q B = (90^{0} - 70^{0}) = 20^{0} ∴ ∠ L Q A = ∠ L Q B = 20^{0} \Rightarrow ∠ A Q B = ∠ L Q A + ∠ L Q B = 40^{0}$

Page No 503:

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle PTO
By using Pythagoras theorem, we have
PO² = OT² + TP²
⇒ PO² = 5² + 10²
⇒ PO² = 25 + 100
⇒ PO² = 125
⇒ PO= $\sqrt{125}$ cm
Hence, the correct answer is option (d).

Page No 503:

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠BPA = 90^∘    (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30^∘
Now, In △ABP
∠PBA + ∠BPA + ∠BAP = 180⁰            [Angle sum property of a triangle]
⇒ 30^∘ + 90⁰ + ∠BAP = 180^∘
⇒ ∠BAP = 60^∘
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 30^∘
Hence, the correct answer is option (b).

Page No 503:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
EK = EM = 9 cm
Now, EK + EM = 18 cm
⇒ ED + DK + EF + FM = 18 cm
⇒ ED + DH + EF + HF = 18 cm (∵DK = DH and FM = FH)
⇒ ED + DF + EF = 18 cm
⇒ Perimeter of △EDF = 18 cm
Hence, the correct answer is option (d)

Page No 503:

Answer:

Suppose PA and PB are two tangents we want to draw which inclined at an angle of 45⁰
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90⁰
Now, In quadrilateral AOBP
∠AOB + ∠OBP + + ∠OAP + ∠APB = 360⁰ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90⁰ + 90⁰ + 45⁰ = 360⁰
⇒ ∠AOB + 225⁰ = 360⁰
⇒ ∠AOB = 135⁰
Hence, the correct answer is option (b).

Page No 504:

Answer:

$(d) 70 ° P Q L i s a tangent O Q is the radius; s o, ∠ O Q L = 90^{0} . ∴ ∠ O Q S = (90^{0} - 50^{0}) = 40^{0} N o w, O Q = O S (Ra d i u s o f t h e s a m e c i r c l e) \Rightarrow ∠ O S Q = ∠ O Q S = 40^{0} S i m i l a r l y, ∠ O R S = (90^{0} - 60^{0}) = 30^{0}, A n d, O R = O S (R a d i u s o f t h e s a m e c i r c l e) \Rightarrow ∠ O S R = ∠ O R S = 30^{0} ∴ ∠ Q S R = ∠ O S Q + ∠ O S R \Rightarrow ∠ Q S R = (40^{0} + 30^{0}) \Rightarrow ∠ Q S R = 70^{0}$

Page No 504:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PS = PU = x
QT = QS = 12 cm
RT = RU = 9 cm
Now,

$Ar (△ PQR) = Ar (△ POR) + Ar (△ QOR) + Ar (△ POQ) \Rightarrow 189 = \frac{1}{2} \times OU \times PR + \frac{1}{2} \times OT \times QR + \frac{1}{2} \times OS \times PQ \Rightarrow 378 = 6 \times (x + 9) + 6 \times (21) + 6 \times (12 + x) \Rightarrow 63 = x + 9 + 21 + x + 12 \Rightarrow 2 x = 21 \Rightarrow x = 10.5 cm$
Now, PQ = QS + SP = 12 + 10.5 = 22.5 cm
Hence, the correct answer is option (c ).

Page No 504:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PT = PQ = 3.8 cm and PT = PR = 3.8 cm
∴ QR = QP + PR = 3.8 + 3.8 = 7.6 cm
Hence, the correct answer is option (d)

Page No 504:

Answer:

(a) 9 cm
$T angents drawn from an external point to a circle are equal . So, A Q = A P = 5 c m C R = C S = 3 c m a n d B R = (B C - C R) \Rightarrow B R = (7 - 3) c m \Rightarrow B R = 4 c m B Q = B R = 4 c m ∴ A B = (A Q + B Q) \Rightarrow A B = (5 + 4) c m \Rightarrow A B = 9 c m$

Page No 504:

Answer:

$(c) 36 cm Given, A P = 6 cm, B P = 5 cm, C Q = 3 cm and D R = 4 cm . T angents drawn from an external points to a circle are equal . So, A P = A S = 6 c m, B P = B Q = 5 c m, C Q = C R = 3 c m, D R = D S = 4 c m . ∴ A B = A P + P B = 6 + 5 = 11 c m B C = B Q + C Q = 5 + 3 = 8 c m C D = C R + D R = 3 + 4 = 7 c m A D = A S + D S = 6 + 4 = 10 c m ∴ Pe r i m e t e r o f q u a drilateral A BCD = AB + BC + CD + DA = (11 + 8 + 7 + 10) c m = 36 c m$

Page No 505:

Answer:

Given: AO and BO are the radius of the circle
Since, AO = BO
∴ △AOB is an isosceles triangle.
Now, in △AOB
∠AOB + ∠OBA + ∠OAB = 180^∘ (Angle sum property of triangle)
⇒ 100^∘ + ∠OAB + ∠OAB = 180^∘ (∠OBA = ∠OAB)
⇒ 2∠OAB = 80^∘
⇒ ∠OAB = 40^∘
We know that the radius and tangent are perperpendular at their point of contact
∵∠OAT = 90^∘
⇒ ∠OAB + ∠BAT = 90^∘
⇒ ∠BAT = 90^∘ − 40^∘ = 50^∘
Hence, the correct answer is option (b).

Page No 505:

Answer:

In right triangle ABC
By using Pythagoras theorem we have
AC² = AB² + BC²
= 5² + 12²
= 25 + 144
= 169
∴ AC² = 169
⇒ AC = 13 cm
Now,

$Ar (△ ABC) = Ar (△ AOB) + Ar (△ BOC) + Ar (△ AOC) \Rightarrow \frac{1}{2} \times AB \times BC = \frac{1}{2} \times OP \times AB + \frac{1}{2} \times OQ \times BC + \frac{1}{2} \times OR \times AC \Rightarrow 5 \times 12 = x \times 5 + x \times 12 + x \times 13 \Rightarrow 60 = 30 x \Rightarrow x = 2 cm$
Hence, the correct answer is option (b).

Page No 505:

Answer:

Construction: Join OR

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
BP = BQ = 27 cm
CQ = CR
Now, BC = 38 cm
⇒ BQ + QC = 38
⇒ QC = 38 − 27 = 11 cm
Since, all the angles in quadrilateral DROS are right angles.
Hence, DROS is a rectangle.
We know that opposite sides of rectangle are equal
∴ OS = RD = 10 cm
Now, CD = CR + RD
= CQ + RD
= 11 + 10
= 21 cm
Hence, the correct answer is option(d)

Page No 505:

Answer:

(a) 2 cm
$Given, A B = 8 cm, B C = 6 cm Now, in ∆ A B C : A C^{2} = A B^{2} + B C^{2} \Rightarrow A C^{2} = (8^{2} + 6^{2}) \Rightarrow A C^{2} = (64 + 36) \Rightarrow A C^{2} = 100 \Rightarrow A C = \sqrt{100} \Rightarrow A C = 10 cm P B Q O is a square . C R = C Q (Since the length s of tangents drawn from an external point are equal) ∴ C Q = (B C - B Q) = (6 - x) cm Similarly, A R = A P = (A B - B P) = (8 - x) cm ∴ A C = (A R + C R) = [(8 - x) + (6 - x)] cm \Rightarrow 10 = (14 - 2 x) cm \Rightarrow 2 x = 4 \Rightarrow x = 2 cm ∴ The radius of the circle is 2 cm .$

Page No 505:

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + DC = AD + BC
⇒6 + 4 = AD + 7
⇒ AD = 3 cm
Hence, the correct answer is option (a).

Page No 505:

Answer:

(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.
$So, P A = P B and therefore, ∠ P A B = ∠ P B A = x (say) . Then, in ∆ P A B : ∠ P A B + ∠ P B A + ∠ A P B = 180^{0} \Rightarrow x + x + 60^{0} = 180^{0} \Rightarrow 2 x = 180^{0} - 60^{0} \Rightarrow 2 x = 120^{0} \Rightarrow x = 60^{0} ∴ Each angle of △ P A B i s 60^{0} and therefore, it is an equilateral triangle . ∴ A B = P A = P B = 5 cm ∴ T he length of the chord A B is 5 cm .$

Page No 506:

Answer:

Construction: Join AF and AE

We know that the radius and tangent are perperpendular at their point of contact
∵∠AED = ∠AFD = 90^∘
Since, in quadrilateral AEDF all the angles are right angles
∴ AEDF is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ AF = DE = 5 cm
Therefore, the radius of the circle is 5 cm
Hence, the correct answer is option (c).

Page No 506:

Answer:

(b) 2 cm

$Given, A B = 5 cm, B C = 7 cm a n d C A = 6 cm . Let, A R = A P = x cm . B Q = B P = y cm C R = C Q = z cm (Since the length of tangents drawn from an external point are equal) Then, A B = 5 cm \Rightarrow A P + P B = 5 cm \Rightarrow x + y = 5 \dots . . (i) Similarly, y + z = 7 \dots \dots \dots (i i) and z + x = 6 \dots . (i i i) Adding (i), (ii) and (iii), we get: (x + y) + (y + z) + (z + x) = 18 \Rightarrow 2 (x + y + z) = 18 \Rightarrow (x + y + z) = 9 \dots \dots . (i v) N o w, (i v) - (i i) : (x + y + z) - (y + z) = (9 - 7) \Rightarrow x = 2 ∴ T he radius of the circle with centre A is 2 cm .$

Page No 506:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
AP = AQ
BP = BD
CQ = CD
Now, AB + BC + AC = 5 + 4 + 6 = 15
⇒AB + BD + DC + AC = 15 cm
⇒AB + BP + CQ + AC = 15 cm
⇒AP + AQ= 15 cm
⇒2AP = 15 cm
⇒AP = 7.5 cm
Hence, the correct answer is option(d)

Page No 506:

Answer:

(c) $4 \sqrt{10} cm$

$Given, O P = 5 cm, P A = 12 cm Now, j oin O and B . Then, O B = 3 cm . Now, ∠ O A P = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) Now, in ∆ O A P : O P^{2} = O A^{2} + P A^{2} \Rightarrow O P^{2} = 5^{2} + 12^{2} \Rightarrow O P^{2} = 25 + 144 \Rightarrow O P^{2} = 169 \Rightarrow O P = \sqrt{169} \Rightarrow O P = 13 Now, in ∆ O B P : P B^{2} = O P^{2} - O B^{2} \Rightarrow P B^{2} = 13^{2} - 3^{2} \Rightarrow P B^{2} = 169 - 9 \Rightarrow P B^{2} = 160 \Rightarrow P B = \sqrt{160} \Rightarrow P B = 4 \sqrt{10} cm$

Page No 506:

Answer:

(d) A circle can have more than two parallel tangents, parallel to a given line.
This statement is false because there can only be two parallel tangents to the given line in a circle.

Page No 507:

Answer:

(d)A straight line can meet a circle at one point only.
This statement is not true because a straight line that is not a tangent but a secant cuts the circle at two points.

Page No 507:

Answer:

(d) A tangent to the circle can be drawn from a point inside the circle.
This statement is false because tangents are the lines drawn from an external point to the circle that touch the circle at one point.

Page No 507:

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

$In ∆ O P Q, ∠ O P Q = 90^{0} . ∴ O Q^{2} = O P^{2} + P Q^{2} \Rightarrow O Q = \sqrt{O P^{2} + P Q^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 cm$

Page No 507:

Answer:

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

Assertion :-
We know that if two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.

Reason:-

Given, a parallelogram ABCD circumscribes a circle with centre O.
$A B = B C = C D = A D$
We know that the tangents drawn from an external point to circle are equal .
$∴ A P = A S \dots \dots . (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = A P + B P + C R + D R = A S + B Q + C Q + D S [from (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = A D + B C Thus, (A B + C D) = (A D + B C) \Rightarrow 2 A B = 2 A D [∵ opposite sides of a parallelogram are equal] \Rightarrow A B = A D ∴ C D = A B = A D = B C$
Hence, ABCD is a rhombus.

Page No 508:

Answer:

(d) Assertion(A) is false and Reasoning(R) is true.

Assertion: In this situation given in the diagram, the sum of opposite sides is always equal.
So, the correct relation should be: AB + CD = AD + CB
Hence, the assertion is false.
Reasoning: We know that in two concentric circles, the chord of the larger circle, which touches( or acts as a tangent to) the smaller circle, is bisected at the point of contact.
Therefore, Reasoning (R) is correct.
Hence, the correct answer is option (d).

Page No 513:

Answer:

$(b) 100 ° Given, ∠QPT = 50^{0} Now, ∠OPT = 90^{0} (S ince tangents drawn from an external point a re perpendicular to t he radius at point of contact) ∴ ∠OPQ = (∠OPT - ∠QPT) = (90^{0} - 50^{0}) = 40^{0} OP = OQ (Radii of the same c ircle) \Rightarrow ∠OPQ = ∠OQP = 40^{0} In △ POQ, ∠ POQ + ∠ OPQ + ∠OQP = 180^{0} \Rightarrow ∠ POQ + 40^{0} + 40^{0} = 180^{0} \Rightarrow ∠ POQ = 180^{0} - (40^{0} + 40^{0}) \Rightarrow ∠POQ = 180^{0} - 80^{0} \Rightarrow ∠POQ = 100^{0}$

Page No 513:

Answer:

$(c) 50 ° OA and OB a re the two r adii of a circle with centre O. Also, AP and BP are the tangents to the c ircle . Given, ∠ AOB = 130^{0} Now, ∠ OAB = ∠ OBA = 90^{0} (S ince tangents drawn from an external point a re perpendicular to the radius at p oint of c ontact) In quadrilateral OAPB, ∠ AOB + ∠ OAB + ∠ OBA + ∠ APB = 360^{0} \Rightarrow 130^{0} + 90^{0} + 90^{0} + ∠ APB = 360^{0} \Rightarrow ∠ APB = 360^{0} - (130^{0} + 90^{0} + 90^{0}) \Rightarrow ∠ APB = 360^{0} - 310^{0} \Rightarrow ∠ APB = 50^{0}$

Page No 513:

Answer:

(b) 50°

$From ∆ O P A and ∆ O P B, O A = O B (R adi i of the same circle) O P (C ommon side) P A = P B (S ince tangents drawn from an external point to a circle are equal) ∴ ∆ O P A ≅ ∆ O P B (SSS rule) ∴ ∠ A P O = ∠ B P O ∴ ∠ A P O = \frac{1}{2} ∠ A P B = 40^{0} A nd ∠ O A P = 90^{0} (S ince tangents drawn from an external point are perpendicular to the radius at point of contact) Now, in ∆ O A P, ∠ A O P + ∠ O A P + ∠ A P O = 180^{0} \Rightarrow ∠ A O P + 90^{0} + 40^{0} = 180^{0} \Rightarrow ∠ A O P = 180^{0} - 130^{0} = 50^{0}$

Page No 513:

Answer:

(b) 10 cm
Since the tangents from an external point are equal, we have:
$A D = A E, C D = C F, B E = B F Perimeter of ∆ A B C = A C + A B + C B = (A D - C D) + (C F + B F) + (A E - B E) = (A D - C F) + (C F + B F) + (A E - B F) = A D + A E = 2 A E = 2 \times 5 = 10 cm$

Page No 513:

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
CR = CQ, AS = AP and BQ = BP
Now, BC = 7 cm
⇒ CQ + BQ = 7
⇒ BQ = 7 − CQ
⇒ BQ = 7 − 3 [∵ CQ = CR = 3]
⇒ BQ = 4 cm
Again, AB = AP + PB
= AP + BQ
= 5 + 4 [∵ AS = AP = 5]
= 9 cm
Hence, the value of x is 9 cm.

Page No 514:

Answer:

$Here, OA = OB And OA ⊥ AP, OA ⊥ BP (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ OAP = 90^{0}, ∠ OBP = 90^{0} ∴ ∠ OAP + ∠ OBP = 90^{0} + 90^{0} = 180^{0} ∴ ∠ AOB + ∠ APB = 180^{0} (S ince, ∠ OAP + ∠ OBP + ∠ AOB + ∠ APB = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Page No 514:

Answer:

We know that tangents drawn from the external point are congruent.
∴ PA = PB
Now, In isoceles triangle APB
∠APB + ∠PBA + ∠PAB = 180^∘            [Angle sum property of a triangle]
⇒ ∠APB + 65^∘ + 65^∘ = 180^∘                          [∵∠PBA = ∠PAB = 65^∘ ]
⇒ ∠APB = 50^∘
We know that the radius and tangent are perperpendular at their point of contact
∴∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘           [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘ + 90^∘ = 360^∘
⇒ 230^∘ + ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Page No 514:

Answer:

$Here, O B is the bisector of ∠ C B D . (T wo tangents are equally inclined to the line segment joining the centre to that point) ∴ ∠ C B O = ∠ D B O = \frac{1}{2} ∠ C B D = 60^{0} ∴ From ∆ B O D, ∠ B O D = 30^{0} Now, from right - angled ∆ B O D, \frac{B D}{O B} = \sin 30^{0} \Rightarrow \frac{B D}{O B} = \frac{1}{2} \Rightarrow O B = 2 B D \Rightarrow O B = 2 B C (S ince tangents from an external point are equal, i . e . B C = B D) ∴ O B = 2 B C$

Page No 514:

Answer:

(i) A line intersecting a circle at two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinite tangents.

Page No 514:

Answer:

Given two tangents AP and AQ are drawn from a point A to a circle with centre O.
$To prove : A P = A Q Join O P, O Q and O A . A P is tangent at P and O P is the radius . ∴ O P ⊥ A P (since tangents drawn from an external point are perpendicular to the radius at the point of contact) Similarly, O Q ⊥ A Q In the right ∆ O P A and ∆ O Q A, we have : O P = O Q [radii of the same circle] ∠ O P A = ∠ O Q A (= 90^{0}) O A = O A [common side] ∴ ∆ O P A ≅ ∆ O Q A [By R . H . S - Congruence] Hence, A P = A Q$

Page No 514:

Answer:

$Here, P T and Q S are the tangents to the circle with centre O and A B is the diameter .$
Now, radius of a circle is perpendicular to the tangent at the point of contact.
$∴ O A ⊥ A T a n d O B ⊥ B S (s i n c e t a n g e n t s d r a w n f r o m a n e x t e r n a l p o i n t a r e p e r p e n d i c u l a r t o t h e r a d i u s a t p o i n t o f c o n t a c t) ∴ ∠ O A T = ∠ O B Q = 90^{0} But ∠ O A T and ∠ O B Q are alternate angles . ∴ A T is parallel to B S .$

Page No 514:

Answer:

$Given, AB = AC$
We know that the tangents from an external point are equal.
$∴ AD = AF, BD = BE and CF = C E \dots \dots .. (i) Now, AB = AC \Rightarrow AD + DB = AF + FC \Rightarrow AF + DB = AF + FC [from (i)] \Rightarrow DB = FC \Rightarrow BE = C E [from (i)] Hence proved .$

Page No 514:

Answer:

Given : A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.
: $To prove : ∠ A O P = ∠ A O Q . Proof : In ∆ A O P and ∆ A O Q, w e have : A P = A Q [tangents from a n external point are equal] O P = O Q [radii of the same circle] O A = O A [common side] ∴ ∆ A O P ≅ ∆ A O Q [by SSS - congruence] Hence, ∠ A O P = ∠ A O Q (c . p . c . t) .$

Page No 514:

Answer:

$Let R A and R B b e two tangents to the circle with centre O and let A B be a chord of the circle . We have to prove that ∠ R A B = ∠ R B A . ∴ Now, R A = R B (S ince tangents drawn from an external point to a circle are equal) ∴ In ∆ R A B, ∠ R A B = ∠ R B A (S ince opposite sides are equal, their base angles are also equal)$

Page No 515:

Answer:

Given, a parallelogram ABCD circumscribes a circle with centre O.
$A B = B C = C D = A D$
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
$∴ A P = A S \dots \dots . (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = A P + B P + C R + D R = A S + B Q + C Q + D S [from (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = A D + B C Thus, (A B + C D) = (A D + B C) \Rightarrow 2 A B = 2 A D [∵ opposite sides of a parallelogram are equal] \Rightarrow A B = A D ∴ C D = A B = A D = B C$
Hence, ABCD is a rhombus.

Page No 515:

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA = 5 cm and OC = 3 cm.
$In ∆ OAC, {OA}^{2} = {OC}^{2} + {AC}^{2} ∴ {AC}^{2} = {OA}^{2} - {OC}^{2} \Rightarrow {AC}^{2} = 5^{2} - 3^{2} \Rightarrow {AC}^{2} = 25 - 9 \Rightarrow {AC}^{2} = 16 \Rightarrow AC = 4 cm ∴ AB = 2 AC (since perpendicular drawn from the centre of the circle bisects the chord) ∴ AB = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Page No 515:

Answer:

We know that the tangents drawn from an external point to a circle are equal.

$∴ A P = A S \dots \dots (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = (A P + B P) + (C R + D R) = (A S + B Q) + (C Q + D S) [using (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = (A D + B C) Hence, (A B + C D) = (A D + B C)$

Page No 515:

Answer:

Given, a quadrilateral ABCD circumscribes a circle with centre O.
$To prove : ∠ A O B + ∠ C O D = 180^{0} and ∠ A O D + ∠ B O C = 180^{0} Join O P, O Q, O R and O S . We know that the tangents drawn from an external point of a circle subtend equal angles at the centre . ∴ ∠ 1 = ∠ 7, ∠ 2 = ∠ 3, ∠ 4 = ∠ 5 and ∠ 6 = ∠ 8 and ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360^{0} [angles at a point] \Rightarrow (∠ 1 + ∠ 7) + (∠ 3 + ∠ 2) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 8) = 360^{0} 2 ∠ 1 + 2 ∠ 2 + 2 ∠ 6 + 2 ∠ 5 = 360^{0} \Rightarrow ∠ 1 + ∠ 2 + ∠ 5 + ∠ 6 = 180^{0} \Rightarrow ∠ A O B + ∠ C O D = 180^{0} and ∠ A O D + ∠ B O C = 180^{0}$

Page No 515:

Answer:

Given, PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.
$T o p r o v e : ∠ A P B + ∠ A O B = 180^{0}$
We know that the tangent to a circle is perpendicular to the radius through the point of contact.
$∴ P A ⊥ O A \Rightarrow ∠ O A P = 90^{0} P B ⊥ O B \Rightarrow ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = (90^{0} + 90^{0}) = 180^{0} \dots \dots . (i)$

$But we know that the sum of all the angles of a quadrilateral is 360^{0} . ∴ ∠ O A P + ∠ O B P + ∠ A P B + ∠ A O B = 360^{0} \dots \dots \dots . (i i)$
From (i) and (ii), we get:
$∠ A P B + ∠ A O B = 180^{0}$

Page No 515:

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 16
⇒ PR + PR = 16
⇒ PR = 8
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 10² = OR² + (8)²
⇒ OR² = 36
⇒ OR = 6
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (8)²
⇒ x² = y² + 64 .....(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y = x² + 64 .....(2)
Solving (1) and (2), we get
x = 10.67
∴ TP = 10.67 cm

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