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#### Question 1:

Consider the figure.
We know that the tangent is perpendicular to the radius of a circle.
So, OPB is a right angled triangle, with $\angle \mathrm{OBP}=90°$
By using pythagoras theorem in $△\mathrm{OPB}$, we get

So, length of the tangent from point P is 21 cm.

#### Question 2:

Consider the figure.
We know that the tangent is perpendicular to the radius of a circle.
So, OPB is a right angled triangle, with $\angle \mathrm{OBP}=90°$
By using pythagoras theorem in $△\mathrm{OPB}$, we get

So, length of the tangent from point P is 21 cm.

#### Question 3:

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.

#### Question 4:

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.

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)
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
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

#### Question 5:

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)
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
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

#### Question 6:

Construction:  Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90âˆ˜
Now, In â–³OCA and â–³OCB
∠OCA = ∠OCB = 90âˆ˜
OA = OB     (Radii of the larger circle)
OC = OC     (Common)
By RHS congruency
â–³OCA ≅ â–³OCB
∴ CA = CB

#### Question 7:

Construction:  Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90âˆ˜
Now, In â–³OCA and â–³OCB
∠OCA = ∠OCB = 90âˆ˜
OA = OB     (Radii of the larger circle)
OC = OC     (Common)
By RHS congruency
â–³OCA ≅ â–³OCB
∴ CA = CB

#### Question 10:

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.

#### Question 11:

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.

#### Question 12:

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,
$\mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right)\phantom{\rule{0ex}{0ex}}⇒54=\frac{1}{2}×\mathrm{BC}×\mathrm{OD}+\frac{1}{2}×\mathrm{AB}×\mathrm{OE}+\frac{1}{2}×\mathrm{AC}×\mathrm{OF}\phantom{\rule{0ex}{0ex}}⇒108=15×3+\left(6+x\right)×3+\left(9+x\right)×3\phantom{\rule{0ex}{0ex}}⇒36=15+6+x+9+x\phantom{\rule{0ex}{0ex}}$

∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

#### Question 13:

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,
$\mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right)\phantom{\rule{0ex}{0ex}}⇒54=\frac{1}{2}×\mathrm{BC}×\mathrm{OD}+\frac{1}{2}×\mathrm{AB}×\mathrm{OE}+\frac{1}{2}×\mathrm{AC}×\mathrm{OF}\phantom{\rule{0ex}{0ex}}⇒108=15×3+\left(6+x\right)×3+\left(9+x\right)×3\phantom{\rule{0ex}{0ex}}⇒36=15+6+x+9+x\phantom{\rule{0ex}{0ex}}$

∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

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

#### Question 14:

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

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 = 90âˆ˜
Now, ∠OQC + ∠POQ = 180âˆ˜       (co-interior angles)
⇒ ∠POQ = 180âˆ˜ − 90âˆ˜ = 90âˆ˜
Similarly, Now, ∠ORA + ∠POR = 180âˆ˜       (co-interior angles)
⇒ ∠POR = 180âˆ˜ − 90âˆ˜ = 90âˆ˜
Now, ∠POR + ∠POQ = 90âˆ˜ + 90âˆ˜ = 180âˆ˜
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180âˆ˜
Hence, QR is a straight line passing through centre O.

#### Question 15:

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 = 90âˆ˜
Now, ∠OQC + ∠POQ = 180âˆ˜       (co-interior angles)
⇒ ∠POQ = 180âˆ˜ − 90âˆ˜ = 90âˆ˜
Similarly, Now, ∠ORA + ∠POR = 180âˆ˜       (co-interior angles)
⇒ ∠POR = 180âˆ˜ − 90âˆ˜ = 90âˆ˜
Now, ∠POR + ∠POQ = 90âˆ˜ + 90âˆ˜ = 180âˆ˜
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180âˆ˜
Hence, QR is a straight line passing through centre O.

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5          [âˆµ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18          [âˆµ 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.

#### Question 16:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5          [âˆµ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18          [âˆµ 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.

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90âˆ˜
By using alternate segment theorem
We have ∠APB = ∠PAT = 30âˆ˜
Now, in â–³APB
∠BAP + ∠APB + ∠BAP = 180âˆ˜      (Angle sum property of triangle)
⇒ ∠BAP = 180âˆ˜ − 90âˆ˜ − 30âˆ˜ = 60âˆ˜
Now, ∠BAP = ∠APT + ∠PTA        (Exterior angle property)
⇒ 60âˆ˜ = 30âˆ˜ + ∠PTA
⇒ ∠PTA = 60âˆ˜ − 30âˆ˜ = 30âˆ˜
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\mathrm{sin}\angle \mathrm{ABP}=\frac{\mathrm{AP}}{\mathrm{BA}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}30°=\frac{\mathrm{AT}}{\mathrm{BA}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{\mathrm{AT}}{\mathrm{BA}}$
∴ BA : AT = 2 : 1

#### Question 1:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90âˆ˜
By using alternate segment theorem
We have ∠APB = ∠PAT = 30âˆ˜
Now, in â–³APB
∠BAP + ∠APB + ∠BAP = 180âˆ˜      (Angle sum property of triangle)
⇒ ∠BAP = 180âˆ˜ − 90âˆ˜ − 30âˆ˜ = 60âˆ˜
Now, ∠BAP = ∠APT + ∠PTA        (Exterior angle property)
⇒ 60âˆ˜ = 30âˆ˜ + ∠PTA
⇒ ∠PTA = 60âˆ˜ − 30âˆ˜ = 30âˆ˜
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\mathrm{sin}\angle \mathrm{ABP}=\frac{\mathrm{AP}}{\mathrm{BA}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}30°=\frac{\mathrm{AT}}{\mathrm{BA}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{\mathrm{AT}}{\mathrm{BA}}$
∴ BA : AT = 2 : 1

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

#### Question 2:

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

Construction: Join OB

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

#### Question 3:

Construction: Join OB

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

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠OTP = ∠OQP = 90âˆ˜
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360âˆ˜            [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90âˆ˜ + 90âˆ˜ + 70âˆ˜ = 360âˆ˜
⇒ 250âˆ˜ + ∠QOT = 360âˆ˜
⇒ ∠QOT = 110âˆ˜
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.
$\therefore \angle \mathrm{TRQ}=\frac{1}{2}\left(\angle \mathrm{QOT}\right)=55°$

#### Question 4:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠OTP = ∠OQP = 90âˆ˜
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360âˆ˜            [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90âˆ˜ + 90âˆ˜ + 70âˆ˜ = 360âˆ˜
⇒ 250âˆ˜ + ∠QOT = 360âˆ˜
⇒ ∠QOT = 110âˆ˜
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.
$\therefore \angle \mathrm{TRQ}=\frac{1}{2}\left(\angle \mathrm{QOT}\right)=55°$

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

#### Question 5:

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

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90âˆ˜
Now, ∠OPQ = ∠OPT − ∠TPQ = 90âˆ˜ − 70âˆ˜ = 20âˆ˜
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20âˆ˜           (Angles opposite to equal sides are equal)
Now, In isosceles â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180âˆ˜       (Angle sum property of a triangle)
⇒ ∠POQ = 180âˆ˜ − 20âˆ˜ − 20âˆ˜ = 140âˆ˜

#### Question 6:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90âˆ˜
Now, ∠OPQ = ∠OPT − ∠TPQ = 90âˆ˜ − 70âˆ˜ = 20âˆ˜
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20âˆ˜           (Angles opposite to equal sides are equal)
Now, In isosceles â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180âˆ˜       (Angle sum property of a triangle)
⇒ ∠POQ = 180âˆ˜ − 20âˆ˜ − 20âˆ˜ = 140âˆ˜

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,
$\mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right)\phantom{\rule{0ex}{0ex}}⇒21=\frac{1}{2}×\mathrm{BC}×\mathrm{OD}+\frac{1}{2}×\mathrm{AB}×\mathrm{OE}+\frac{1}{2}×\mathrm{AC}×\mathrm{OF}\phantom{\rule{0ex}{0ex}}⇒42=7×2+\left(4+x\right)×2+\left(3+x\right)×2\phantom{\rule{0ex}{0ex}}⇒21=7+4+x+3+x\phantom{\rule{0ex}{0ex}}$

∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

#### Question 7:

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,
$\mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right)\phantom{\rule{0ex}{0ex}}⇒21=\frac{1}{2}×\mathrm{BC}×\mathrm{OD}+\frac{1}{2}×\mathrm{AB}×\mathrm{OE}+\frac{1}{2}×\mathrm{AC}×\mathrm{OF}\phantom{\rule{0ex}{0ex}}⇒42=7×2+\left(4+x\right)×2+\left(3+x\right)×2\phantom{\rule{0ex}{0ex}}⇒21=7+4+x+3+x\phantom{\rule{0ex}{0ex}}$

∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

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.

The length of the chord of the larger circle is 8 cm.

#### Question 8:

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.

The length of the chord of the larger circle is 8 cm.

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 âˆ¥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90âˆ˜ (OP ⊥ AB)
⇒ ∠ORP  = 90âˆ˜
∠ROP + ∠OPA = 180âˆ˜ (Co interior angles)
⇒∠ROP + 90âˆ˜ = 180âˆ˜
⇒∠ROP = 90âˆ˜
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.

#### Question 9:

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 âˆ¥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90âˆ˜ (OP ⊥ AB)
⇒ ∠ORP  = 90âˆ˜
∠ROP + ∠OPA = 180âˆ˜ (Co interior angles)
⇒∠ROP + 90âˆ˜ = 180âˆ˜
⇒∠ROP = 90âˆ˜
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.

Construction: Join PO and OQ
In â–³POR and â–³QOR
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, â–³POR ≅ â–³QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120âˆ˜
⇒ ∠PRO = 60âˆ˜
Now, In â–³POR
$\mathrm{cos}{60}^{°}=\frac{\mathrm{PR}}{\mathrm{OR}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{\mathrm{PR}}{\mathrm{OR}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=2\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=\mathrm{PR}+\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=\mathrm{PR}+\mathrm{RQ}$

#### Question 10:

Construction: Join PO and OQ
In â–³POR and â–³QOR
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, â–³POR ≅ â–³QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120âˆ˜
⇒ ∠PRO = 60âˆ˜
Now, In â–³POR
$\mathrm{cos}{60}^{°}=\frac{\mathrm{PR}}{\mathrm{OR}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{\mathrm{PR}}{\mathrm{OR}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=2\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=\mathrm{PR}+\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=\mathrm{PR}+\mathrm{RQ}$

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)
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

#### Question 11:

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)
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

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

#### Question 12:

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

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 = $\frac{\mathrm{AB}}{2}$ = 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.

#### Question 13:

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 = $\frac{\mathrm{AB}}{2}$ = 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.

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90âˆ˜
Now, ∠OPQ = ∠OPT − ∠QPT = 90âˆ˜ − 60âˆ˜ = 30âˆ˜
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30âˆ˜          (Angles opposite to equal sides are equal)
Now, In isosceles â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180âˆ˜       (Angle sum property of a triangle)
⇒ ∠POQ = 180âˆ˜ − 30âˆ˜ − 30âˆ˜ = 120âˆ˜
Now, ∠POQ + reflex ∠POQ = 360âˆ˜         (Complete angle)
⇒ reflex ∠POQ = 360âˆ˜ − 120âˆ˜ = 240âˆ˜
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.
$\therefore \angle \mathrm{PRQ}=\frac{1}{2}\left(\mathrm{reflex}\angle \mathrm{POQ}\right)=120°$

#### Question 14:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90âˆ˜
Now, ∠OPQ = ∠OPT − ∠QPT = 90âˆ˜ − 60âˆ˜ = 30âˆ˜
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30âˆ˜          (Angles opposite to equal sides are equal)
Now, In isosceles â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180âˆ˜       (Angle sum property of a triangle)
⇒ ∠POQ = 180âˆ˜ − 30âˆ˜ − 30âˆ˜ = 120âˆ˜
Now, ∠POQ + reflex ∠POQ = 360âˆ˜         (Complete angle)
⇒ reflex ∠POQ = 360âˆ˜ − 120âˆ˜ = 240âˆ˜
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.
$\therefore \angle \mathrm{PRQ}=\frac{1}{2}\left(\mathrm{reflex}\angle \mathrm{POQ}\right)=120°$

Construction: Join OB

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

#### Question 15:

Construction: Join OB

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

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

In âˆ†OPB and âˆ†OPA
OB = OA = a                    (Radii of the circle)
$\angle \mathrm{OBP}=\angle \mathrm{OAP}=90°$        (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, âˆ†OPB â‰Œ âˆ†OPA           (SAS Congruence Axiom)
$\therefore \angle \mathrm{OPB}=\angle \mathrm{OPA}=30°$    (CPCT)
Now,
In âˆ†OPB,
$\mathrm{sin}30°=\frac{\mathrm{OB}}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{a}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OP}=2a$
Thus, the length of OP is 2a.
Disclaimer: The answer given in the book is incorrect.

#### Question 1:

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

In âˆ†OPB and âˆ†OPA
OB = OA = a                    (Radii of the circle)
$\angle \mathrm{OBP}=\angle \mathrm{OAP}=90°$        (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, âˆ†OPB â‰Œ âˆ†OPA           (SAS Congruence Axiom)
$\therefore \angle \mathrm{OPB}=\angle \mathrm{OPA}=30°$    (CPCT)
Now,
In âˆ†OPB,
$\mathrm{sin}30°=\frac{\mathrm{OB}}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{a}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OP}=2a$
Thus, the length of OP is 2a.
Disclaimer: The answer given in the book is incorrect.

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

Hence, the correct answer is option (b)

#### Question 2:

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

Hence, the correct answer is option (b)

We know that the radius and tangent are perperpendular at their point of contact
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).

#### Question 3:

We know that the radius and tangent are perperpendular at their point of contact
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).

(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.

#### Question 4:

(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.

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

#### Question 5:

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

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).

#### Question 6:

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).

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).

#### Question 7:

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).

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).

#### Question 8:

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).

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

#### Question 9:

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

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

#### Question 10:

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

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

#### Question 11:

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

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).

#### Question 12:

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).

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).

#### Question 13:

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).

(b) 50°

#### Question 14:

(b) 50°

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

#### Question 15:

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

(c)

(c)

(c) 70°

#### Question 17:

(c) 70°

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).

#### Question 18:

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).

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

#### Question 19:

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

#### Question 20:

Construction: Join OB

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

#### Question 21:

Construction: Join OB

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

(c)

#### Question 22:

(c)

We know that the radius and tangent are perperpendular at their point of contact
Now, In â–³PQA
∠PQA + ∠QAP + ∠APQ = 180âˆ˜            [Angle sum property of a triangle]
⇒ 90âˆ˜ + ∠QAP + 27âˆ˜ = 180âˆ˜                          [âˆµ∠OAB = ∠OBA ]
⇒ ∠QAP = 63âˆ˜
In â–³PQA and â–³PRA
PQ = PR              (Tangents draw from same external point are equal)
QA = RA             (Radii of the circle)
AP = AP              (common)
By SSS congruency
â–³PQA ≅ â–³PRA
∠QAP = ∠RAP = 63âˆ˜
∴∠QAR = ∠QAP + ∠RAP = 63âˆ˜ + 63âˆ˜ = 126âˆ˜
Hence, the correct answer is option (c).

#### Question 23:

We know that the radius and tangent are perperpendular at their point of contact
Now, In â–³PQA
∠PQA + ∠QAP + ∠APQ = 180âˆ˜            [Angle sum property of a triangle]
⇒ 90âˆ˜ + ∠QAP + 27âˆ˜ = 180âˆ˜                          [âˆµ∠OAB = ∠OBA ]
⇒ ∠QAP = 63âˆ˜
In â–³PQA and â–³PRA
PQ = PR              (Tangents draw from same external point are equal)
QA = RA             (Radii of the circle)
AP = AP              (common)
By SSS congruency
â–³PQA ≅ â–³PRA
∠QAP = ∠RAP = 63âˆ˜
∴∠QAR = ∠QAP + ∠RAP = 63âˆ˜ + 63âˆ˜ = 126âˆ˜
Hence, the correct answer is option (c).

Construction: Join CA and CB

We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠CAP = ∠CBP = 90âˆ˜
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).

#### Question 24:

Construction: Join CA and CB

We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠CAP = ∠CBP = 90âˆ˜
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).

(b) 50°

#### Question 25:

(b) 50°

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠QPR = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
By using alternate segment theorem
We have ∠APQ = ∠PRQ = 58âˆ˜
Now, In â–³PQR
∠PQR + ∠PRQ + ∠QPR = 1800            [Angle sum property of a triangle]
⇒ ∠PQR + 58âˆ˜ +  900 = 180âˆ˜
⇒ ∠PQR= 32âˆ˜
Hence, the correct answer is option (a).

#### Question 26:

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠QPR = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
By using alternate segment theorem
We have ∠APQ = ∠PRQ = 58âˆ˜
Now, In â–³PQR
∠PQR + ∠PRQ + ∠QPR = 1800            [Angle sum property of a triangle]
⇒ ∠PQR + 58âˆ˜ +  900 = 180âˆ˜
⇒ ∠PQR= 32âˆ˜
Hence, the correct answer is option (a).

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠DPC = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
Now, In â–³CDP
∠CDP + ∠DCP + ∠DPC = 180âˆ˜            [Angle sum property of a triangle]
⇒ ∠CDP + ∠DCP +  90âˆ˜ = 180âˆ˜
⇒ ∠CDP + ∠DCP = 90âˆ˜
By using alternate segment theorem
We have ∠CDP = ∠CPB
∴∠CPB + ∠ACP = 90âˆ˜
Hence, the correct answer is option (b).

#### Question 27:

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠DPC = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
Now, In â–³CDP
∠CDP + ∠DCP + ∠DPC = 180âˆ˜            [Angle sum property of a triangle]
⇒ ∠CDP + ∠DCP +  90âˆ˜ = 180âˆ˜
⇒ ∠CDP + ∠DCP = 90âˆ˜
By using alternate segment theorem
We have ∠CDP = ∠CPB
∴∠CPB + ∠ACP = 90âˆ˜
Hence, the correct answer is option (b).

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠BAC = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
By using alternate segment theorem
We have ∠PAB = ∠ACB = 67âˆ˜
Now, In â–³ABC
∠ABC + ∠ACB + ∠BAC = 180âˆ˜            [Angle sum property of a triangle]
⇒ ∠ABC + 67âˆ˜ +  90âˆ˜ = 180âˆ˜
⇒ ∠ABC= 23âˆ˜
Now, ∠BAQ = 180âˆ˜ − ∠PAB          [Linear pair angles]
= 180âˆ˜ − 67âˆ˜
= 113âˆ˜
Now, In â–³ABQ
∠ABQ + ∠AQB + ∠BAQ = 180âˆ˜            [Angle sum property of a triangle]
⇒ 23âˆ˜ + ∠AQB + 113âˆ˜ = 180âˆ˜
⇒ ∠AQB = 44âˆ˜
Hence, the correct answer is option (d).

#### Question 28:

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠BAC = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
By using alternate segment theorem
We have ∠PAB = ∠ACB = 67âˆ˜
Now, In â–³ABC
∠ABC + ∠ACB + ∠BAC = 180âˆ˜            [Angle sum property of a triangle]
⇒ ∠ABC + 67âˆ˜ +  90âˆ˜ = 180âˆ˜
⇒ ∠ABC= 23âˆ˜
Now, ∠BAQ = 180âˆ˜ − ∠PAB          [Linear pair angles]
= 180âˆ˜ − 67âˆ˜
= 113âˆ˜
Now, In â–³ABQ
∠ABQ + ∠AQB + ∠BAQ = 180âˆ˜            [Angle sum property of a triangle]
⇒ 23âˆ˜ + ∠AQB + 113âˆ˜ = 180âˆ˜
⇒ ∠AQB = 44âˆ˜
Hence, the correct answer is option (d).

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 = 180âˆ˜           [Linear pair angles]
⇒ ∠NBC + ∠NCB + ∠NAC + ∠NCA= 180âˆ˜      [Exterior angle property]
⇒ 2∠NCB + 2∠NCA= 180âˆ˜
⇒ 2(∠NCB + ∠NCA) = 180âˆ˜
⇒ ∠ACB = 90âˆ˜
Hence, the correct answer is option (c).

#### Question 29:

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 = 180âˆ˜           [Linear pair angles]
⇒ ∠NBC + ∠NCB + ∠NAC + ∠NCA= 180âˆ˜      [Exterior angle property]
⇒ 2∠NCB + 2∠NCA= 180âˆ˜
⇒ 2(∠NCB + ∠NCA) = 180âˆ˜
⇒ ∠ACB = 90âˆ˜
Hence, the correct answer is option (c).

(a) 60 cm2

(a) 60 cm2

(c) 40°

#### Question 31:

(c) 40°

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).

#### Question 32:

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).

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠BPA = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30âˆ˜
Now, In â–³ABP
∠PBA + ∠BPA + ∠BAP = 1800            [Angle sum property of a triangle]
⇒ 30âˆ˜ +  900 + ∠BAP = 180âˆ˜
⇒ ∠BAP = 60âˆ˜
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60âˆ˜ = 30âˆ˜ +  ∠PTA
⇒ ∠PTA = 30âˆ˜
Hence, the correct answer is option (b).

#### Question 33:

We know that a chord passing through the centre is the diameter of the circle.
âˆµ∠BPA = 90âˆ˜    (Angle in a semi circle is 90âˆ˜)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30âˆ˜
Now, In â–³ABP
∠PBA + ∠BPA + ∠BAP = 1800            [Angle sum property of a triangle]
⇒ 30âˆ˜ +  900 + ∠BAP = 180âˆ˜
⇒ ∠BAP = 60âˆ˜
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60âˆ˜ = 30âˆ˜ +  ∠PTA
⇒ ∠PTA = 30âˆ˜
Hence, the correct answer is option (b).

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           (âˆµDK = DH and FM = FH)
⇒ ED + DF + EF = 18 cm
⇒ Perimeter of â–³EDF = 18 cm
Hence, the correct answer is option (d)

#### Question 34:

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           (âˆµDK = DH and FM = FH)
⇒ ED + DF + EF = 18 cm
⇒ Perimeter of â–³EDF = 18 cm
Hence, the correct answer is option (d)

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
âˆµ∠OBP = ∠OAP = 900
∠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).

#### Question 35:

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
âˆµ∠OBP = ∠OAP = 900
∠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).

#### Question 36:

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,

Now, PQ = QS + SP = 12 + 10.5 = 22.5 cm
Hence, the correct answer is option (c ).

#### Question 37:

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,

Now, PQ = QS + SP = 12 + 10.5 = 22.5 cm
Hence, the correct answer is option (c ).

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)

#### Question 38:

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)

(a) 9 cm

(a) 9 cm

#### Question 40:

Given: AO and BO are the radius of the circle
Since, AO = BO
∴ â–³AOB is an isosceles triangle.
Now, in â–³AOB
∠AOB + ∠OBA + ∠OAB = 180âˆ˜        (Angle sum property of triangle)
⇒ 100âˆ˜  + ∠OAB + ∠OAB = 180âˆ˜      (∠OBA = ∠OAB)
⇒ 2∠OAB = 80âˆ˜
⇒ ∠OAB = 40âˆ˜
We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠OAT = 90âˆ˜
⇒ ∠OAB  + ∠BAT = 90âˆ˜
⇒ ∠BAT = 90âˆ˜ − 40âˆ˜ = 50âˆ˜
Hence, the correct answer is option (b).

#### Question 41:

Given: AO and BO are the radius of the circle
Since, AO = BO
∴ â–³AOB is an isosceles triangle.
Now, in â–³AOB
∠AOB + ∠OBA + ∠OAB = 180âˆ˜        (Angle sum property of triangle)
⇒ 100âˆ˜  + ∠OAB + ∠OAB = 180âˆ˜      (∠OBA = ∠OAB)
⇒ 2∠OAB = 80âˆ˜
⇒ ∠OAB = 40âˆ˜
We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠OAT = 90âˆ˜
⇒ ∠OAB  + ∠BAT = 90âˆ˜
⇒ ∠BAT = 90âˆ˜ − 40âˆ˜ = 50âˆ˜
Hence, the correct answer is option (b).

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

Hence, the correct answer is option (b).

#### Question 42:

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

Hence, the correct answer is option (b).

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)

#### Question 43:

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)

(a) 2 cm

#### Question 44:

(a) 2 cm

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
Hence, the correct answer is option (a).

#### Question 45:

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
Hence, the correct answer is option (a).

(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.

#### Question 46:

(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.

Construction: Join AF and AE

We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠AED = ∠AFD = 90âˆ˜
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).

#### Question 47:

Construction: Join AF and AE

We know that the radius and tangent are perperpendular at their point of contact
âˆµ∠AED = ∠AFD = 90âˆ˜
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).

(b) 2 cm

#### Question 48:

(b) 2 cm

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)

#### Question 49:

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)

(c)

#### Question 50:

(c)

(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.

#### Question 51:

(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.

(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.

#### Question 52:

(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.

(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.

#### Question 53:

(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.

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

#### Question 54:

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

(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.
$AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}$
We know that the
tangents drawn from an external point to circle are equal .

Hence, ABCD is a rhombus.

#### Question 55:

(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.
$AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}$
We know that the
tangents drawn from an external point to circle are equal .

Hence, ABCD is a rhombus.

(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).

#### Question 1:

(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).

(b) 50°

#### Question 4:

(b) 50°

(b) 10 cm
Since the tangents from an external point are equal, we have:

#### Question 5:

(b) 10 cm
Since the tangents from an external point are equal, we have:

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          [âˆµ CQ = CR = 3]
⇒ BQ = 4 cm
Again, AB = AP + PB
= AP + BQ
= 5 + 4                           [âˆµ AS = AP = 5]
= 9 cm
Hence, the value of x is 9 cm.

#### Question 6:

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          [âˆµ CQ = CR = 3]
⇒ BQ = 4 cm
Again, AB = AP + PB
= AP + BQ
= 5 + 4                           [âˆµ AS = AP = 5]
= 9 cm
Hence, the value of x is 9 cm.

#### Question 7:

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

#### Question 8:

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

#### Question 9:

(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.

#### Question 10:

(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.

Given two tangents AP and AQ are drawn from a point A to a circle with centre O.

#### Question 11:

Given two tangents AP and AQ are drawn from a point A to a circle with centre O.

Now, radius of a circle is perpendicular to the tangent at the point of contact.

#### Question 12:

Now, radius of a circle is perpendicular to the tangent at the point of contact.

$\mathrm{Given},\mathit{AB}=\mathit{AC}$
We know that the tangents from an external point are equal.

#### Question 13:

$\mathrm{Given},\mathit{AB}=\mathit{AC}$
We know that the tangents from an external point are equal.

Given : A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.
:

#### Question 14:

Given : A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.
:

#### Question 15:

Given, a parallelogram ABCD circumscribes a circle with centre O.
$AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}$
We know that the lengths of tangents drawn from an exterior point to a circle
are equal.

Hence, ABCD is a rhombus.

#### Question 16:

Given, a parallelogram ABCD circumscribes a circle with centre O.
$AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}$
We know that the lengths of tangents drawn from an exterior point to a circle
are equal.

Hence, ABCD is a rhombus.

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.

The length of the chord of the larger circle is 8 cm.

#### Question 17:

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.

The length of the chord of the larger circle is 8 cm.

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

#### Question 18:

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

Given, a quadrilateral ABCD circumscribes a circle with centre O.

#### Question 19:

Given, a quadrilateral ABCD circumscribes a circle with centre O.

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.

We know that the tangent to a circle is perpendicular to the radius through the point of contact.

From (i) and (ii), we get:
$\angle APB+\angle AOB={180}^{0}$

#### Question 20:

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.

We know that the tangent to a circle is perpendicular to the radius through the point of contact.

From (i) and (ii), we get:
$\angle APB+\angle AOB={180}^{0}$

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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