RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 8 - Circles

Question 1:

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
$$\begin{gathered} \Rightarrow {\mathrm{OB}}^2+{\mathrm{PB}}^2={\mathrm{OP}}^2 \\ \Rightarrow (20{)}^2+{\mathrm{PB}}^2=(29{)}^2 \\ \Rightarrow 400+{\mathrm{PB}}^2=841 \\ \Rightarrow {\mathrm{PB}}^2=841-400=441 \\ \Rightarrow \mathrm{PB}=\sqrt{441}=21 \end{gathered}$$
So, length of the tangent from point P is 21 cm.

Question 3:

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

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.

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)
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

Question 6:

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

Question 7:

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 â–³OCA and â–³OCB
∠OCA = ∠OCB = 90^∘
OA = OB     (Radii of the larger circle)
OC = OC     (Common)
By RHS congruency
â–³OCA ≅ â–³OCB
∴ CA = CB

Question 8:

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

Question 9:

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

Question 10:

$$\begin{gathered} \text{Let the circle touch the sides of the quadrilateral }\mathit{\text{AB,BC,CD}}\text{ and }\mathit{\text{DA}}\text{ at }\mathit{\text{P,Q,R and}}\mathit{ }\mathit{\text{ S}}\text{ respectively}\text{.} \\ \text{Given, }AB=6\text{ cm, }BC=7\text{ cm and }CD=4\text{ cm}\text{.} \\ \text{Tangents drawn from an external point are equal.} \\ \therefore AP=AS,BP=BQ,CR=CQ\text{ and }DR=DS \\ \begin{array}{l}\text{Now, }AB+CD=(AP+BP)+(CR+DR)\\ =>AB+CD=(AS+BQ)+(CQ+DS)\\ =>AB+CD=(AS+DS)+(BQ+CQ)\\ =>AB+CD=AD+BC\\ =>AD=(AB+CD)-BC\\ =>AD=(6+4)-7\\ =>AD=3\text{ cm}\text{.}\end{array} \\ \therefore \text{The length of }AD\text{ is 3 cm}\text{.} \end{gathered}$$

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:

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

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,
$$\begin{gathered} \mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right) \\ \Rightarrow 54=\frac{1}{2}\times \mathrm{BC}\times \mathrm{OD}+\frac{1}{2}\times \mathrm{AB}\times \mathrm{OE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OF} \\ \Rightarrow 108=15\times 3+\left(6+x\right)\times 3+\left(9+x\right)\times 3 \\ \Rightarrow 36=15+6+x+9+x \end{gathered}$$

$$\begin{gathered} \Rightarrow 36=30+2x \\ \Rightarrow 2x=6 \\ \Rightarrow x=3 \mathrm{cm} \end{gathered}$$
 ∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 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
 

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 = 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.

Question 16:

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.

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 = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘ 
Now, in â–³APB
∠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
$$\begin{gathered} \sin \angle \mathrm{ABP}=\frac{\mathrm{AP}}{\mathrm{BA}} \\ \Rightarrow \sin 30°=\frac{\mathrm{AT}}{\mathrm{BA}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{AT}}{\mathrm{BA}} \end{gathered}$$
∴ BA : AT = 2 : 1

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
⇒ AD = 5 cm

Question 3:

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^∘

Question 4:

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.
$\therefore \angle \mathrm{TRQ}=\frac{1}{2}\left(\angle \mathrm{QOT}\right)=55°$

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

Question 6:

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 â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

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,
$$\begin{gathered} \mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right) \\ \Rightarrow 21=\frac{1}{2}\times \mathrm{BC}\times \mathrm{OD}+\frac{1}{2}\times \mathrm{AB}\times \mathrm{OE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OF} \\ \Rightarrow 42=7\times 2+\left(4+x\right)\times 2+\left(3+x\right)\times 2 \\ \Rightarrow 21=7+4+x+3+x \end{gathered}$$

$$\begin{gathered} \Rightarrow 21=14+2x \\ \Rightarrow 2x=7 \\ \Rightarrow x=3.5 \mathrm{cm} \end{gathered}$$
 ∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 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.  
$$\begin{gathered} \mathrm{In}\mathrm{ }∆OAC, O{A}^2=O{C}^2+A{C}^2 \\ \therefore A{C}^2=O{A}^2-O{C}^2 \\ \Rightarrow A{C}^2=5^2-3^2 \\ \Rightarrow A{C}^2=25-9 \\ \Rightarrow A{C}^2=16 \\ \Rightarrow AC=4\mathrm{ }\mathrm{cm} \\ \therefore AB=2AC \left(\mathrm{since}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{drawn}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{centre}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{circle} \\ \mathrm{bisects}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{chord}\right) \\ \therefore AB=2\times 4=8 \mathrm{cm} \end{gathered}$$
The length of the chord of the larger circle is 8 cm.

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 ∥ 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.

Question 10:

Construction: Join PO and OQ
In â–³POR and â–³QOR
OP = OQ   (Radii)
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 = 120^∘
⇒ ∠PRO = 60^∘
Now, In â–³POR
$$\begin{gathered} \cos {60}^{°}=\frac{\mathrm{PR}}{\mathrm{OR}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{PR}}{\mathrm{OR}} \\ \Rightarrow \mathrm{OR}=2\mathrm{PR} \\ \Rightarrow \mathrm{OR}=\mathrm{PR}+\mathrm{PR} \\ \Rightarrow \mathrm{OR}=\mathrm{PR}+\mathrm{RQ} \end{gathered}$$

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)
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

Question 12:

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.

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.

Question 14:

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 â–³POQ
∠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.
$\therefore \angle \mathrm{PRQ}=\frac{1}{2}\left(\mathrm{reflex}\angle \mathrm{POQ}\right)=120°$

Question 15:

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^∘

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 ∆OPB and ∆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, ∆OPB ≌ ∆OPA           (SAS Congruence Axiom)  
$\therefore \angle \mathrm{OPB}=\angle \mathrm{OPA}=30°$    (CPCT)
Now,
In ∆OPB,
$$\begin{gathered} \sin 30°=\frac{\mathrm{OB}}{\mathrm{OP}} \\ \Rightarrow \frac{1}{2}=\frac{a}{\mathrm{OP}} \\ \Rightarrow \mathrm{OP}=2a \end{gathered}$$
Thus, the length of OP is 2a.
Disclaimer: The answer given in the book is incorrect.

Question 2:

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


Hence, the correct answer is option (b)

Question 3:

We know that the radius and tangent are perperpendular at their point of contact
$\mathrm{OQ}=\frac{1}{2}\mathrm{QS}=3 \mathrm{cm}$          [∵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).

Question 4:

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

Question 5:

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

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).

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).

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).

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 = 180^∘            [Angle sum property of a triangle]
⇒  2∠OQP + 90^∘ = 180^∘
⇒ ∠OQP = 45^∘
Hence, the correct answer is option (b).

Question 10:

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).

Question 11:

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).

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).

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).

Question 14:

(b) 50°
$$\begin{gathered} \angle ABC={90}^0\left(\text{A}\mathrm{ngle}\mathrm{ }\mathrm{in}\mathrm{ }\mathrm{a}\mathrm{ }\mathrm{semicircle}\right) \\ \mathrm{I}\mathrm{n}\mathrm{ }△ABC, \mathrm{w}\mathrm{e}\mathrm{ }\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{e}: \angle ACB+\angle CAB+\angle ABC={180}^0 \\ \Rightarrow {50}^0+\angle CAB+{90}^0={180}^0 \\ \Rightarrow \angle CAB=\left({180}^0-{140}^0\right) \\ \Rightarrow \angle CAB={40}^0 \\ \mathrm{Now,} \angle CAT={90}^0(\text{T}\mathrm{angents}\mathrm{ }\mathrm{drawn}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{an}\mathrm{ }\mathrm{external}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{are}\mathrm{ }\mathrm{perpendicular} \\ \mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{radius}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{contact}) \\ \therefore \angle CAB+\angle BAT={90}^0 \\ \Rightarrow {40}^0+\angle BAT={90}^0 \\ \Rightarrow \angle BAT=\left({90}^0-{40}^0\right) \\ \Rightarrow \angle BAT={50}^0 \end{gathered}$$

Question 15:

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).

Question 16:

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

$$\begin{gathered} OA\perp AT \\ \mathrm{So},\frac{AT}{OT}=\cos {30}^0 \\ \Rightarrow \frac{AT}{4}=\frac{\sqrt{3}}{2} \\ \Rightarrow AT=\left(\frac{\sqrt{3}}{2}\times 4\right) \\ \Rightarrow AT=2\sqrt{3 } \end{gathered}$$

Question 17:

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

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).

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 =  180^∘
⇒∠BOC =  180^∘ − 135^∘ = 45^∘ 
Hence, the correct answer is option (b).

Question 20:

$$\begin{gathered} \left(\mathrm{a}\right) 100° \\ \mathrm{Given}, \angle QPT={50}^0 \\ \mathrm{and} \angle OPT={90}^0\left(\text{T}\mathrm{angents}\mathrm{ }\mathrm{drawn}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{an}\mathrm{ }\mathrm{external}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{are}\mathrm{ }\mathrm{perpendicular} \\ \mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{radius}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{contact}\right) \\ \therefore \angle OPQ=\left(\angle OPT-\angle QPT\right)=\left({90}^0-{50}^0\right)={40}^0 \\ OP=OQ \left(\text{R}\mathrm{adius}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{same}\mathrm{ }\mathrm{circle}\right) \\ \Rightarrow \angle OQP=\angle OPQ={40}^0 \\ \mathrm{In} ∆POQ, \angle POQ+\angle OQP+\angle OPQ={180}^0 \\ \therefore \angle POQ={180}^0-\left({40}^0+{40}^0\right)={100}^0 \end{gathered}$$

Question 21:

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).

Question 22:

(c) $3\sqrt{3} \mathrm{cm}$

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

Question 23:

We know that the radius and tangent are perperpendular at their point of contact
Now, In â–³PQA
∠PQA + ∠QAP + ∠APQ = 180^∘            [Angle sum property of a triangle]
⇒ 90^∘ + ∠QAP + 27^∘ = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠QAP = 63^∘
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 = 63^∘
∴∠QAR = ∠QAP + ∠RAP = 63^∘ + 63^∘ = 126^∘
Hence, the correct answer is option (c).

Question 24:

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).

Question 25:

(b) 50°
$$\begin{gathered} \mathrm{Given}, PA \mathrm{and}\mathrm{ }PB \mathrm{are}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{tangents} \mathrm{to}\mathrm{ }\mathrm{a}\mathrm{ }\mathrm{circle}\mathrm{ }\mathrm{with}\mathrm{ }\mathrm{centre} O\mathrm{ }\mathrm{and}\mathrm{ }\angle APB={80}^0. \\ \therefore \angle APO=\frac{1}{2}\angle APB={40}^0 \\ \left[S\mathrm{ince}\mathrm{ }\mathrm{they} \mathrm{are} \mathrm{equally} \mathrm{inclined} \mathrm{to} \mathrm{the} \mathrm{line} \mathrm{segment} \mathrm{joining} \mathrm{the} \mathrm{centre} \mathrm{to} \mathrm{that} \mathrm{point}\right] \\ \mathrm{and}\mathrm{ }\angle OAP={90}^0 \\ \left[S\mathrm{ince}\mathrm{ }\mathrm{tangents}\mathrm{ }\mathrm{drawn}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{an}\mathrm{ }\mathrm{external}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{are}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{radius}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{contact}\right] \\ \mathrm{Now},\mathrm{ }\mathrm{in}\mathrm{ }\mathrm{triangle} AOP: \\ \angle AOP+\angle OAP+\angle APO={180}^0 \\ \Rightarrow \angle AOP+{90}^0+{40}^0={180}^0 \\ \Rightarrow \angle AOP={180}^0-{130}^0 \\ \Rightarrow \angle AOP={50}^0 \end{gathered}$$

Question 26:

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 â–³PQR
∠PQR + ∠PRQ + ∠QPR = 180⁰            [Angle sum property of a triangle]
⇒ ∠PQR + 58^∘ +  90⁰ = 180^∘                      
⇒ ∠PQR= 32^∘
Hence, the correct answer is option (a).

Question 27:

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 â–³CDP
∠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).

Question 28:

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 â–³ABC
∠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 â–³ABQ
∠ABQ + ∠AQB + ∠BAQ = 180^∘            [Angle sum property of a triangle]
⇒ 23^∘ + ∠AQB + 113^∘ = 180^∘                      
⇒ ∠AQB = 44^∘
Hence, the correct answer is option (d).

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 = 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).

Question 30:

(a) 60 cm²

$$\begin{gathered} \mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n},\mathrm{ } \\ OQ=OR=5 \mathrm{cm}, OP=13\mathrm{ }\mathrm{cm}. \\ \angle OQP=\angle ORP={90}^0\left(\text{T}\mathrm{angents}\mathrm{ }\mathrm{drawn}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{an}\mathrm{ }\mathrm{external}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{are}\mathrm{ }\mathrm{perpendicular}\mathrm{ } \\ \mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{radius}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{contact}\right) \\ \mathrm{From}\mathrm{ }\mathrm{right}-\mathrm{angled}\mathrm{ }∆POQ: \\ P{Q}^2=\left(O{P}^2-O{Q}^2\right) \\ \Rightarrow P{Q}^2= {13}^2-5^2 \\ \Rightarrow P{Q}^2=169-25 \\ \Rightarrow P{Q}^2=144 \\ \Rightarrow PQ=\sqrt{144} \\ \Rightarrow PQ=12\mathrm{ }\mathrm{cm} \\ \therefore ar\left(△OQP\right)=\frac{1}{2}\times PQ\times OQ \\ \Rightarrow ar\left(△OQP\right)=\left(\frac{1}{2}\times 12\times 5\right){\mathrm{cm}}^2 \\ \Rightarrow ar\left(△OQP\right)=30 {\mathrm{cm}}^2 \\ \mathrm{S}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{l}\mathrm{y},\mathrm{ }ar\left(△ORP\right)=30 {\mathrm{cm}}^2 \\ \therefore \mathrm{a}\mathrm{r}\left(quad.PQOR\right)=\left(30+30\right) {\mathrm{cm}}^2=60\mathrm{ }{\mathrm{cm}}^2 \end{gathered}$$