The number of tangents that can be drawn from an external point to a circle is ____[CBSE 2011, 12]

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જવાબ : 2

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to _____ [CBSE 2014]

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જવાબ : 5 cm

In a circle of radius 7 cm, tangent *PT* is drawn from a point *P,* such that *PT* = 24 cm. If *O* is the centre of the circle, then *OP* = ?

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જવાબ : 25 cm

Which of the following pair of lines in a circle cannot be parallel? [CBSE 2011]

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જવાબ : two diameters

The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is ____ [CBSE 2014]

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જવાબ : 10√2 cm

In the given figure, PT is a tangent to a circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is _____

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જવાબ : 8 cm

In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then the radius of the circle is ______ [CBSE 2011, 12]

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જવાબ : 10 cm

PQ is a tangent to a circle with centre O at the point P. If △OPQ is an isosceles triangle, then ∠OQP is equal to ______ [CBSE 2014]

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જવાબ : 45^{∘}

In the given figure, AB and AC are tangents to a circle with centre O such that ∠BAC = 40^{∘} .Then ∠BOC is equal to _____ [CBSE 2011, 14]

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જવાબ : 140^{∘}

In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of AB is _____

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જવાબ : 16 cm

If a chord AB subtends an angle of 60^{∘} at the centre of a circle, then the angle between the tangents to the circle drawn from A and B is to _____ [CBSE 2013C]

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જવાબ : 120^{∘}

In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA = 17 cm, then the length of AC (in cm) is ________

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જવાબ : 15 cm

In the given figure, *O* is the centre of a circle. *AOC* is its diameter, such that ∠*ACB* = 50°. If *AT* is the tangent to the circle at the point *A,* then ∠*BAT* = ?

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જવાબ : 50°

In the given figure, O is the centre of the circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70^{∘} , then ∠TPQ is equal to _____ [CBSE 2011]

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જવાબ : 35^{∘}

In the given figure, *AT* is a tangent to the circle with centre *O*, such that *OT* = 4 cm and ∠*OTA* = 30°. Then, AT = ?

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જવાબ : 2√3cm

If *PA* and *PB* are two tangents to a circle with centre *O*, such that ∠*AOB* = 110°, find ∠*APB*.

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જવાબ : 70°

In the given figure, the length of BC is _____ [CBSE 2012, '14]

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જવાબ : 10 cm

In the given figure, ∠AOD = 135^{∘} then ∠BOC is equal to _____

^{}

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જવાબ : 45^{∘}

If *O *is the centre of a circle and* PT* is the tangent to the circle. If *PQ* is a chord, such that ∠*QPT* = 50° then ∠*POQ* = ?

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જવાબ : 100°

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60^{∘} then ∠OAB is ____ [CBSE 2011]

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જવાબ : 30^{∘}

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is ____

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જવાબ : 3√3cm

In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27^{∘} then ∠QAR equals ____ [CBSE 2012]

^{}

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જવાબ : 126^{∘}

In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB

then the length of each tangent ______ [CBSE 2013]

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then the length of each tangent ______ [CBSE 2013]

જવાબ : 4 cm

If *PA* and *PB* are two tangents to a circle with centre *O,* such that ∠*APB* = 80°, then ∠*AO*P = ?

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જવાબ : 50°

In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58^{∘} then the measue of ∠PQB is ____ [CBSE 2014]

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જવાબ : 32^{∘}

In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30^{∘} then ∠CPB + ∠ACP is equal to _____

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જવાબ : 90^{∘}

જવાબ : 44^{∘}

જવાબ : 90^{∘}

જવાબ : 60 cm^{2}

In the given figure, *PQR* is a tangent to the circle at *Q*, whose centre is *O* and *AB* is a chord parallel to *PR,* such that ∠*BQR* = 70°. Then, ∠*AQB* = ?

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જવાબ : 40°

The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is

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જવાબ : √125 cm

In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30^{∘} then ∠PTA = ?

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જવાબ : 30^{∘}

In the given figure, a circle touches the side DF of △EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of △EDF is ____

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જવાબ : Perimeter of △EDF = 18 cm

To draw a pair of tangents to a circle, which are inclined to each other at angle of 45^{0} , we have to draw the tangents at the end points of those two radii, the angle between which is ______ [CBSE 2011]

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જવાબ : ⇒ ∠AOB = 135^{0}

In the given figure,* O* is the centre of a circle; *PQL* and *PRM* are the tangents at the points *Q* and* R* respectively, and *S *is a point on the circle, such that ∠*SQL* = 50° and ∠*SRM* = 60°. Find ∠QSR.

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જવાબ : 70°

In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 12 cm and 9 cm respectively. If the area of △PQR = 189 cm^{2} then the length of side PQ is _____ [CBSE 2011]

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જવાબ : 22.5 cm

In the given figure, QR is a common tangent to the given circle, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is _____ [CBSE 2014]

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જવાબ : 7.6 cm

In the given figure, quadrilateral* ABCD* is circumscribed, touching the circle at *P*, *Q*, *R* and *S*. If *AP* = 5 cm, *BC* = 7 cm and *CS* = 3 cm, *A*B = ?

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જવાબ : 9 cm

In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If ∠AOB = 100^{∘} then ∠BAT is equal to _____ [CBSE 2011]

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જવાબ : 50

In the given figure, quadrilateral *ABCD* is circumscribed, touching the circle at *P*, *Q*, *R* and *S*. If *AP *= 6 cm, *BP *= 5 cm, *CQ* = 3 cm and *DR* = 4 cm, then the perimeter of quadrilateral *ABCD* is _____

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જવાબ : 36 cm

In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is ____ [CBSE 2014]

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જવાબ : 2 cm

In the given figure, Δ*ABC* is right-angled at *B,* such that *BC* = 6 cm and *AB* = 8 cm. A circle with centre *O* has been inscribed in the triangle. *OP* ⊥ *AB*, *OQ* ⊥ *BC* and *OR* ⊥* AC*.

If*OP* = *OQ* = *OR* = *x* cm, then *x* = ?

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If

જવાબ : 2 cm

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is _____ [CBSE 2013]

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જવાબ : 21 cm

Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then the length of AD is _____ [CBSE 2012]

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જવાબ : 3 cm

In the given figure, *PA* and *PB* are tangents to the given circle, such that *PA* = 5 cm and ∠*APB* = 60°. The length of chord *AB* is _____

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જવાબ : 5 cm

In the given figure, DE and DF are two tangents drawn from an external point D to a circle with centre A. If DE = 5 cm. and DE ⊥ DF then the radius of the circle is ____ [CBSE 2013]

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જવાબ : 5 cm

In the given figure, three circles with centres *A*, *B,* *C*, respectively, touch each other externally. If *AB* = 5 cm, *BC* = 7 cm and *CA *= 6 cm, the radius of the circle with centre *A* is _____

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જવાબ : 2 cm

In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is _____ [CBSE 2012]

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જવાબ : 7.5 cm

In the given figure, *O* is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point* P* tangents *PA* and *PB* are drawn to these circles. If *PA* = 12 cm then *PB *is equal to

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જવાબ : 4√10 cm

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^{∘} , prove that BA : AT = 2 : 1 [CBSE 2015]

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જવાબ : AB is the diameter

∴∠APB = 90^{∘} (angle in a semi circle is a right angle)

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= AP/BA
⇒sin30°=AT/BA
⇒1/2=AT/BA

∴ BA : AT = 2 : 1

A point *P* is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn from *P* to the circle. [CBSE 2017]

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જવાબ :

OPB is a right angled triangle, with ∠OBP=90° ……..{ the tangent is perpendicular to the radius

By using pythagoras theorem in △OPB, we get

⇒OB^{2}+PB^{2}=OP^{2}

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

A point *P* is 25 cm away from the centre of a circle and the length of tangent drawn from *P *to the circle is 24 cm. Find the radius of the circle.

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જવાબ :

let P be a point such that OP=25 cm.

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle [CBSE 2011]

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જવાબ :

In right triangle AOP

AO^{2} = OP^{2} + PA^{2}

⇒ (6.5)^{2} = (2.5)^{2} + PA^{2}

⇒ PA^{2} = 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.

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

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જવાબ : we have

AD = AF, BD = BE and CE = CF

Now, AD + BD = 12 cm .....(A)

AF + FC = 10 cm

⇒ AD + FC = 10 cm .....(B)

BE + EC = 8 cm

⇒ BD + FC = 8 cm .....(C)

Adding all these we get

AD + BD + AD + FC + BD + FC = 30

⇒2(AD + BD + FC) = 30

⇒AD + BD + FC = 15 cm .....(D)

Solving (A) and (D), we get

FC = 3 cm

Solving (B) and (D), we get

BD = 5 cm

Solving (C) and (D), we get

and AD = 7 cm

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

From an external point *P,* tangents *PA* and *PB* are drawn to a circle with centre *O*. If *CD* is the tangent to the circle at a point *E* and *PA* = 14 cm, find the perimeter of Δ*PCD*.

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જવાબ : 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−CE)+(CE+DE)+(PB−DE) =(PA+PB) =2PA (∵PA=PB) =(2×14) cm =28 cm ∴Perimeter of △PCD =28 cm.In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

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જવાબ :

We know that

∴ ∠OCA = ∠OCB = 90^{∘} (the radius and tangent are perperpendular at their point of contact)

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

A circle is inscribed in Δ*ABC,* touching *AB, BC* and *AC* at *P*, *Q* and *R*, respectively. If *AB* = 10 cm, *AR* = 7 cm and *CR* = 5 cm, find the length of *BC*.

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જવાબ : 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)=> BC=3+5 = 8 ∴The length of BC is 8 cm.In the given figure, *PA* and *PB* are the tangents to a circle with centre *O.* Show that the points *A,* *O,* *B,* *P* are concyclic.

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જવાબ : Here, OA=OB And OA⊥AP, OA⊥BP

∴∠OAP=90ᵒ, ∠OBP=90ᵒ ∴∠OAP+∠OBP= 90ᵒ +90ᵒ =180ᵒ ∴∠AOB+∠APB=180ᵒ (Since,∠OAP+∠OBP+∠AOB+∠APB=360ᵒ) Sum of opposite angle of a quadrilateral is 180°. Hence, A,O,B and P are concyclic.In the given figure, a circle touches all the four sides of a quadrilateral *ABCD* whose three sides are *AB* = 6 cm, *BC* = 7 cm and *CD* = 4 cm. Find *AD*.

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જવાબ :

Given, AB=6 cm, BC=7 cm and CD=4 cm.

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC [CBSE 2012]

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જવાબ : AR = AQ, BR = BP and CP = CQ ……………..( tangent segments to a circle from the same external point are congruent.)

Now, AB = AC

⇒ AR + RB = AQ + QC

⇒ AR + RB = AR + QC

⇒ RB = QC

⇒ BP = CP

Hence, P bisects BC at P.

In the given figure, common tangents AB and CD to the two circle with centres O_{1} and O_{2} intersect at E. Prove that AB = CD [CBSE 2014]

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જવાબ : EA = EC for the circle having centre O_{1 }(tangent segments to a circle from the same external point are congruent)

ED = EB for the circle having centre O

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

EA + ED = EC + ED

⇒EA + EB = EC + ED

⇒AB = CD

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70^{∘} then ∠TRQ [CBSE 2015]

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જવાબ :

∠OTP = ∠OQP = 90^{∘ }(radius and tangent are perperpendular)

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^{∘}

∴∠TRQ= ½ (∠QOT) =55° (angle subtended by an arc at the centre is double the angle subtended by the arc)

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50^{∘} then what is the measure of ∠OAB is [CBSE 2015]

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જવાબ :

∠OBP = ∠OAP = 90^{∘ (}radius and tangent are perperpendular)

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

⇒ 130

⇒ ∠OAB = 25

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]

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જવાબ : AB + CD = AD + BC (In quadrilateral circumscribes a circle, sum of opposites sides is equal to the sum of other opposite sides.)

⇒6 + 8 = AD + 9

⇒ AD = 5 cm

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^{∘} , prove that BA : AT = 2 : 1 [CBSE 2015]

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જવાબ : AB is the chord passing through the centre

So, AB is the diameter

∴∠APB = 90ᵒ (angle in a semi circle is a right angle)

By using alternate segment theorem

We have ∠APB = ∠PAT = 30ᵒ

Now, in △APB

∠BAP + ∠APB + ∠BAP = 180ᵒ (Angle sum property of triangle)

⇒ ∠BAP = 180ᵒ − 90ᵒ − 30ᵒ = 60ᵒ

Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)

⇒ 60ᵒ = 30ᵒ + ∠PTA

⇒ ∠PTA = 60ᵒ − 30ᵒ = 30ᵒ

We know that sides opposite to equal angles are equal.

∴ AP = AT

In right triangle ABP

sin∠ABP=AP/BA ⇒sin30°=AT/BA ⇒12=AT/BA sin∠ABP=AP/BA ⇒sin30°=AT/BA ⇒12=AT/BA

∴ BA : AT = 2 : 1

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90^{∘} and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]

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જવાબ : DS = DR, AR = AQ (tangent segments to a circle from the same external point are congruent)

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.

Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]

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જવાબ :

Suppose CD and AB are two parallel tangents of a circle with centre O

∠OQC = ∠ORA = 90^{ᵒ }(radius and tangent are perperpendular)

Now, ∠OQC + ∠POQ = 180^{ᵒ} (co-interior angles)

⇒ ∠POQ = 180^{ᵒ} − 90^{ᵒ} = 90^{ᵒ}

Similarly, Now, ∠ORA + ∠POR = 180^{ᵒ} (co-interior angles)

⇒ ∠POR = 180^{ᵒ} − 90^{ᵒ} = 90^{ᵒ}

Now, ∠POR + ∠POQ = 90^{ᵒ} + 90^{ᵒ} = 180^{ᵒ}

Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^{ᵒ}

Hence, QR is a straight line passing through centre O.

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

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જવાબ :

Let TR = *y* and TP = *x*

Now, PR + RQ = 4.8

⇒ PR + PR = 4.8

⇒ PR = 2.4

Now, in right triangle POR

By Using Pyhthagoras theorem, we have

PO

⇒ 3

⇒ OR

⇒ OR = 1.8

Now, in right triangle TPR

By Using Pyhthagoras theorem, we have

TP

⇒

⇒

Again, in right triangle TPQ

By Using Pyhthagoras theorem, we have

TO

⇒ (

⇒

⇒

Solving (A) and (B), we get

∴ TP = 4 cm

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm^{2} then find the lengths of sides of AB and AC. [CBSE 2011, '15]

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જવાબ :

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)

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

In the given figure, *O* is the centre of two concentric circles of radii 4 cm and 6 cm respectively. *PA *and *PB* are tangents to the outer and inner circles, respectively. If *PA* = 10 cm, find the length of *PB* up to one decimal place.

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જવાબ : Given- O is the centre of two concentric circles of radii OA=6 cm and OB=4 cm.

∠OAP=∠OBP=90ᵒ ∴From right angled △OAP, OPજવાબ :

1-C, 2-A, 3-B, 4-D

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-C, 2-A, 3-B, 4-D

Math

Chapter 10 : Circles

- Math Book for CBSE Class 10
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- Geography Book for CBSE Class 10
- Political Science Book for CBSE Class 10
- Chemistry Book for CBSE Class 10
- Economics Book for CBSE Class 10
- Biology Book for CBSE Class 10
- Physics Book for CBSE Class 10

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.