GSEB std 10 science solution for Gujarati check Subject Chapters Wise::
જવાબ : 2
જવાબ : 5 cm
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જવાબ : 25 cm
જવાબ : two diameters
જવાબ : 10√2 cm
જવાબ : 8 cm
જવાબ : 10 cm
જવાબ : 45∘
જવાબ : 140∘
જવાબ : 16 cm
જવાબ : 120∘
જવાબ : 15 cm
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જવાબ : 50°
જવાબ : 35∘
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જવાબ : 2√3cm
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જવાબ : 70°
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જવાબ : 10 cm
જવાબ : 45∘
જવાબ : 100°
જવાબ : 30∘
જવાબ : 3√3cm
જવાબ : 126∘
then the length of each tangent ______ [CBSE 2013]
જવાબ : 4 cm
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જવાબ : 50°
જવાબ : 32∘
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જવાબ : 90∘
જવાબ : 44∘
જવાબ : 90∘
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જવાબ : 60 cm2
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જવાબ : 40°
જવાબ : √125 cm
જવાબ : 30∘
જવાબ : Perimeter of △EDF = 18 cm
જવાબ : ⇒ ∠AOB = 1350
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જવાબ : 70°
જવાબ : 22.5 cm
જવાબ : 7.6 cm
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જવાબ : 9 cm
જવાબ : 50
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જવાબ : 36 cm
જવાબ : 2 cm
If OP = OQ = OR = x cm, then x = ?
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જવાબ : 2 cm
જવાબ : 21 cm
જવાબ : 3 cm
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જવાબ : 5 cm
જવાબ : 5 cm
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જવાબ : 2 cm
જવાબ : 7.5 cm
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જવાબ : 4√10 cm
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
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
⇒OB2+PB2=OP2
So, length of the tangent from point P is 21 cm.
જવાબ : .png)
let P be a point such that OP=25 cm.
જવાબ : 
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.
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
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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.
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
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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..png)
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..png)
જવાબ : .png)
Given, AB=6 cm, BC=7 cm and CD=4 cm.
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.
જવાબ : EA = EC for the circle having centre O1 (tangent segments to a circle from the same external point are congruent)
Similarly,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
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)
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જવાબ : .png)
∠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∘ [Angle sum property of a triangle]
⇒ 130∘ + 2∠OAB = 1800 [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25∘
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
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
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.
જવાબ : 
Suppose CD and AB are two parallel tangents of a circle with centre O
∠OQC = ∠ORA = 90ᵒ (radius and tangent are perperpendular)
Now, ∠OQC + ∠POQ = 180ᵒ (co-interior angles)
⇒ ∠POQ = 180ᵒ − 90ᵒ = 90ᵒ
Similarly, Now, ∠ORA + ∠POR = 180ᵒ (co-interior angles)
⇒ ∠POR = 180ᵒ − 90ᵒ = 90ᵒ
Now, ∠POR + ∠POQ = 90ᵒ + 90ᵒ = 180ᵒ
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180ᵒ
Hence, QR is a straight line passing through centre O.
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જવાબ : .png)
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
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 .....(A)
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 .....(B)
Solving (A) and (B), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm
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
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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, OP2=OA2+PA2 =>OP=√(OA2+PA2 ) =>OP=√(62+102) =>OP=√136 cm. ∴From right-angled △OAP, OP2 =OB2 +PB2 =>PB=√(OP2−OB2) =>PB=√(136−16) =>PB=√120 cm =>PB=10.9 cm. ∴The length of PB is 10.9 cm.There are No Content Availble For this Chapter
જવાબ :
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
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Circles
Math
Chapter 10 : Circles
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