# Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics

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## Solution.

Considering  the figure, We know that the tangent is perpendicular to the radius of a circle.
So, OPB is a right angled triangle, with ∠OBP=90°
By using pythagoras theorem in △OPB, we get
⇒ OB2+PB2=OP2
⇒ (20)2+PB2=(29)2

⇒ 400+PB2=841

⇒ PB2=841-400=441⇒PB=√441=21

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

## Solution.

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from to the circle is 24 cm.So, TP here is the tangent , OT is the radius So, OTP is a right angled triangle, with ∠OTP=90°
By using pythagoras theorem in △OTP, we get

⇒ OT2+TP2=OP2
⇒ (OT)2+(24)2=(25)2

⇒ OT2+576=625

⇒ OT2=625 – 576=49⇒OT=√49=7 cm.

### YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics

Hence, the radius OT is 7 cm.

## Solution.

In this given figure, 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 (AB)  is 12 cm.

## Solution.

Construction:  Join OA, OC and OB
We know that the radius and tangent are perpendicular at their point of contact
∴ ∠OCA = ∠OCB = 90
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90
OA = OB     (Radii of the larger circle)
OC = OC     (Common)

△OCA ≅ △OCB  (By RHS congruency
∴ CA = CB ( Proved).

## Solution.

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 as △ ABC is isosceles triangle
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

## Solution.

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.

## Solution.

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

Hence, the length of AD is 5 cm.

## Solution.

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360            [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90 + 90 + 70 = 360
⇒ 250 + ∠QOT = 360
⇒ ∠QOT = 110
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.

∴ , ∠TRQ = ½ (∠QOT) = 55 .

## Solution.

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

Then the value of ∠BOC is 140 .

## Solution.

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

## Solution.

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

## Solution.

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

## Solution.

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

## Solution.

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

Hence, the perimeter of △EDF = 18 cm.

## Solution.

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 length of QR is 7.6 cm.

## Solution.

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

### YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics

Then the value of ∠BAT is 50 .

## Solution.

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.

Then the length of AP is 7.5 cm.

## Solution.

We know that tangents drawn from the external point are congruent.
∴ PA = PB

Now, In isoceles triangle APB
∠APB + ∠PBA + ∠PAB = 180                      [Angle sum property of a triangle]
⇒ ∠APB + 65 + 65 = 180                          [∵∠PBA = ∠PAB = 65 ]
⇒ ∠APB = 50

We know that the radius and tangent are perperpendular at their point of contact
∴∠OBP = ∠OAP = 90

Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360           [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90 + 50 + 90 = 360
⇒ 230 + ∠BOC = 360
⇒ ∠AOB = 130

Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180                 [Angle sum property of a triangle]
⇒ 130 +  2∠OAB = 1800                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25

## Solution.

Construction: Join AF and AE.
We know that the radius and tangent are perpendicular at their point of contact
∵∠AED = ∠AFD = 90
Since, in quadrilateral AEDF all the angles are right angles
∴ AEDF is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ AF = DE = 5 cm
Therefore, the radius of the circle is 5 cm.

## Solution.

We know that a chord passing through the centre is the diameter of the circle.
∵∠BAC = 90    (Angle in a semi circle is 90)
By using alternate segment theorem
We have ∠PAB = ∠ACB = 67

Now, In △ABC
∠ABC + ∠ACB + ∠BAC = 180            [Angle sum property of a triangle]
⇒ ∠ABC + 67 +  90 = 180
⇒ ∠ABC= 23

Now, ∠BAQ = 180 − ∠PAB          [Linear pair angles]
= 180 − 67
= 113

Now, In △ABQ
∠ABQ + ∠AQB + ∠BAQ = 180            [Angle sum property of a triangle]
⇒ 23 + ∠AQB + 113 = 180
⇒ ∠AQB = 44

Then the value of ∠AQB is 44