# CBSE Solutions for Class 10 Maths

#### Select CBSE Solutions for class 10 Subject & Chapters Wise :

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

Answer :

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] Answer :

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 = ? Answer :

25 cm

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

Answer :

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]

Answer :

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 _____ Answer :

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] Answer :

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]

Answer :

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] Answer :

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 _____ Answer :

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]

Answer :

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 ________ Answer :

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 = ? Answer :

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] Answer :

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 = ? Answer :

2√3cm

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

70°

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

10 cm

In the given figure, AOD = 135 then BOC is equal to _____ Answer :

45

If 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 = ?

Answer :

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] Answer :

30

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]

Answer : 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

(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.

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. Find the radius of the circle.

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

Let, TP be the tangent, so that TP=24 cm.

Join OT,where OT is radius.

Now, tangent drawn from an external point is perpendicular to the radius at  the point of contact.

OTPT In the right OTP, we have:

OP2=OT2+TP2   [By Pythagoras' theorem:]

OT=√ [OP2−TP2]= √(252−242) =√(625−576)  =√49 =7 cm

The length of the radius is 7 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]

Answer : 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.

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] Answer :

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 ΔPCD. Answer :

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. Answer : 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 ΔABC, touching AB, BC and AC at PQ and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC. Answer :

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. Answer :

Here, OA=OB And  OAAP,  OABP

∴∠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. Answer : Given, AB=6 cm, BC=7 cm and CD=4 cm.

AP=AS, BP=BQ, CR=CQ  and DR=DS ……..( Tangents drawn from an external point are equal.)

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 cm.

The length of AD is 3 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] Answer :

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. 1 TP A 7cm 2 TO B 25cm 3 OP C 24cm 4 ∠T D 90°

Answer :

1-C, 2-A, 3-B, 4-D 1 ∠ T A 6cm 2 TP B 90° 3 TO C 10cm 4 OP D 8cm

Answer :

1-B, 2-D, 3-A, 4-C 1 ∠A A OC 2 ∠O B 40° 3 AB C 50° 4 OB D AC

Answer :

1-B, 2-C, 3-D, 4-A 1 ∠O A AC 2 ∠C B OA 3 OB C 30° 4 BC D 60°

Answer :

1-D, 2-C, 3-B, 4-A 1 OB A 90° 2 AB B <90° 3 ∠B C AC 4 ∠A D OC

Answer :

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

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### Chapter 10 : Circles

Class 10 Maths Chapter 10 deals with the existence of the tangents to a circle and some of the properties of circle. Students are introduced to some complex terms such as tangents, tangents to a circle, number of tangents from a point on the circle. This chapter seems very interesting due to the diagrams and involvement of geometrical calculations.

### Browse & Download CBSE Books For Class 10 - All Subjects

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.