# 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]

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] 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 = ? 25 cm

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

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]

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 _____ 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] 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]

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] 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 _____ 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]

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 ________ 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 = ? 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] 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 = ? 2√3cm

If PA and PB are two tangents to a circle with centre O, such that AOB = 110°, find APB. 70°

In the given figure, the length of BC is _____     [CBSE 2012, '14] 10 cm

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

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

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

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] 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] 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)
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

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. 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.  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. 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. 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.  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)

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] 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°

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

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

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

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

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