CBSE Solutions for Class 10 Maths

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.

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.

∴OT⊥PT 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.

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.

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

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.

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

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.

Answer :

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.

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.

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.

Answer :

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

Answer :

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

Answer :

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

Answer :

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

Answer :

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