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**Q.1. Prove that the tangents drawn at the ends of a chord of circle make equal angles with the chord.Ans.** Given: A circle with centre O, PA and PB are tangents drawn at the ends A and B on chord AB.

To prove: ∠PAB = ∠PBA

Construction: Join OA and OB.

Proof: In ΔOAB we have

OA = OB ...(i) (Radii of the same circle)

∠2 = ∠1 ...(ii) (Angles opposite to equal sides of a Δ)

Also (∠2 + ∠3 = ∠1 + ∠4) ...(iii) (Both 90° as Radius ⊥ Tangent)

Subtracting (ii) from (iii), we have

∴ ∠3 = ∠4 ⇒ ∠PAB = ∠PBA

Ans.

∠CBA = ?

∠ACB = 90° [Angle in a semicircle is right angle]

∠PCB = 110°- 90° = 20°

∠PCB = ∠CAB = 20°

[Alternate segment theorem]

In ΔABC,

∠ABC + ∠CAB + ∠BCA = 180°

[∵ Sum of angles of a Δ is 180°.]

⇒ ∠ABC = 180° - 110° = 70°.

Ans.

Construction: Join AD and BC

Proof: The tangents drawn from an internal point to a circle are equal in length.

If A is external point for circle having center O'

AB = AD ...(i)

If C is external point then

BC = CD ...(ii)

Now, B is external point for circle having center O

AB = BC ...(iii)

So, from (i), (ii) and (iii), we get

AB = BC = CD

So, AB = CD Hence proved.

Ans.

PA = 10 cm.

PA = PB [If P is external point] ... (i)

[∵ From an external point tangents drawn to a circle are equal in length]

If C is external point, then CA = CE

If D is external point, then

DB = DE ...(ii)

Perimeter of triangle ΔPCD

= PC + CD + PD

= PC + CE + ED + PD

= PC + CA + DB + PD

= PA + PB

= PA + PA = 2 PA

= 2 x 10 = 20 cm. [From (i)]

Ans.

AC is diameter, ∠B = 90°

[In a semicircle there is always a right angle]

So, ∠ACB + ∠CAB = 90° ...(i)

[∵ sum of angles of a Δ is 180°]

OA ⊥ AT [Radius and tangent are ⊥ to each other at the point of contact]

∠OAT = 90°

∠OAB + ∠BAT = 90° ...(ii)

From (i) and (ii),

∠ACB = ∠BAT

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