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RD Chapter 10 Circles Ex10.2 Solutions

Question - 51 : - In the figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOP is a diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2. [CBSE 2017]

Answer - 51 : -

Construction : Join RT.
 
Given, ∠POR = 130°
∠POQ = 180°- (∠POR) = 180° – 130° = 50°
Since, PQ is a tangent
∠PQO = 90°
Now, In ∆POQ,
∠POQ + ∠PQO + ∠QPO = 180°
=> 50° + 90° + ∠1 = 180°
=> ∠1 = 180° – 140°
=> ∠1 = 40°
Now, In ∆RST
∠RST = 1/2 ∠ROT
[Angle which is subtended by on arc at the centre of a circle is double the size of the angle subtended at any point on the circumference]
=> ∠2 = 1/2 x 130° = 65°
Now ∠1 + ∠2 = 40° + 65° = 105°

Question - 52 : - In the figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC = PD. [CBSE 2017]

Answer - 52 : -

PA = PB = 12 cm …(i)
QC = AC = 3cm …(ii)
QD = BD = 3 cm …(iii)
[Tangents drawn from an external point are equal]
To find : PC + PD
= (PA – AC) + (PB – BD)
= (12 – 3) + (12 – 3) [From (i), (ii), and (iii)]
= 9 + 9 = 18 cm

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