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In Figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and\angle BQR = 70°,then \angle AQB is equal to

(A) 20°                        (B) 40°                        (C) 35°                        (D) 45°

Answer(B) 40^{\circ}                       
Solution
Given :\angle BQR= 70^{\circ}, AB\parallel PR
Since DQ perpendicular to PR
\angle DQR= 90^{\circ}
\angle DQR= \angle DQB+\angle BQR
90^{\circ}= \angle DQB+70^{\circ}
\angle DQB= 20^{\circ}
Since DQ bisect \angle AQB
Hence , \angle DQB=\angle DQA= 20^{\circ}
\angle AQB= \angle DQB+\angle DQA
= 20^{\circ}+20^{\circ}= 40^{\circ}
\angle AQB= 40^{\circ}

        

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If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm,then length of each tangent is equal to

(A) \frac{3}{2}\sqrt{3}cm                 (B) 6 cm                      (C) 3 cm                      (D) 3\sqrt{3}cm

Answer (D) 3\sqrt{3}cm
Solution
According to question
          

Given OQ = OR = 3 cm (Radius)
\angle P= 60^{\circ}
Draw line OP which bisect \angle P . That is \angle OPQ= 30^{\circ}

\angle OPR= 30^{\circ}
In \bigtriangleup OPQ
\tan \theta = \frac{P}{B}
\tan 30^{\circ} = \frac{3}{PQ}
\frac{1}{\sqrt{3}}= \frac{3}{PQ}
PQ= 3\sqrt{3}
Here PQ = PR
Hence , PQ = PR= 3\sqrt{3}
 

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In Figure, if PA and PB are tangents to the circle with centre O such that \angleAPB = 50°, then \angleOAB is equal to


(A) 25°                        (B) 30°                        (C) 40°                        (D) 50°

Answer (A) 25°
Solution 
Given : \angleAPB = 50°
  We know that length of tangents drawn from an external point is equal
Hence, PA = PB
Since, PA = PB
Let  \anglePAB = \angle PBA = x0     
In \bigtriangleupPAB                 
\anglep+\angleA+\angleB = 180^{\circ}     (\because  Sum of interior angles of a tangent is 180^{\circ})
50^{\circ}+x^{\circ}+x^{\circ}= 180^{\circ}
2x^{\circ}= 130^{\circ}
x^{\circ}= 65^{\circ}

\anglePAB = \anglePBA = 65^{\circ}
\anglePAO = 90^{\circ} (\mathbb{Q}  tangent is perpendicular to radius)
\anglePAO = \anglePAB +\angleOAB
90^{\circ}= 65^{\circ}+\angle OAB
\angle OAB= 90^{\circ}-65^{\circ}
\angle OAB= 25^{\circ}

 

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In Figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to


(A) 100°          (B) 80°              (C) 90°             (D) 75°

Answer(A) 100^{\circ}
Solution
Given :\angle QPR= 50^{\circ} 
We know that tangent is perpendicular to radius.
Hence PR\perp PO
\angleOPR = 90^{\circ}
\angleOPR = \angleOPQ + \angleQPR
90^{\circ} = \angle OPQ + 50^{\circ}
\angleOPQ = 40^{\circ}
\angleOPQ = \angleOQD         \left ( \because \, OP= OQ \, radius\, of\, circle\right )
Hence , \angleOPQ = 40^{\circ}
We know that the sum of interior angles of a triangle is 180^{\circ}
In \bigtriangleup OPQ
\angleO+\angleQ +\angleP = 180^{\circ}
\angleO +40+40 = 180
\angleo = 100^{\circ}
Hence \angleOPQ = 100^{\circ}

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In Figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4 cm                      (B) 2 cm                      (C) 2\sqrt{3}cm                  (D) 4\sqrt{3}cm

Answer(C) 2\sqrt{3}cm
Solution
      

Given ∠OTA = 30°  , OT = 4cm
Join OA, since AT is tangent.                        
Hence it is perpendicular to OA
In \bigtriangleupOAT
\cos \theta = \frac{B}{H}
\cos 30^{\circ} = \frac{AT}{OT}
\frac{\sqrt{3}}{2}= \frac{AT}{4}\; \; \left ( \because \cos 30^{\circ}= \frac{\sqrt{3}}{2} \right )
AT= 4\times \frac{\sqrt{3}}{2}
AT= 2\sqrt{3}cm

 

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At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

(A) 4 cm                      (B) 5 cm                                  (C) 6 cm                      (D) 8 cm

Answer (D) 8 cm
Solution 
According to questions
                

Given AO = OB = 5 cm
Distance between XY and CD = 8 cm
Since D is the center of the circle and CD is chord. If we join OD it become the radius of circle that is OD = 5cm
AZ = 8 cm (given)
AZ = AO + OZ
8 = 5 + OZ
OZ = 3cm
In \bigtriangleupODZ, use Pythagoras theorem
\left ( OD \right )^{2}= \left ( OZ \right )^{2}+\left ( ZD \right )^{2}
\left ( 5 \right )^{2}= \left ( 3 \right )^{2}+\left ( ZD \right )^{2}
\left ( ZD \right )^{2}= 25-9
ZD= \sqrt{16}= 4cm
CD= CZ+ZD
CD= ZD+ZD  \left ( \because CZ= ZD \right )
CD= 2ZD= 2\left ( 4 \right )= 8cm
Length of chord CD = 8 cm

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From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(A) 60 cm2                   (B) 65 cm2                   (C) 30 cm2                   (D) 32.5 cm2

Answer (A) 60 cm2
Solution
According to question

Given OP = 13 cm
OQ = OR = radius = 5 cm
  In \bigtriangleupPOQ, \angle Q= 90^{\circ}          (\because  PQ is tangent)                                    
Using Pythagoras theorem in \bigtriangleupPOQ       

\left ( PO \right )^{2}= \left ( PQ \right )^{2}+\left ( QO \right )^{2}      ( because H2 = B2 + P2)
\left ( 13 \right )^{2}= \left ( PQ \right )^{2}+\left ( 5 \right )^{2}
\left ( PQ\right )^{2}= 169-25
PQ= \sqrt{144}= 12
PQ= 12
Area of \bigtriangleupPOQ = \frac{1}{2}  × perpendicular × base
= \frac{1}{2}\times PQ\times OQ
= \frac{1}{2}\times 12\times 5= 30cm^{2}
Area of quadrilateral = 2 × area of \bigtriangleupPOQ
= 2 × 30 = 60 cm2

      

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In Figure, AB is a chord of the circle and AOC is its diameter such that < ACB = 50°. If AT is the tangent to the circle at the point A, then \angle BAT is equal to

(A) 65°                        (B) 60°                        (C) 50°                        (D) 40°

Answer  (C)
Solution
 
Given \angle ACB = 50°.
We know that the angle subtended by a diameter is right.
Hence \angleB = 90°
We know that sum of interior angle of a triangle is 180°.
In \bigtriangleupABC
\angle A+\angle B+\angle C= 180^{\circ} 
\angle A+90^{\circ}+50^{\circ}= 180^{\circ}
\angle A=180^{\circ}-140^{\circ}
\angle A= 40^{\circ}
That is \angle CAB= 40^{\circ}
\angle CAT= \angle CAB+\angle BAT
90^{\circ}= 40^{\circ}+\angle BAT   \left (\because CA \perp AT \right )
\angle BAT= 90^{\circ}-40^{\circ}
\angle BAT= 50^{\circ}

                        

                                                

                        

 

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In Figure, if ∠AOB = 125°, then ∠COD is equal to

(A) 62.5°                     (B) 45°                        (C) 35°            (D) 55°

Answer (D)
Solution

Given : ∠AOB = 125°
Let       ∠COD = x°
As we know that the sum of opposite sides angles of a quadrilateral circumscribing a circle is equal to 180°.
Hence
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
x° = 180° – 125°        (\mathbb{Q}  ∠COD = x°)
x° = 55°
Hence, ∠COD = 55°

 

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If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is
(A) 3 cm                      (B) 6 cm                      (C) 9 cm                      (D) 1 cm

Answer (B) 6 cm
Solution
According to question

Here A is the center and AB = 4 cm radius of small circle and AD = 5 cm radius of large circle. 
We have to find the length of CD
\DeltaABD is a right angle triangle.
 Hence use Pythagoras theorem in \DeltaABD
\left ( AD \right )^{2}= \left ( AB \right )^{2}+\left ( BD \right )^{2}
\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( BD \right )^{2}
25-16= \left ( BD \right )^{2}
BD= \sqrt{9}= 3
BD= 3cm
CD= CB+BD
CD= CB+BD
CD= BD+BD= 2BD\; \; \left ( \because CB= BD \right )
CD= 2\times 3= 6cm
CD= 6cm
Hence length of chord is 6 cm.



              

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