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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 $\angle$APB = 50°, then $\angle$OAB is equal to (A) 25°                        (B) 30°                        (C) 40°                        (D) 50°

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

$\angle$PAB = $\angle$PBA = $65^{\circ}$
$\angle$PAO = $90^{\circ}$ ($\mathbb{Q}$  tangent is perpendicular to radius)
$\angle$PAO = $\angle$PAB +$\angle$OAB
$90^{\circ}= 65^{\circ}+\angle OAB$
$\angle OAB= 90^{\circ}-65^{\circ}$
$\angle OAB= 25^{\circ}$

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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 $PR\perp PO$
$\angle$OPR = $90^{\circ}$
$\angle$OPR = $\angle$OPQ + $\angle$QPR
$90^{\circ}$ = $\angle$ OPQ + $50^{\circ}$
$\angle$OPQ = $40^{\circ}$
$\angle$OPQ = $\angle$OQD         $\left ( \because \, OP= OQ \, radius\, of\, circle\right )$
Hence , $\angle$OPQ = $40^{\circ}$
We know that the sum of interior angles of a triangle is $180^{\circ}$
In $\bigtriangleup OPQ$
$\angle$O+$\angle$Q +$\angle$P = $180^{\circ}$
$\angle$O +40+40 = 180
$\angle$o = $100^{\circ}$
Hence $\angle$OPQ = $100^{\circ}$

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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 $\bigtriangleup$OAT
$\cos \theta = \frac{B}{H}$
$\cos 30^{\circ} = \frac{AT}{OT}$
$\frac{\sqrt{3}}{2}= \frac{AT}{4}\; \; \left ( \because \cos 30^{\circ}= \frac{\sqrt{3}}{2} \right )$
$AT= 4\times \frac{\sqrt{3}}{2}$
$AT= 2\sqrt{3}cm$

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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 $\bigtriangleup$ODZ, 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 $\bigtriangleup$POQ, $\angle Q= 90^{\circ}$          ($\because$  PQ is tangent)
Using Pythagoras theorem in $\bigtriangleup$POQ

$\left ( PO \right )^{2}= \left ( PQ \right )^{2}+\left ( QO \right )^{2}$      ( 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 $\bigtriangleup$POQ = $\frac{1}{2}$  × perpendicular × base
$= \frac{1}{2}\times PQ\times OQ$
$= \frac{1}{2}\times 12\times 5= 30cm^{2}$
Area of quadrilateral = 2 × area of $\bigtriangleup$POQ
= 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 $\angle$B = 90°
We know that sum of interior angle of a triangle is 180°.
In $\bigtriangleup$ABC
$\angle A+\angle B+\angle C= 180^{\circ}$
$\angle A+90^{\circ}+50^{\circ}= 180^{\circ}$
$\angle A=180^{\circ}-140^{\circ}$
$\angle A= 40^{\circ}$
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
$\Delta$ABD is a right angle triangle.
Hence use Pythagoras theorem in $\Delta$ABD
$\left ( AD \right )^{2}= \left ( AB \right )^{2}+\left ( BD \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( BD \right )^{2}$
$25-16= \left ( BD \right )^{2}$
$BD= \sqrt{9}= 3$
$BD= 3cm$
$CD= CB+BD$
$CD= CB+BD$
$CD= BD+BD= 2BD\; \; \left ( \because CB= BD \right )$
$CD= 2\times 3= 6cm$
$CD= 6cm$
Hence length of chord is 6 cm.

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