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#9.1

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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#9.1

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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#9.1

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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#9.1

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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#9.1

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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#9.1

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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#9.1

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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#9.1

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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#9.1

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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#9.1

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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Get Answers to all your Questions

All Questions

In Figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and BQR = 70°,then AQB is equal to (A) 20° (B) 40° (C) 35° (D) 45°

Posted by

infoexpert27

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) (B) 6 cm (C) 3 cm (D)

Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

In Figure, if PA and PB are tangents to the circle with centre O such that APB = 50°, then OAB is equal to (A) 25° (B) 30° (C) 40° (D) 50°

Posted by

infoexpert27

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°

Posted by

infoexpert27

JEE Main high-scoring chapters and topics

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) (D)

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°

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) $3\sqrt{3}cm$

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°

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°

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) $4\sqrt{3}cm$

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

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

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°

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

(A) 62.5° (B) 45° (C) 35° (D) 55°

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