M.C.Q (1 Marks)

MCQ 11 Mark

A chord of a circle of radius $10 \ cm$ subtends a right angle at its centre. The length of the chord $($in $cm )$ is

✓
$5 \sqrt{2}$
B
$10 \sqrt{2}$
C
$\frac{5}{\sqrt{2}}$
D
$10 \sqrt{3}$

Answer

Correct option: A.

$5 \sqrt{2}$

The figure represents the circle of radius $10 \ cm$ with the chord that subtends right angle at the centre.

Consider $\triangle P X S$
$\cos 45^{\circ}=\frac{XS}{PX}$
$\Rightarrow \cos 45^{\circ}=\frac{X S}{10}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{X S}{10}$
$\Rightarrow X S=\frac{10}{\sqrt{2}}=5 \sqrt{2} \ cm$
Thus the correct answer is $(a).$

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MCQ 21 Mark

In figure, $P Q$ and $P R$ are two tangents to a circle with centre $O$. If $\angle P Q R=46^{\circ},$ then $\angle Q O R$ equals:

A
$67^{\circ}$
✓
$134^{\circ}$
C
$44^{\circ}$
D
$46^{\circ}$

Answer

Correct option: B.

$134^{\circ}$

Given: $\angle Q P R=46^{\circ}$
$P Q$ and $P R$ are tangents.
Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
So, we have $O Q \perp P Q$ and $O R \perp R P$.
$\angle O Q P=\angle O R P=90^{\circ}$
So, in quadrilateral $P Q O R$, we have
$\angle O Q P+\angle Q P R+\angle P R O+\angle R O Q=360^{\circ}$
$($Angle sum property$)$
$\Rightarrow 90^{\circ}+46^{\circ}+90^{\circ}+\angle R O Q=360^{\circ}$
$=\angle R O Q=360^{\circ}-226^{\circ}=134^{\circ}$

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MCQ 31 Mark

In a right triangle $\text{ABC}$, right $-$ angled at $B, BC =12 \ cm$ and $AB =5 \ cm$. The radius of the circle inscribed in the triangle $($in $\ cm )$ is

A
$4$
B
$3$
✓
$2$
D
$1$

Answer

Correct option: C.

$2$

Consider the triangle $\text{ABC},$ right angled at $B$ where $AB =5 \ cm$ and $BC =12 \ cm$.
$T$ is the centre of the circle and $T P, T R$ and $T Q$ are the radii of the inscribed circle.

$\text{ABC}$ is a right angle triangle with an inscribed circle centered at $T$.
Let $r$ be the radius of the circle, so $T Q=T R=T P$
$A B, B C$ and $A C$ are tangents to the circle at $R, Q$ and $P$.
Using Pythagor as theorem in $\triangle A B C$
$A C^2=A B^2+B C^2$
$\Rightarrow A C^2=5^2+12^2$
$\Rightarrow A C^2=25+144$
$\Rightarrow A C^2=169$
$\Rightarrow A C=\sqrt{169}=13 \ cm$
Area of $\triangle ABC =\frac{1}{2} \times AB \times BC$
$\Rightarrow$ Area of $\triangle ABC =\frac{1}{2} \times 5 \times 12=30 \ cm^2$
Also, Area of $\triangle A B C=$ Area $\triangle T A B+$ Area $\triangle T B C+$ Area $\triangle T C A$
$\Rightarrow 30=\frac{1}{2} \times TR \times AB+\frac{1}{2} \times TQ \times BC+\frac{1}{2} \times TP \times AC$
$\Rightarrow 30=\frac{1}{2} \times r \times AB+\frac{1}{2} \times r \times BC+\frac{1}{2} \times r \times AC\ ( T P, T R$ and $T Q$ are the radii of the circle$)$
$\Rightarrow 30=\frac{1}{2} \times r(AB+BC+AC)$
Substitute the value for the length of the sides
$\Rightarrow 30=\frac{1}{2} \times r (5+12+13)$
$\Rightarrow 30=\frac{1}{2} \times r \times 30$
$\Rightarrow T=\frac{30 \times 2}{30}=2 \ cm$

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MCQ 41 Mark

Two circles touch each other externally at $\text{P.AB}$ is a common tangent to the circles touching them at $A$ and $B$. The value of $\angle \text{APB}$ is

✓
$30^{\circ}$
B
$45^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

The diagram is represented as follows:

In $\triangle \text{TAP}$
$\text{TA = TP} ($Tangents from an external point are equal$)$
So $\angle \text{TAP}=\angle \text{TPA}=x ($Corresponding angles are equal$)$
In $\triangle \text{TBP}$
$\text{TB = TP} ($Tangents from an external point are equal$)$
So $\angle \text{TBP}=\angle \text{TPB}=y ($Corresponding angles are equal$)$
In $\triangle \text{PAB,} \angle \text{PAB}+\angle \text{PBA}+\angle \text{APB} ($Sum of angles of a triangle$)$
$\Rightarrow x+y+x+y=180^{\circ}$
$\Rightarrow 2 x+2 y=180^{\circ}$
$\Rightarrow 2(x+y)=180^{\circ}$
$\Rightarrow x+y=\frac{180^{\circ}}{2}$
Therefore, $\angle \text{APB}=x+y=90^{\circ}$
The correct answer is $(d).$

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MCQ 51 Mark

In figure, $\text{QR}$ is a common tangent to the given circles, touching externally at the point $T$. The tangent at $T$ meets $\text{QR}$ at $P$. If $\text{QP} =3.8$, then the length of $\text{QR} ($in $cm )$ is :

A
$3.8$
✓
$7.6$
C
$5.7$
D
$1.9$

Answer

Correct option: B.

$7.6$

It is known that the length of the tangents drawn from an external point to a circle is equal.
$\text{QP =PT}=3.8 \ cm \ldots \ldots(i)$
$\text{PR =PT}=3.8 \ cm \ldots \ldots(ii)$
From equations $(i)$ and $(ii),$ we get:
$\text{QP=PR}=3.8 \ cm$
Now,
$\text{QR=Q P+PR}$
$=3.8 \ cm+3.8 \ cm=7.6 \ cm$

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MCQ 61 Mark

In Fig. $2,$ a circle with centre $O$ is inseribed in a quadrilateral $\text{ABCD}$ such that, it touches the sides $B C, A B, A D$ and $C D$ at points $P, Q, R$ and $S$ respectively, If $A B=29 \ cm, A D=23 \ cm, \angle B=90^{\circ}$ and $D S=5 \ cm$, then the radius of the circle $($in $\ cm )$ is:

✓
$11$
B
$18$
C
$6$
D
$15$

Answer

Correct option: A.

$11$

Here it is given that $A B, B C, C D$ and $A D$ are tangents to the circle with centre $O$ and touch the circle at $Q, P, S$ and $R$ respectively.
And also,
$B A=29 \ cm, A D=23 \ cm, \angle B=90^{\circ} $ and $ D S=5 \ cm$
We know that the lengths of the tangents drawn from an external point to a circle are equal.
$D S=D R=5 \ cm$
$A R=A D-D R$
$=23-5=18 \ cm \ \ A D=23 \ cm)$
$\Rightarrow A Q=A R=18 \ cm$
$\therefore Q B=A B-A Q$
$=29-18=11 \ cm\ \ (A B=29 \ cm)$
And, $Q B=B P=11 \ cm$
$\angle P B Q=90^{\circ} \ \ [$Given$]$
We know that, angle between the tangent and the radius at the point of contact is a right angle.
Thus, $\angle O P B=90^{\circ}$ and $\angle O Q B=90^{\circ}$
Now, In quadrilateral $\text{OPBQ}$,
$\angle P B Q+\angle O P B+\angle O Q B+\angle P O Q=360^{\circ}$
$[$Angle sum property of a quadrilateral$]$
$\Rightarrow 90^{\circ}+90^{\circ}+90^{\circ}=\angle P O Q=360^{\circ}$
$\Rightarrow 270^{\circ}+\angle P O Q=360^{\circ}$
$\Rightarrow \angle P O Q=360^{\circ}-270^{\circ}$
$\Rightarrow \angle P O Q=90^{\circ}$
Here all sides of $\text{OPBQ}$ are equal and all angles are
$\therefore O P B Q$ is a square.
$\therefore O Q=Q B=B P=P O=r=11 \ cm$
Thus the radius of the circle is $" 11 \ cm "$

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MCQ 71 Mark

In Fig. $P A$ and $P B$ are two tangents drawn from an external point $P$ to a circle with centre $C$ and radius $4 \ cm$ . If $P A \perp P B,$ then the length of each tangent is:

A
$3 \ cm$
✓
$4 \ cm$
C
$5 \ cm$
D
$6 \ cm$

Answer

Correct option: B.

$4 \ cm$

It is given that $A P \perp P B$ and radius of the circle is $4 \ cm$
Construction : Join $C A$ and $C B$.
It is known fact that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\therefore A C \perp A P$ and $B C \perp P B$
$\Rightarrow \angle C A P=\angle C B P=90^{\circ}$
Also, In quadrilateral $\text{ACBP},$
$\angle C A P+\angle A P B+\angle P B C+\angle B C A=360^{\circ}$
$($Angle sum property$)$
$\Rightarrow 90^{\circ}+90^{\circ}+90^{\circ}+\angle B C A=360^{\circ}$
$\Rightarrow 270^{\circ}+\angle B C A=360^{\circ}$
$\Rightarrow \angle B C A=360^{\circ}-270^{\circ}$
$\Rightarrow \angle B C A=90^{\circ}$
And also, $A C=C B$
$($Radius of the circle$)$
Here all sides of $\text{APBC}$ are equal and all angles are $90^{\circ}$
$\therefore \text{APBC}$ is a square.
$\therefore A C=C B=B P=P A=4 \ cm$
Thus the length of each tangent is $" 4 \ cm "$

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MCQ 81 Mark

In Figure $ 1, A P, A Q$ and $B C$ are tangents to the circle. If $AB =5, BC =4$ and $AC =6 \ cm,$ then the length of $AP$ (in $\ cm )$ is

✓
$7.5$
B
$15$
C
$10$
D
$9$

Answer

Correct option: A.

$7.5$

Let us suppose that the tangent $B C$ touches the circle at point $O$.
Now, we know that, tangents drawn from an exterior point to a circle are equal in length.
$\therefore B P=B O \ [$ From external point $B]$
$C Q=C O\ [$ From external point $ C]$
$A P=A Q\ [$ From external point $ A]$
Now, In $\triangle A B C$,
$A B+B C+A C=5 \ cm+4 \ cm+6 \ cm$
$\Rightarrow A B+(B O+C O)+A C=15 \ cm$
$\Rightarrow(A B+B P)+(C Q+A C)=15 \ cm$
$[\because B P=B O$ and $C Q=C O]$
$\Rightarrow A P+A Q=15 \ cm$
$\Rightarrow 2 A P=15 \ cm$
$\Rightarrow A P=7.5 \ cm \quad[\because A P=A Q]$
$\Rightarrow A P$
Therefore, length of the tangent $A P$ is $7.5 \ cm .$

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MCQ 91 Mark

From a point $Q, 13 \ cm$ away from the centre of a circle, the length of tangent $PQ$ to the circle is $12 \ cm.$ The radius of the circle $($ in $cm )$ is

A
$25$
B
$\sqrt{313}$
✓
$5$
D
$1$

Answer

Correct option: C.

$5$

We know that, tangent at a point is perpendicular to the radius through that point.Therefore, $OP$ is perpendicular to $PQ$.
In right triangle $\text{OPQ}$, Using pythagoras theorem, we have
$\ce{OQ^2=O P^2 + P Q^2}$
$\Rightarrow(13)^2=OP^2+(12)^2$
$\Rightarrow OP^2=169-144$
$\Rightarrow OP^2=25 \Rightarrow O P=5$
Therefore, the length of the radius of the circle, $O P$ is $5 \ cm .$
The correct option is $(c).$

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MCQ 101 Mark

The radii of two circles are $4 \ cm$ and $3 \ cm$ respectively. The diameter of the circle having area equal to the sum of the areas of the two circles $($in $cm)$ is

A
$5$
B
$7$
✓
$10$
D
$14$

Answer

Correct option: C.

$10$

Let $r_1$ and $r_2$ denote the radii of two circles and $R$ be the radius of the bigger circle.
$\therefore r_1=4 \ cm$ and $r_1=3 \ cm$
According to the question,
Sum of the areas of two smaller circles $=$ Area of the bigger circle
$\pi r_1{ }^2+\pi r_2{ }^2=\pi R^2$
Substituting the values of $r_1$ and $r_2$ in the above equation.
$\Rightarrow \pi(4)^2+\pi(3)^2=\pi R^2$
$\Rightarrow(4)^2+(3)^2=R^2$
$\Rightarrow 16+9=R^2$
$\Rightarrow R^2=6+9$
$\Rightarrow R^2=25$
$\Rightarrow R=\sqrt{25}$
$\Rightarrow R=5$
Diameter $=2 \times$ Radius
Therefore, the required diameter of the circle is
Hence, the correct option is $(c).$

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MCQ 111 Mark

In the given figure, $O$ is the centre of a circle, $A B$ is a chord and $A T$ is the tangent at $A$. If $\angle AOB = 110^{\circ}$, then $\angle BAT$ is equal to

A
$100^{\circ}$
B
$40^{\circ}$
✓
$50^{\circ}$
D
$90^{\circ}$

Answer

Correct option: C.

$50^{\circ}$

Consider the given diagram

Here, in $\triangle A O B, \angle A O B=100^{\circ}, $
$O A=O B=$ Radius of the circle
$\Rightarrow \triangle A O B$ is an isosceles triangle.
$\therefore \angle O B A=\angle O A B=x$
In $\triangle A O B$ apply angle sum property.
$\therefore \angle A O B+\angle O B A+\angle O A B=180^{\circ}$
$\Rightarrow 100^{\circ}+x+x=180^{\circ} $
$\Rightarrow 100^{\circ}+2 x=180^{\circ}$
$\Rightarrow 2 x=180^{\circ}-100^{\circ}$
$ \Rightarrow 2 x=80^{\circ}$
$\Rightarrow x=40^{\circ}$
$\therefore \angle O B A=\angle O A B=40^{\circ}$
Clearly $\angle O A T=90^{\circ} \ ($as tangents are perpendicular to the radius at the point of contact.$)$
From the diagram,
$\angle O A T=\angle O A B+\angle B A T$
$\Rightarrow 90^{\circ}=40^{\circ}+\angle B A T$
$\Rightarrow \angle B A T=90^{\circ}-40^{\circ}$
$=50^{\circ}$

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MCQ 121 Mark

The diameter of a car wheel is $42 \ cm$ . The number of complete revolutions it will make in moving $132 \ km$ is

A
$10^4$
✓
$10^5$
C
$10^6$
D
$10^3$

Answer

Correct option: B.

$10^5$

Diameter of wheel $=42 \ cm$
Radius of the wheel $=\frac{42}{2}=21 \ cm$
Distance in $1$ revolution
$=$ Circumference of the wheel $=2 \pi r$
$=2 \times \frac{22}{7} \times 3$
$=132 \ cm$
Number of revolutions
Total distance covered by wheel
Distance covered in $1$ revolution
$=\frac{13200000}{132}$
$=100000$
$=10^5$

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MCQ 131 Mark

Points $A (-1, y)$ and $B (5,7)$ lie on a circle with centre $O (2,-3 y)$. The values of $y$ are

A
$1,-7$
✓
$-1,7$
C
2,7
D
$-2,-7$

Answer

Correct option: B.

$-1,7$

(b)
As point A and B lie on the circle and O is the centre.
AO and BO will be the radii of the circle.

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MCQ 141 Mark

In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any point on the circle. Then the measure of $\angle AQB$ is

✓
$621 / 2^{\circ}$
B
$125^{\circ}$
C
$55^{\circ}$
D
$90^{\circ}$

Answer

Correct option: A.

$621 / 2^{\circ}$

(a)
Angle at the remaining part of the circle is half the angle subtended by the arc at the centre.

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MCQ 151 Mark

The circumferences of two circles are in the ratio $4: 5$. What is the ratio of their radii?

A
$16: 25$
B
$25: 16$
C
$2: \sqrt{5}$
✓
$4: 5$

Answer

Correct option: D.

$4: 5$

(d)
Circumference of a circle
$
\begin{aligned}
& =2 \pi r \\
\frac{2 \pi R_1}{2 \pi R_2} & =\frac{4}{5} \\
\frac{R_1}{R_2} & =\frac{4}{5} \\
R_1: R_2 & =4: 5
\end{aligned}
$

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MCQ 161 Mark

A statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option :
Assertion (A) : A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R): The lengths of tangents drawn from an external point to a circle are equal.

A
Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
✓
Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
C
Assertion (A) is true but Reason (R) is false.
D
Assertion (A) is false but Reason (R) is true.

Answer

Correct option: B.

Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).

(b)
Both Assertion and reason are correct but reason does not give the correct explanation of assertion.

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MCQ 171 Mark

The length of the tangent from an external point A to a circle, of radius 3 cm , is 4 cm . The distance of A from the centre of the circle is :

A
7 cm
✓
5 cm
C
$\sqrt{7} cm$
D
25 cm

Answer

Correct option: B.

5 cm

(b)
Given that $AB =4 cm, OB =3 cm$
To find OA
Applying Pythagoras theorem to triangle OAB
$
\begin{aligned}
OB^2+AB^2 & =OA^2 \\
3^2+4^2 & =OA \\
OA^2 & =25 \\
OA & =5 cm
\end{aligned}
$

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MCQ 181 Mark

How many tangents can be drawn to a circle from a point on it?

✓
One
B
Two
C
Infinite
D
Zero

Answer

Correct option: A.

One

(a)
If a point lies on the circle, only one tangent can be drawn from this point and it is perpendicular to the radius of the circle.

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MCQ 191 Mark

In the given circle in Figure, number of tangents parallel to tangent PQ is

A
$0$
B
many
C
2
✓
1

Answer

Correct option: D.

(d)
1

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MCQ 201 Mark

In the given figure, $\text{PA}$ and $\text{PB}$ are tangents to the given circle such that $\text{PA} =5 \ cm$ and $\angle \text{APB} =60^{\circ}$. The length of chord $\text{AB}$ is

A
$5 \sqrt{2} \ cm$
✓
$5 \ cm$
C
$5 \sqrt{3} \ cm$
D
$7.5 \ cm$

Answer

Correct option: B.

$5 \ cm$

$\because \text{PA = PB}\ [$Tangent from external point $]$
$\therefore \angle \text{PBA} =\angle \text{PAB} = x ^{\circ} \text { (say) }$
In $\triangle \text{APB} ,$
$\angle \text{APB + PBA} +\angle \text{BAP} =180^{\circ}$
$\Rightarrow 60^{\circ}+ x ^{\circ}+ x ^{\circ}=180^{\circ}$
$\Rightarrow 2 x ^{\circ}=180^{\circ}-60^{\circ}$
$\Rightarrow 2 x ^{\circ}=120^{\circ}$
$\Rightarrow x =\frac{120}{2}=60^{\circ}$
$\therefore \triangle \text{APB}$ is equilateral triangle
$\therefore \text{AB = AP = PB} =5 \ cm$

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MCQ 211 Mark

In a right triangle $A B C$, right $-$ angled at $B$, $BC =12 \ cm$ and $AB =5 \ cm$. The radius of the circle inscribed in the triangle is

A
$1 \ cm$
✓
$2 \ cm$
C
$3 \ cm$
D
$4 \ cm$

Answer

Correct option: B.

$2 \ cm$

In right $\triangle A B C$, we have
$ AC =\sqrt{( AB )^2+( BC )^2}$
$=\sqrt{(5)^2+(12)^2}$
$=\sqrt{25+144}=\sqrt{169}=13 \ cm$
$\text { or }(\triangle ABC )=\frac{1}{2} \times(\text { perimeter of } \triangle ABC ) \times r$
$\Rightarrow \frac{1}{2} \times 5 \times 12=\frac{1}{2} \times(5+12+13) \times r$
$\Rightarrow r =\frac{60}{30}=2 \ cm$

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MCQ 221 Mark

In the given figure, quadrilateral $\text{ABCD}$ is circumscribed, touching the circle at $\text{P, Q, R}$ and S . If $AP =5 \ cm, BC =7 \ cm$ and $CS =3 \ cm$. Then, the length $AB =$ ?

✓
$9 \ cm$
B
$10 \ cm$
C
$12 \ cm$
D
$8 \ cm$

Answer

Correct option: A.

$9 \ cm$

Since the length of tangents drawn from an external point to a circle are equal,
We have $\ce{AQ = AP}=5 \ cm$
$\ce{CR=CS=}3 \ cm$ and $\ce{BR=(BC-CR)}$
$=7-3=4 \ cm$
$\ce{BQ=BR}=4 \ cm$
$\therefore \ce{AB=(AQ+BQ)}=5+4=9 \ cm$

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MCQ 231 Mark

To draw a pair of tangents to a circle, which are inclined to each other at an angle of $45^{\circ}$, We have to draw tangents at the end points of those two radii, the angle between which is

A
$105^{\circ}$
✓
$135^{\circ}$
C
$140^{\circ}$
D
$145^{\circ}$

Answer

Correct option: B.

$135^{\circ}$

(b)
Let PA and PB be the desired tangents to a circle with centre $O$ from an exterior point $P$.

The, $\angle APB =45^{\circ}$
BOAP must be a cyclic quadrilateral.
$
\begin{aligned}
\therefore & \angle AOB+\angle APB=180^{\circ} \Rightarrow \angle AOB=180^{\circ}-45^{\circ} \\
& =135^{\circ}
\end{aligned}
$
So, the angle between the two radii must be $135^{\circ}$

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MCQ 241 Mark

The length of the tangent from an external point $P$ to a circle of radius 5 cm is 10 cm . The distance of the point from the centre of the circle is.

A
8 cm
B
$\sqrt{104} cm$
C
12 cm
✓
$\sqrt{125} cm$

Answer

Correct option: D.

$\sqrt{125} cm$

(d)
In right angled $\triangle OAP$,
By pythagoras theorem

$\begin{aligned} OP & =\sqrt{( OA )^2+( AP )^2} \\ & =\sqrt{(5)^2+(10)^2} \\ & =\sqrt{25+100}=\sqrt{125} cm\end{aligned}$

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MCQ 251 Mark

If two tangents inclined at an angle of $60^{\circ}$ are drawn to a circle of radius $3 \ cm$ then the length of each tangent is

A
$3 \ cm$
B
$\frac{3 \sqrt{3}}{2} \ cm$
✓
$3 \sqrt{3} \ cm$
D
$6 \ cm$

Answer

Correct option: C.

$3 \sqrt{3} \ cm$

Let $P Q$ and $P R$ be tangents to a circle with centre $O$ rom a point $P$ such that $\angle Q P R=60^{\circ}$
Then $\angle QPO =\angle RPO $
$=\frac{1}{2} \angle QPR =30^{\circ}$

$[\because$ tangents from an external point are equally inclined to the line joining the point and the centre.$]$
In right $\triangle OPQ$, we have,
$\tan \ (\angle OPQ)=\frac{OQ}{PQ} $
$\Rightarrow \tan 30^{\circ}=\frac{3}{PQ}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3}{PQ} $
$\Rightarrow PQ=3 \sqrt{3} \ cm$
Now, $PR = PQ $
$=3 \sqrt{3} \ cm$.

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MCQ 261 Mark

In the given figure, $PA$ and $PB$ are two tangents to the circle with centre $O$. If $\angle APB =60^{\circ}$ then $\angle OAB$ is

A
$15^{\circ}$
✓
$30^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

$\because \angle APB =2 \angle OAB \text { (By theorem) }$
$\therefore 60^{\circ}=2 \angle OAB$
$\Rightarrow \angle OAB $
$=\frac{60^{\circ}}{2}$
$=30^{\circ}$

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MCQ 271 Mark

If $PA$ and $PB$ are two tangents to a circle with centre $O$ such that $\angle A O B=110^{\circ}$ then $\angle APB$ is equal to

A
$55^{\circ}$
B
$60^{\circ}$
✓
$70^{\circ}$
D
$90^{\circ}$

Answer

Correct option: C.

$70^{\circ}$

$\because \square OAPB$ is a cyclic quadrilateral
$\therefore \angle AOB+\angle APB=180^{\circ} ($Opposite angle of a cyclic quadrilateral$)$
$\Rightarrow 110^{\circ}+\angle APB=180^{\circ}$
$\angle APB=180^{\circ}-110^{\circ}$
$=70^{\circ}$

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MCQ 281 Mark

In the given figure, $AT$ is a tangent to the circle with centre $O$ such that $O T=4 \ cm$ and $\angle O T A=$ $30^{\circ}$. Then, $AT$ = ?

A
$4 \ cm$
B
$2 \ cm$
✓
$2 \sqrt{3} \ cm$
D
$4 \sqrt{3} \ cm$

Answer

Correct option: C.

$2 \sqrt{3} \ cm$

In right $\triangle OAT ($having right angle at $A ),$ we have
$\cos (\angle OTA)=\frac{AT}{OT} $
$\Rightarrow \cos 30^{\circ}=\frac{AT}{4}$
$\Rightarrow AT=4 \times \cos 30^{\circ}$
$=4 \times \frac{\sqrt{3}}{2}$
$=2 \sqrt{3} \ cm$

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MCQ 291 Mark

In a given figure, $AB$ and $AC$ are tangents to a circle with centre $O$ and radius $8 \ cm .$ If $OA =17 \ cm ,$ then the length of $AC ($in $cm )$ is

A
$9$
✓
$15$
C
$\sqrt{353}$
D
$25$

Answer

Correct option: B.

$15$

$\triangle OBA$ us a right angled triangle.
$\therefore AB=\sqrt{(OA)^2-(OB)^2}$
$=\sqrt{(17)^2-(8)^2}$
$=\sqrt{289-64}$
$=\sqrt{225}$
$=15$
$\therefore AB=AC$
$(\because \text { tangents from an external point are equal) }$
$\therefore AC=15 \ cm$

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MCQ 301 Mark

If a chord $A B$ subtends an angle of $60^{\circ}$ at the centre of a circle, then the angle between the tangents to the circle drawn from $A$ and $B$ is

A
$30^{\circ}$
B
$60^{\circ}$
C
$90^{\circ}$
✓
$120^{\circ}$

Answer

Correct option: D.

$120^{\circ}$

$\because \square \ce{OAPB}$ is a cyclic quadrilateral

$\therefore \angle AOB+\angle APB=180^{\circ}$
$\Rightarrow \angle APB=180^{\circ}-60^{\circ}$
$=120^{\circ}$

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MCQ 311 Mark

In the gives figure, $AB$ and $AC$ are tangents to the circle with centre $O$ such that $\angle BAC =40^{\circ}$ Then $\angle BOC$ is equal to

A
$80^{\circ}$
✓
$140^{\circ}$
C
$120^{\circ}$
D
$100^{\circ}$

Answer

Correct option: B.

$140^{\circ}$

$\because \angle OBA=OCA=90^{\circ}$
$[\because$ Tangent at any point is perpendicular to the radius through the point of contact$]$
$\therefore \angle ABO+\angle BOC+\angle OCA+\angle CAB=360^{\circ}$
$\Rightarrow 90^{\circ}+\angle BOC+90^{\circ}+40^{\circ}=360^{\circ}$
$\Rightarrow \angle BOC=360^{\circ}-220^{\circ}=140^{\circ}$

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MCQ 321 Mark

$PQ$ is a tangent to a circle with centre $O$ at the point $P .$ If $\triangle OPQ$ is an isosceles triangle, then $\angle O Q P$ is equal to

A
$30^{\circ}$
✓
$45^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: B.

$45^{\circ}$

We have $OP \perp PQ$, i.e $\angle OPQ =90^{\circ}$
$\because \triangle OPQ$ is isosceles

$OP=PQ$
$\therefore \angle OQP=\angle POQ=45^{\circ} [\because$ In a triangle, angle opposite equal sides are equal$]$

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MCQ 331 Mark

In the given figure, $PT$ is a tangent to the circle with centre $O$ . If $OT =6 \ cm$ and $OP =10 \ cm$, then the length of tangent $PT$ is

✓
$8 \ cm$
B
$10 \ cm$
C
$12 \ cm$
D
$16 \ cm$

Answer

Correct option: A.

$8 \ cm$

In right $\triangle \text{OTP}$
By pythagoras theorem,
$\ce{(OP)^2=(OT)^2 + (PT)^2}$
$\Rightarrow PT=\sqrt{(10)^2-(6)^2}$
$=\sqrt{100-36}$
$=\sqrt{64}$
$=8 \ cm$

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MCQ 341 Mark

Two chord of a circle of radius $10 \ cm$ subtends a right angle at its centre. The length of the chord $($in $cm )$ is

A
$\frac{5}{\sqrt{2}}$
B
$5 \sqrt{2}$
✓
$10 \sqrt{2}$
D
$10 \sqrt{3}$

Answer

Correct option: C.

$10 \sqrt{2}$

Length of the chord

$AB=\sqrt{(10)^2+(10)^2}$
$=\sqrt{100+100}$
$=\sqrt{200}$
$=\sqrt{100 \times 2}$
$=10 \sqrt{2} \ cm$

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MCQ 351 Mark

Which of the following pairs of lines in a circle cannot be parallel?

A
two chords
B
a chord and a tangents
C
two tangents
✓
two diameters.

Answer

Correct option: D.

two diameters.

(d)
Two diameters in a circle cannot be parallel.

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MCQ 361 Mark

In a given figure, $R Q$ is a tangent to the circle with centre $O$. If $S Q=6 \ cm$ and $Q R=4 \ cm$, then $OR$ is equal to

A
$2.5 \ cm$
B
$3 \ cm$
✓
$5 \ cm$
D
$8 \ cm$

Answer

Correct option: C.

$5 \ cm$

$\because$ We know that tangent at any point is perpendicular to the radius through the point of contact.
$\therefore \angle OQR=90^{\circ}$
and $OS = OQ =3 \ cm ($radius of circle$)$
In $\triangle OQR$
By Pythagoras theorem,
$(OR)^2=(OQ)^2+(QR)^2$
$\Rightarrow OR=\sqrt{9+16}$
$=\sqrt{25}$
$=5 \ cm$

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MCQ 371 Mark

The number of tangents that can be drawn from an external point to a circle is

A
1
✓
2
C
3
D
4

Answer

Correct option: B.

(b)
There are only 2 tangents that can be drawn from an external point to a circle

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Maths M.C.Q (1 Marks)

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