M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

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M.C.Q (1 Marks)

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MCQ 11 Mark

In the given figure, $O$ is the centre of a circle, $P Q$ is a chord and $P T$ is the tangent at $P, \angle P O Q=70^{\circ}$, then $\angle T P Q$ is equal to

A
$55^{\circ}$
B
$70^{\circ}$
C
$45^{\circ}$
✓
$35^{\circ}$

Answer

Correct option: D.

$35^{\circ}$

(d) : In $\triangle O P Q, O P=O Q \quad$ (Radii of same circle)
$
\Rightarrow \angle O Q P=\angle O P Q
$
(Angles opposite to equal sides are equal)
$
\Rightarrow \angle O Q P=\angle O P Q=55^{\circ}
$
[By using angle sum property]
Also, $\angle O P T=90^{\circ}$
$[\because$ Tangent is perpendicular to the radius through the point of contact.]
$
\Rightarrow \angle T P Q=90^{\circ}-55^{\circ}=35^{\circ}
$
[From (i) and (ii)]

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MCQ 21 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

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

Answer

Correct option: B.

$10 \sqrt{2}$

Let $A B$ is a chord of circle which subtends right angle at its centre.
$\therefore$ In $\triangle O A B$, by Pythagoras theorem, we have
$(A B)^2=(O A)^2+(O B)^2$
$\Rightarrow(A B)^2=(10)^2+(10)^2$
$\Rightarrow(A B)^2=200$
$\Rightarrow A B=10 \sqrt{2} \ cm$

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MCQ 31 Mark

In the given figure, there are two concentric circles of radii $6 \ cm$ and $4 \ cm$ with centre $O$. If $A P$ is a tangent to the larger circle and $B P$ to the smaller circle and length of $A P$ is $8 \ cm$, then the length of $B P$ is

A
$21 cm$
B
26
✓
$2 \sqrt{21} cm$
D
None of these

Answer

Correct option: C.

$2 \sqrt{21} cm$

In right $\triangle A O P, O P^2=A P^2+O A^2$
$=8^2+6^2=100$
In right $\triangle B O P, O P^2=B P^2+O B^2$
$\Rightarrow 100=B P^2+4^2$
$\Rightarrow B P^2=100-16=84$
$\Rightarrow B P=2 \sqrt{21} \ cm$

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MCQ 41 Mark

The length of the tangent drawn from a point $8 cm$ away from the centre of circle of radius $6 cm$ is

A
$\sqrt{7} cm$
✓
$2 \sqrt{7} cm$
C
$10 cm$
D
$5 cm$

Answer

Correct option: B.

$2 \sqrt{7} cm$

(b) : Since tangent to a circle is perpendicular to the radius through the point of contact.
$
\therefore \angle O T P=90^{\circ}
$
In $\triangle O T P$, by Phthagoras theorem, we have
$
\begin{aligned}
& O P^2=O T^2+P T^2 \\
\Rightarrow & (8)^2=(6)^2+P T^2 \\
\Rightarrow & P T^2=64-36=28 \\
\Rightarrow & P T=\sqrt{28}=2 \sqrt{7} cm
\end{aligned}
$

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MCQ 51 Mark

In the given figure, $P A$ and $P B$ are tangents to the circle from an external point $P . C D$ is another tangent touching the circle at $Q$. If $P A=12 cm , Q C=Q D=3 cm$ then the value of $P C$ is

A
$6 cm$
✓
$9 cm$
C
$12 cm$
D
$3 cm$

Answer

Correct option: B.

$9 cm$

(b) : As we know that tangents drawn from an external point are equal in length.
$
\therefore \quad Q C=C A ; Q D=B D \text { and } P A=P B
$
Since $Q C=Q D=3 cm$
(Given)
$
\Rightarrow C A=B D=3 cm
$
Also, $P C=P A-A C=(12-3) cm =9 cm$

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MCQ 61 Mark

In figure, $\text{PQ}$ is tangent to the circle with centre at $O$, at the point $B$. If $\angle AOB=100^{\circ}$, then $\angle ABP$ is equal to

✓
$50^{\circ}$
B
$40^{\circ}$
C
$60^{\circ}$
D
$80^{\circ}$

Answer

Correct option: A.

$50^{\circ}$

In $\triangle \text{OAB, OA=OB}$
$( \because$ Radii of same circle$)$
$\therefore \angle O A B=\angle O B A$
$(\because$ Angles opposite to equal sides are equal $)$
Now, by applying angle sum property in $\triangle O A B$
$\angle OAB+\angle ABO+\angle A O B=180^{\circ}$
$\Rightarrow \angle ABO=40^{\circ}$
Here, $\angle OBP=90^{\circ}$
$($Radius is perpendicular to tangent at point of contact$)$
$\Rightarrow \angle OBA+\angle A B P=90^{\circ}$
$\Rightarrow 40^{\circ}+\angle A B P=90^{\circ}$
$\Rightarrow \angle A B P=90^{\circ}-40^{\circ}=50^{\circ}$

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MCQ 71 Mark

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

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

Answer

Correct option: B.

$134^{\circ}$

We have, $O Q \perp P Q$ and $O R \perp R P$
$[\because$ Radius is perpendicular to the tangent through the point of contact$]$
$\Rightarrow \angle O Q P=\angle O R P=90^{\circ}
$In quadrilateral $\text{PQOR}$,
we have $\angle O Q P+\angle Q P R+\angle P R O+\angle R O Q=360^{\circ}$
$\Rightarrow 90^{\circ}+46^{\circ}+90^{\circ}+\angle R O Q=360^{\circ}$
$\Rightarrow \angle R O Q=360^{\circ}-226^{\circ}=134^{\circ}$

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MCQ 81 Mark

From a point $Q$, the length of the tangent to a circle is $12 cm$ and the distance of $Q$ from the centre is $15 cm$. The radius of the circle is

✓
$9 cm$
B
$12 cm$
C
$15 cm$
D
$24.5 cm$

Answer

Correct option: A.

$9 cm$

(a) : $Q P$ is a tangent at $P$
$
\therefore \angle P=90^{\circ}
$In $\triangle O P Q$, by Phythagoras theorem
$
\begin{aligned}
& O Q^2=O P^2+P Q^2 \\
\Rightarrow & (15)^2=O P^2+12^2 \\
\Rightarrow & O P^2=225-144=81 \Rightarrow O P=9 cm
\end{aligned}
$

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MCQ 91 Mark

Two concentric circles of radii $a$ and $b$ where $a>b$, are given, the length of a chord of the larger circle which touches the other circle is

A
$\sqrt{a^2-b^2}$
✓
$2 \sqrt{a^2-b^2}$
C
$\sqrt{a^2+b^2}$
D
$2 \sqrt{a^2+b^2}$

Answer

Correct option: B.

$2 \sqrt{a^2-b^2}$

(b) : Radius of larger circle $=a$Radius of smaller circle $=b$
$\because \quad$ In $\triangle O A M$, we have
$
\begin{aligned}
& O A^2=O M^2+A M^2 \\
\Rightarrow & a^2=b^2+A M^2 \\
\Rightarrow & A M^2=a^2-b^2 \\
\Rightarrow & A M=\sqrt{a^2-b^2}
\end{aligned}
$
Now, length of chord of larger circle is $A B=2 A M$
$
=2 \sqrt{a^2-b^2}
$

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MCQ 101 Mark

In the given figure, $\text{A P, A Q}$ and $B C$ are tangents to the circle. If $A B=5 \ cm , A C=6 \ cm$ and $B C=4 \ cm$, then the length of $A P ($ in $cm )$ is

✓
$7.5$
B
$15$
C
$10$
D
$9$

Answer

Correct option: A.

$7.5$

As, length of tangents drawn from an external point to a circle are equal.
$\therefore \ce{AP=AQ \ldots (i), PB=BR \ldots (ii), CQ=CR \ldots (iii)}$
$\text { Now }, \ce{2 AP=AP + AP}$
$\Rightarrow \ce{2 AP=AP + AQ}$
$\Rightarrow \ce{2 AP=(AB + PB) + (AC + CQ)}$
$\Rightarrow \ce{2 AP=(AB + BR) + (AC + CR)} [$ Using $(ii)$ and $(iii)]$
$\Rightarrow \ce{2 AP=AB + BC + AC=5 + 4 + 6}$
$\Rightarrow \ce{AP}=7.5 \ cm [$Using $(i)]$

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MCQ 111 Mark

In the given figure, a circle with centre $O$ is inscribed in a quadrilateral $A B C D$ such that, it touches the sides $B C, A B, A D S$ 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

(a): We know that the length of the tangents drawn from an external point to a circle are equal.
$
\begin{aligned}
\therefore \quad & A Q=A R, D R=D S, B Q=B P, C S=C P \\
& S o, D S=D R=5 cm \\
& A Q=A R=A D-D R=23-5=18 cm \\
& B P=Q B=A B-A Q=29-18=11 cm
\end{aligned}
$
Since, $O Q B P$ is a square
$\Rightarrow$ Radius of circle $(O P)=11 cm$.

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MCQ 121 Mark

Two concentric circles are of radii $5 cm$ and $3 cm$. Find the length of the chord of the larger circle which touches the smaller circle.

✓
$8 cm$
B
$4 cm$
C
$10 cm$
D
$6 cm$

Answer

Correct option: A.

$8 cm$

(a): Here, $O A^2=O D^2+A D^2$
$
\Rightarrow A D=\sqrt{25-9}=4 cm
$As $O D$ bisects $A B$, then
$
A B=2 A D=2 \times 4=8 cm
$

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MCQ 131 Mark

In the given figure, point $P$ is $26 cm$ away from the centre $O$ of a circle and the length PT of the tangent drawn from $P$ to the circle is $24 cm$. Then the radius of the circle is

A
$25 cm$
B
$26 cm$
C
$24 cm$
✓
$10 cm$

Answer

Correct option: D.

$10 cm$

(d) : Let us join $O T$
We have, $O P=26 cm , P T=24 cm$
Since, radius is perpendicular to the tangent at the point of contact.

$
\therefore \angle P T O=90^{\circ}
$In right $\triangle P T O$, using Pythagoras theorem, we get
$
\begin{aligned}
& O P^2=P T^2+O T^2 \\
\Rightarrow & (26)^2=(24)^2+O T^2 \\
\Rightarrow & O T^2=676-576=100 \\
\Rightarrow & O T=10 cm
\end{aligned}
$
Hence, radius of the circle is $10 cm$.

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MCQ 141 Mark

Two concentric circles of radii $13 cm$ and $5 cm$ are given. The length of the chord of the larger circle which touches the smaller circle is

A
$16 cm$
B
$4 cm$
C
$24 cm$
D
$10 cm$

Answer

(c) : In $\triangle A D O$, by Pythagoras theorem
$\begin{aligned} & O A^2=A D^2+O D^2 \\ & \Rightarrow(13)^2=A D^2+(5)^2 \\ & \Rightarrow 169-25=A D^2 \\ & \Rightarrow A D^2=144 \\ & \Rightarrow A D=12 \mathrm{~cm}\end{aligned}$

Since, the perpendicular drawn from the centre to the chord bisects the chord.
$
\therefore A B=2 \times A D=2 \times 12=24 cm
$

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MCQ 151 Mark

In figure, if $\angle A O B = 125^{\circ}$, then $\angle C O D$ is equal to

A
$62.5^{\circ}$
B
$45^{\circ}$
C
$35^{\circ}$
✓
$55^{\circ}$

Answer

Correct option: D.

$55^{\circ}$

We know that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
$\Rightarrow \angle A O B+\angle C O D=180^{\circ}$
$\Rightarrow \angle C O D=180^{\circ}-\angle A O B=180^{\circ}-125^{\circ}=55^{\circ}$

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MCQ 161 Mark

In the given circle, $O$ is a centre and $\angle B D C=42^{\circ}$, then $\angle A C B$ is equal to

A
$42^{\circ}$
B
$45^{\circ}$
✓
$48^{\circ}$
D
$60^{\circ}$

Answer

Correct option: C.

$48^{\circ}$

(c) : $B D$ is a diameter of circle
$
\therefore \angle B C D=90^{\circ}
$
[angle in a semicircle]
In $\triangle O C D$,
$
O D=O C
$
[radii of circle]
$\angle O D C=\angle O C D=42^{\circ}$
Now, $\angle O C D+\angle O C B=90^{\circ}$
$\Rightarrow \angle O C B=90^{\circ}-42^{\circ}=48^{\circ}$
$\Rightarrow \angle A C B=\angle O C B=48^{\circ}$

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MCQ 171 Mark

What is the distance between two parallel tangents of a circle of radius $4 cm$ ?

A
$2 cm$
✓
$8 cm$
C
$6 cm$
D
None of these

Answer

Correct option: B.

$8 cm$

(b)

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MCQ 181 Mark

Two tangents, drawn at the end points of diameter of a given circle are always

✓
parallel
B
perpendicular
C
intersect each other
D
None of these

Answer

Correct option: A.

parallel

(a): Two tangents drawn at the end points of diameter are always parallel.

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MCQ 191 Mark

In the given figure, three circles with centres $A, B, C$ respectively touch each other externally. If $A B=5 cm , B C=7 cm$ and $C A=6 cm$, then the radius of the circle with centre $A$ is

A
$1.5 cm$
✓
$2 cm$
C
$2.5 cm$
D
$3 cm$

Answer

Correct option: B.

$2 cm$

(b) : Let the radii of the three circles with centre $A$, $B$ and $C$ be $x, y, z$ respectively. Then,
$
x+y=5, y+z=7 \text { and } z+x=6
$Adding all three equations, we get
$
\begin{aligned}
& 2(x+y+z)=18 \Rightarrow x+y+z=9 \\
\therefore \quad & x=(x+y+z)-(y+z)=(9-7)=2 cm
\end{aligned}
$

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MCQ 201 Mark

In the given figure, $P Q$ and $P R$ are tangents drawn from point $P$ to a circle with centre $O$. If $\angle O P Q=35^{\circ}$, then value of $a$ and $b$ are

A
$30^{\circ}, 60^{\circ}$
✓
$35^{\circ}, 55^{\circ}$
C
$40^{\circ}, 50^{\circ}$
D
$55^{\circ}, 45^{\circ}$

Answer

Correct option: B.

$35^{\circ}, 55^{\circ}$

(b) : Since $P Q$ is a tangent.
$
\therefore \angle O Q P=90^{\circ}
$In $\triangle P Q O, \angle O Q P+\angle O P Q+\angle Q O P=180^{\circ}$
[By angle sum property]
$
\Rightarrow 90^{\circ}+35^{\circ}+b=180^{\circ} \Rightarrow b=180^{\circ}-125^{\circ}=55^{\circ}
$
Since the tangents from an external point $P$ to the circle are equally inclined to $O P$.
$
\therefore \quad \angle R P O=\angle O P Q \Rightarrow a=35^{\circ}
$

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MCQ 211 Mark

A tangent to a circle is a line that touches the circle at exactly

A
two points
B
three points
✓
one point
D
None of these

Answer

Correct option: C.

one point

(c) : A tangent to a circle is a line that intersects or touches the circle at exactly one point.

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MCQ 221 Mark

If the angle between two radii of a circle is $140^{\circ}$, then the angle between the tangents at the ends of the radii is

A
$90^{\circ}$
B
$50^{\circ}$
C
$70^{\circ}$
✓
$40^{\circ}$

Answer

Correct option: D.

$40^{\circ}$

$\angle OPA=90^{\circ}$ and $\angle OQA=90^{\circ}$
$[\because$ Angle between tangent and radius through the point of contact is $90^{\circ}$ ] In quadrilateral $\text{OPAQ}$,
$\angle POQ+\angle OPA+\angle OQA+\angle PAQ=360^{\circ}$
$[$Angle sum property of a quadrilateral$]$
$\Rightarrow 140^{\circ}+90^{\circ}+90^{\circ}+\angle PAQ=360^{\circ}$
$\Rightarrow \angle PAQ=360^{\circ}-320^{\circ}=40^{\circ}$

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MCQ 231 Mark

In the given figure, $A D=8 cm , A C=6 cm$ and $T B$ is the tangent at $B$ to the circle with centre $O$. Find $O T$, if $B T$ is $4 cm$.

✓
$\sqrt{41} cm$
B
$\sqrt{43} cm$
C
$\sqrt{39} cm$
D
$\sqrt{47} cm$

Answer

Correct option: A.

$\sqrt{41} cm$

(a): Clearly, $\angle C A D=90^{\circ}$ [angle in a semi-circle]
$
\therefore \text { In } \triangle A C D, C D^2=A C^2+A D^2=36+64=100
$
[by Pythagoras theorem]
$
\Rightarrow C D=10 cm
$Therefore, $O C=O D=O B=\frac{10}{2} cm =5 cm$
So, in right $\triangle O B T, O T^2=O B^2+B T^2=25+16=41$
[by Pythagoras theorem]
$
\Rightarrow O T=\sqrt{41} cm
$

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MCQ 241 Mark

$O$ is the centre of the circle. $P Q$ is tangent to the circle and secant $P A B$ passes through the centre $O$. If $P Q=5 cm$ and $P A=1 cm$, then radius of the circle is

A
$8 cm$
✓
$12 cm$
C
$10 cm$
D
$6 cm$

Answer

Correct option: B.

$12 cm$

(b) : $O Q=O A=r$
[radii of circle]In
$\triangle O P Q, \angle Q=90^{\circ}$
$O P^2=O Q^2+P Q^2$
$(r+1)^2=r^2+(5)^2$
$\Rightarrow r^2+2 r+1=r^2+25$
$\Rightarrow 2 r=24 \Rightarrow r=12 cm$

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MCQ 251 Mark

In the given figure, $O$ is the centre of two concentric circles of radii $5 \ cm$ and $3 \ cm$. From an external point $P$, tangents $P A$ and $P B$ are drawn to these circles. If $P A=12 \ cm$, then $P B=$

A
$5\sqrt{2} \ cm$
B
$3 \sqrt{5} \ cm$
✓
$4 \sqrt{10} \ cm$
D
$5 \sqrt{10} \ cm$

Answer

Correct option: C.

$4 \sqrt{10} \ cm$

In right $\triangle P A O, P A=12 \ cm$ and $O A=5 \ cm$.
$\therefore$ By Pythagoras theorem,
$O P^2=O A^2+P A^2=5^2+(12)^2=25+144=169$
$\Rightarrow O P=\sqrt{169}=13 \ cm$
In right $\triangle P B O, P B^2=O P^2-O B^2$
$=13^2-3^2=169-9=160$
$\Rightarrow P B=\sqrt{160} \ cm =4 \sqrt{10} \ cm$

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MCQ 261 Mark

If two tangents inclined at an angle $60^{\circ}$ 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
$3 cm$
✓
$3 \sqrt{3} cm$

Answer

Correct option: D.

$3 \sqrt{3} cm$

(d) : As $O P$ is a bisector of $\angle A P C$.

$
\therefore \angle A P O=\angle C P O=30^{\circ}
$Also, $O A \perp A P$
$[\because$ Tangent at any point of a circle is perpendicular to the radius through the point of contact.]
$
\therefore \angle O A P=90^{\circ}
$
In right angled $\triangle O A P, \tan 30^{\circ}=\frac{O A}{A P}=\frac{3}{A P}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3}{A P} \Rightarrow$ Length of tangent $A P=3 \sqrt{3} cm$

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MCQ 271 Mark

How many tangents can a circle have from a point lying inside the circle ?

A
2
B
infinitely many
C
1
✓
None of these

Answer

Correct option: D.

None of these

(d) : There is no tangent to a circle passing through a point lying inside the circle.

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MCQ 281 Mark

Two circles with centres $O$ and $P$, and radii $8 \ cm$ and $4 \ cm$ touch each other externally. Find the length of their common tangent $Q R$.

A
$16 \ cm$
✓
$8 \sqrt{2} \ cm$
C
$4 \ cm$
D
$4 \sqrt{2} \ cm$

Answer

Correct option: B.

$8 \sqrt{2} \ cm$

As $S P=Q R$, as they are opposite sides of rectangle $P R Q S$.

$O P=8 \ cm +4 \ cm =12 \ cm$
$O S=8 \ cm -4 \ cm =4 \ cm$
Now, in $\triangle O S P, O P^2=O S^2+S P^2$
$\Rightarrow Q R=S P=\sqrt{O P^2-O S^2}=\sqrt{12^2-4^2} \ cm =8 \sqrt{2} \ cm$

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MCQ 291 Mark

There are two concentric circles with centre $O$ and of diameters $10 cm$ and $6 cm$ respectively. $A B$, a chord of outer circle touches the inner circle at $T$. The length of $B T$ is

A
$6 cm$
B
$7 cm$
✓
$4 cm$
D
$10 cm$

Answer

Correct option: C.

$4 cm$

(c) : In $\triangle O B T, \angle O T B=90^{\circ}$
[Tangent of a circle is perpendicular to the radius]
$\therefore \quad O B^2=O T^2+B T^2$
[By Pythagoras theorem]
$
\Rightarrow\left(\frac{10}{2}\right)^2=\left(\frac{6}{2}\right)^2+B T^2 \Rightarrow 25=9+B T^2
$
$
\Rightarrow B T^2=16 \Rightarrow B T=4 cm
$

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MCQ 301 Mark

Two circles touch internally at point $Q$. From an external point $R$, two tangents $R M$ and $R N$ are drawn to the two circles. Then,

✓
cannot be determined
B
$R M=R N$
C
$R M>R N$
D
$R M< R N$

Answer

Correct option: A.

cannot be determined

(a) : Join RQ.Since tangents drawn from an external point to a circle are equal in length.

$
\therefore R Q=R N
$
$
\text { (i) and } R Q=R M
$
$
\Rightarrow R N=R M
$

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MCQ 311 Mark

In the given figure, $Q R$ is a common tangent to the given circles, touching externally at the point $T$. The tangent at $T$ meets $Q R$ at $P$. If $P T=3.8 cm$, then the length of $Q R($ in $cm )$ is

A
1.9
✓
3.8
C
7.6
D
5.7

Answer

Correct option: B.

3.8

(b) : It is known that the length of the tangents drawn from an external point to a circle are equal.
$\therefore \quad Q P=P T=3.8 cm$ and $P R=P T=3.8 cm$
Now, $Q R=Q P+P R=3.8 cm +3.8 cm =7.6 cm$

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MCQ 321 Mark

In the given figure, a quadrilateral $A B C D$ is drawn to circumscribe a circle such that its sides $A B, B C, C D$ and $A D$ touch the circle at $P, Q, R$ and $S$ respectively. If $A B=x cm$, $B C=7 cm , C R=3 cm$ and $A S=5 cm$, find $x$.

A
$7 cm$
✓
$10 cm$
C
$9 cm$
D
$8 cm$

Answer

Correct option: B.

$10 cm$

(b) : $A P=A S, B P=B Q, C Q=C R, D R=D S$
[Tangents drawn from an external point to the circle are equal in Length]
So, $C R=C Q \Rightarrow C Q=3 cm$
Now, $B C=7 cm \Rightarrow C Q+B Q=7 cm$
$\Rightarrow B Q=(7-3) cm =4 cm$
Also, $B Q=B P \Rightarrow B P=4 cm$
Also, $A S=A P$ and $A S=5 cm \Rightarrow A P=5 cm$
$
\therefore \quad A B=A P+P B=(5+4) cm =9 cm
$So, $x=9$

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MCQ 331 Mark

Two tangents $B C$ and $B D$ are drawn to a circle with centre $O$ such that $\angle C B D=120^{\circ}$. Then $O B=$

✓
$BC / 2$
B
$2 B C$
C
$B C$
D
$3 B C$

Answer

Correct option: A.

$BC / 2$

Since, tangents from an external point $B$ to a circle are equally inclined to $O B$
$\therefore \angle C B O=\frac{1}{2} \angle C B D$
$=\frac{1}{2} \times 120^{\circ}=60^{\circ}$
Also, $\angle O C B=90^{\circ}[\because O C \perp C B]$
In $\triangle O C B, \frac{B C}{O B}=\cos 60^{\circ}=\frac{1}{2}$
$\Rightarrow O B=2 B C$

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MCQ 341 Mark

Two parallel lines touch the circle at points $A$ and $B$. If area of the circle is $16 \pi \ cm ^2$, then $A B$ is equal to

A
$16 \ cm$
✓
$5 \ cm$
C
$8 \ cm$
D
$10 \ cm$

Answer

Correct option: B.

$5 \ cm$

Let the radius of the circle be $r \ cm$.

Area of circle $=16 \pi$
$\Rightarrow \pi r^2=16 \pi$
$\Rightarrow r^2=16 $
$\Rightarrow r=4$
$\therefore A B=2 O A=2 r=8 \ cm$
$[$Given$]$

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MCQ 351 Mark

If four sides of a quadrilateral $A B C D$ are tangent to a circle, then

A
$A C+A D=B C+D B$
✓
$A C+A D=B D+C D$
C
$A B+C D=B C+A D$
D
$A B+C D=A C+B C$

Answer

Correct option: B.

$A C+A D=B D+C D$

Since the lengths of tangents to a circle from an external point are equal.

$\therefore A P=A S$
$B P=B Q$
$C R=C Q$
$D R=D S$
Adding $(i), (ii), (iii)$ and $(iv),$ we get
$(A P+B P)+(C R+D R)=A S+B Q+C Q+D S$
$\Rightarrow A B+C D=B C+A D$

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MCQ 361 Mark

If $A B$ is chord of a circle with centre $O$ and $P Q$ is a tangent to the circle at $B$ with reflex $\angle A O B=210^{\circ}$, then $\angle O B A=$

✓
$15^{\circ}$
B
$75^{\circ}$
C
$150^{\circ}$
D
$210^{\circ}$

Answer

Correct option: A.

$15^{\circ}$

Since, angle about a point is $360^{\circ}$.
$\therefore \angle A O B=360^{\circ}- $ reflex $ \angle A O B=360^{\circ}-210^{\circ}=150^{\circ}$
In $ \triangle A O B, \angle O A B+\angle A B O+\angle A O B=180^{\circ}$
$\Rightarrow \angle A B O+\angle A B O=180^{\circ}-150^{\circ}$
$\Rightarrow 2 \angle O B A=30^{\circ} [\angle O A B=\angle O B A $ and $ \angle A O B=150^{\circ}]$
$\Rightarrow 2 O B A=15^{\circ}$

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MCQ 371 Mark

In the given diagram, $\text{PQ}$ and $\text{RS}$ are common tangents to the two circles with centres $C$ and $D$. The value of $\text{RS}$ is

A
$12 \ cm$
B
$9 \ cm$
C
$5 \ cm$
✓
$15 \ cm$

Answer

Correct option: D.

$15 \ cm$

$DS=9 \ cm$ and $CD=13 \ cm$
Draw $A C \| R S$

$\therefore \text{DA=DS-SA[SA=CR]}$
$=9-4=5 \ cm$
In $\triangle C A D$, right angled at $D$
$CA^2=CD^2-DA^2$
$=13^2-5^2$
$=144$
$\Rightarrow C A=\sqrt{144}=12 \ cm$
$\Rightarrow C A=R S=12 \ cm$

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MCQ 381 Mark

In the diagram, $P Q$ and $Q R$ are tangents to the circle with centre $O$. Find the value of $x$.

A
$55^{\circ}$
B
$25^{\circ}$
✓
$35^{\circ}$
D
$65^{\circ}$

Answer

Correct option: C.

$35^{\circ}$

(c) : $\angle P O R+\angle P Q R=180^{\circ}$
$
\angle O P Q=\angle O R Q=90^{\circ}
$
$
\therefore \angle P O R=180^{\circ}-50^{\circ}=130^{\circ}
$Also, $\angle P S R=\frac{1}{2} \angle P O R$

$[\because$ Angle made by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle]
$
\therefore \quad \angle P S R=\frac{1}{2} \times 130^{\circ}=65^{\circ}
$

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MCQ 391 Mark

In the adjoining figure, if $P C$ is the tangent at $A$ of the circle with $\angle P A B$ $=72^{\circ}$ and $\angle A O B=132^{\circ}$, then $\angle A B C=$

A
can't be determined
✓
$18^{\circ}$
C
$30^{\circ}$
D
$60^{\circ}$

Answer

Correct option: B.

$18^{\circ}$

(b) : Here, $\angle P A B=72^{\circ}$
$
\therefore \angle O A B=90^{\circ}-72^{\circ}=18^{\circ}
$Also, $\angle A O B=132^{\circ}$
[Given]
Now, in $\triangle O A B, \angle A B C=180^{\circ}-132^{\circ}-18^{\circ}=30^{\circ}$

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MCQ 401 Mark

In the given figure, $P Q$ is the common tangent to both the circles. $S R$ and $P T$ are also tangents. If $S R=4 cm , P T=7 cm$, then find $R P$.

✓
$2 cm$
B
$3 cm$
C
$5 cm$
D
$3.5 cm$

Answer

Correct option: A.

$2 cm$

(a) : Since tangents drawn from an external point to a circle are equal in length.
$
\therefore P Q=P T=7 cm \text { and } R Q=R S=4 cm
$
Now, $R P=P Q-R Q=(7-4) cm =3 cm$

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MCQ 411 Mark

A circle is inscribed in a $\triangle A B C$ having sides $8 cm , 10 cm$ and $12 cm$ as shown in given figure. The length of $A D, B E$ and $C F$ (in $cm$ ) respectively are

✓
$2,8,4$
B
$7,5,3$
C
$8,4,2$
D
$6,6,4$

Answer

Correct option: A.

$2,8,4$

(a) : Let $A D=x cm$
Then $A F=x cm$
$
\begin{aligned}
\therefore \quad F C & =A C-A F=(10-x) cm \\
\text { and } C E & =F C=(10-x) cm \\
E B & =B C-C E=8-(10-x)=8-10+x=(x-2) cm \\
D B & =A B-A D=(12-x) cm
\end{aligned}
$
Now, $D B=E B \Rightarrow 12-x=x-2 \Rightarrow x=7$
$
\therefore A D=7 cm , B E=7-2=5 cm \text { and } C F=10-7=3 cm
$

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MCQ 421 Mark

From an external point $P$, tangents $P A$ and $P B$ are drawn to a circle with centre $O$. If $C D$ is the tangent to the circle at a point $E$ and $P A=14 \ cm$, find the perimeter of $\triangle P C D$.

A
$26 \ cm$
✓
$14 \ cm$
C
$28 \ cm$
D
$21 \ cm$

Answer

Correct option: B.

$14 \ cm$

Since the tangents drawn from an external point to a circle are equal.
$\therefore P A=P B, C A=C E $ and $ D B=D E$
Perimeter of $\triangle P C D=P C+C D+P D$
$=(P A-C A)+(C E+D E)+(P B-D B)$
$=(P A-C E)+(C E+D E)+(P B-D E)$
$=P A+P B=2 P A$
$=(2 \times 14) \ cm $
$=28 \ cm$

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MCQ 431 Mark

In the following figure, find the length of the chord $A B$ if $P A=6 cm$ and $\angle P A B=60^{\circ}$.

A
$5 cm$
B
$2 cm$
✓
$6 cm$
D
$4 cm$

Answer

Correct option: C.

$6 cm$

(c) : $P B=P A=6 cm \quad[\because$ Tangents drawn from an external point to a circle are equal in length]
$
\therefore \quad \angle P B A=\angle P A B=60^{\circ}
$
(Angles opposite to sides are equal)
$
\therefore \quad \angle A P B=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ}
$
[Angle sum property of a triangle]
$\therefore \quad \triangle A P B$ is an equilateral triangle.
$
\therefore \quad A B=P A=P B=6 cm
$

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MCQ 441 Mark

$AP$ and $AQ$ are two perpendicular tangents to the circle with centre $O$. The length of each tangent is 5 $cm$. The radius of the circle is .......... $cm$.

A
10
B
7.5
✓
5
D
2.5

Answer

Correct option: C.

5

5

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MCQ 451 Mark

Two concentric circles are of radii $17 cm$ and $15 cm$. The length of the chord of the larger circle which touches the smaller circle is ........... $cm$.

A
8
B
12
✓
16
D
20

Answer

Correct option: C.

16

16

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MCQ 461 Mark

If the angle between radii of the circle is $130^{\circ}$, then angle between two tangent drawn at the ends points of radii is ............

A
$90^{\circ}$
✓
$50^{\circ}$
C
$70^{\circ}$
D
$40^{\circ}$

Answer

Correct option: B.

$50^{\circ}$

$50^{\circ}$

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MCQ 471 Mark

In $\triangle ABC , \angle B =90^{\circ}, BC =12 cm$ and $AB =5 cm$. The in radius of $\triangle ABC$ is $\ldots \ldots \ldots cm$.

A
4
B
3
✓
2
D
1

Answer

Correct option: C.

2

2

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MCQ 481 Mark

The angle between two tangents to a circle with centre $O$ is $60^{\circ}$. If the radius of the circle is $3 cm$ then the length of each tangent is ........... $cm$.

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

Answer

Correct option: D.

$3 \sqrt{3}$

$3 \sqrt{3}$

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MCQ 491 Mark

$O$ is the centre of the circle and its radius is $9 cm , A$ and $AQ$ are tangents to the circle from the exterior point $A$. If $OA =15 cm$ then $AP + AQ =\ldots \ldots . . cm$.

A
12
B
18
✓
24
D
36

Answer

Correct option: C.

24

24

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MCQ 501 Mark

In the given figure, $AB =12 cm , BC =8 cm$ and $AC =10 cm$, then $AD =\ldots \ldots \ldots cm$.

A
5
B
4
C
6
✓
7

Answer

Correct option: D.

7

7

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M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip