M.C.Q - Maths STD 9 Questions - Vidyadip

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MCQ 11 Mark

If $A B=12 cm, B C=16 cm$ and $A B$ is perpendicular to $B C$, then the radius of the circle passing through the points A, B and C is

A
6 cm
B
8 cm
✓
10 cm
D
12 cm

Answer

Correct option: C.

10 cm

(c) 10 cm
Perpendicular from the centre to a chord bisects the chord. Therefore, $L$ and $M$ are mid-points of $A B$ and $B C$ respectively. Thus, in right triangle $O L B$, we have
$O L=B M=\frac{1}{2} B C=8 cm \text { and } B L=\frac{1}{2} A B=6 cm$
Applying Pythagoras theorem in $\triangle O L B$, we obtain
$O B^2=O L^2+L B^2 \Rightarrow O B=\sqrt{8^2+6^2}=\sqrt{100}=10 cm$

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MCQ 21 Mark

In Fig. two congruent circles have centres $O$ and $O^{\prime}$. Arc AXB subtends an angle of $75^{\circ}$ at the centre $O$ and arc $A^{\prime} Y B^{\prime}$ subtends an angle of $25^{\circ}$ at the centre $O^{\prime}$. Then the ratio of arcs $A X B$ and $A^{\prime} Y B^{\prime}$ is

A
$2: 1$
B
$1: 2$
✓
$3: 1$
D
$1: 3$

Answer

Correct option: C.

$3: 1$

(c) $3: 1$
$\frac{\operatorname{arc} A X B}{\operatorname{arc} A^{\prime} Y^{\prime} B}=\frac{m(\widehat{A X B})}{m\left(A^{\prime} X B^{\prime}\right)}=\frac{75^{\circ}}{25^{\circ}}=\frac{3}{1}$

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MCQ 31 Mark

In Fig. if $\angle O A B=40^{\circ}$, then $\angle A C B$ is equal to

✓
$50^{\circ}$
B
$40^{\circ}$
C
$60^{\circ}$
D
$70^{\circ}$

Answer

Correct option: A.

$50^{\circ}$

(a) $50^{\circ}$
In $\triangle O A B$, we have
$O A=O B \Rightarrow \angle O B A=\angle O A B \Rightarrow \angle O B A=40^{\circ}$
Using angle sum property in $\triangle A O B$, we obtain
$\begin{array}{ll}& \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\
\Rightarrow & 40^{\circ}+40^{\circ}+\angle A O B=180^{\circ} \Rightarrow \angle A O B=100^{\circ}\end{array}$
Thus, $\operatorname{arc} A B$ subtends $\angle A O B=100^{\circ}$ at the centre and $\angle A C B$ at a point on the circumference.
$\therefore \quad \angle A C B=\frac{1}{2} \angle A O B=50^{\circ}$

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MCQ 41 Mark

In Fig. if $\angle D A B=60^{\circ}$ and $\angle A B D=50^{\circ}$, then $\angle A C B=$

A
$60^{\circ}$
B
$50^{\circ}$
✓
$70^{\circ}$
D
$80^{\circ}$

Answer

Correct option: C.

$70^{\circ}$

(c) $70^{\circ}$
In $\triangle A B D$, we have
$\begin{array}{l}\angle D A B=60^{\circ} \text { and } \angle A B D=50^{\circ} \\
\angle A D B=180^{\circ}-60^{\circ}-50^{\circ}=70^{\circ}\end{array}$
We find that $\angle A D B$ and $\angle A C B$ are angles made by the are $A B$ in the same segment.
$
∴\quad$ \angle A C B=\angle A D B \Rightarrow \angle A C B=70^{\circ}$

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MCQ 51 Mark

In Fig. if $\angle A O B$ is a diameter of the circle and $A C=B C$, then $\angle C A B$ is equal to

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

Answer

Correct option: D.

$45^{\circ}$

(d) $45^{\circ}$
We find that angle in a semi-circle is a right angle. Therefore, $\angle A C B=90^{\circ}$.
It is given that
$A C=B C \Rightarrow \angle C A B=\angle C B A$
Using angle sum property in $\triangle A B C$, we obtain
$\angle A C B+\angle C A B+\angle C B A=180^{\circ} \Rightarrow 90^{\circ}+2 \angle C A B=180^{\circ} \Rightarrow\angle C A B=45^{\circ}$

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MCQ 61 Mark

In Fig. if $\angle A O B=90^{\circ}$ and $\angle A B C=30^{\circ}$, then $\angle C A O$ is equal to
[Each equal to radius]

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

Answer

Correct option: D.

$60^{\circ}$

(d) $60^{\circ}$
In $\triangle A O B$, we find that
$\begin{array}{ll}& O A=O B \\
\Rightarrow \quad & \angle O A B=\angle O B A\end{array}$
Using angle sum property in $\triangle O A B$, we obtain
$\begin{array}{ll}& \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\
\Rightarrow \quad & 2 \angle O A B+90^{\circ}=180^{\circ} \Rightarrow \angle O A B=45^{\circ} \Rightarrow \angle O BA=45^{\circ}\end{array}$
Arc $A B$ subtends $\angle A O B=90^{\circ}$ at the centre $O$ and $\angle A C B$ at a point on the remaining part of the circle.
$\therefore \quad \angle A C B=\frac{1}{2} \angle A O B=\frac{1}{2} \times 90^{\circ}=45^{\circ}$
Thus, in $\triangle A B C$, we have $\angle A B C=30^{\circ}$ and $\angle A C B=45^{\circ}$.
$\begin{array}{ll}\therefore & \angle B A C=180^{\circ}-30^{\circ}-45^{\circ}=105^{\circ} \quad \text { [Using angle sum property in } \triangle A B C \text { ] } \\
\Rightarrow & \angle C A O+\angle O A B=105^{\circ} \Rightarrow \angle C A O+45^{\circ}=105^{\circ} \Rightarrow
\angle C A O=60^{\circ}\end{array}$

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MCQ 71 Mark

In Fig. if $\angle A B C=20^{\circ}$, then $\angle A O C$ is equal to

A
$20^{\circ}$
✓
$40^{\circ}$
C
$60^{\circ}$
D
$10^{\circ}$

Answer

Correct option: B.

$40^{\circ}$

(b) $40^{\circ}$
We observe that arc AC makes angle $\angle A B C=20^{\circ}$ at a point B on the circumference of the circle and $\angle A O C$ at the centre O.
$\therefore \quad \angle A O C=2 \angle A B C=2 \times 20^{\circ}=40^{\circ}$

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MCQ 81 Mark

In Fig. BC is a diameter of the circle and $\angle B A O=60^{\circ}$. Then, $\angle A D C$ is equal to

A
$30^{\circ}$
B
$45^{\circ}$
✓
$60^{\circ}$
D
$120^{\circ}$

Answer

Correct option: C.

$60^{\circ}$

(c) $60^{\circ}$
We observe that $\angle B A C$ is the angle in a semi-circle. Therefore, $\angle B A C=90^{\circ}$.
$\therefore \quad \angle O A C=\angle B A C-\angle O A B=90^{\circ}-60^{\circ}-30^{\circ}$
In $\triangle O A C$, we have
$O A=O C \Rightarrow \angle O A C=\angle O C A \Rightarrow \angle O C A=30^{\circ}$
Thus, in $\triangle O A C$, we have
$\begin{array}{ll}& \angle O A C=\angle O C A=30^{\circ} \\
\therefore \quad & \angle A O C=180^{\circ}-(\angle O A C+\angle O C A)=180^{\circ}
\left(30^{\circ}+30^{\circ}\right)=120^{\circ}\end{array}$
Clearly, arc $A C$ makes angle $\angle A O C=120^{\circ}$ at the centre $O$ and $\angle A D C$ at point on the remaining part of the circle.
$\therefore \quad \angle A D C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

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MCQ 91 Mark

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is

A
17 cm
B
15 cm
C
4 cm
✓
8 cm

Answer

Correct option: D.

8 cm

(d) 8 cm
Let O be the centre of the circle and $O L \perp A B$. Then, $L$ is the midpoint of $A B$. We have, $O A=17 cm$ and $A L=15 cm$.
Applying Pythagoras theorem in $\triangle O L A$, we obtain
$\begin{array}{ll}& O A^2=O L^2+A L^2 \\
\Rightarrow \quad & O L=\sqrt{O A^2-O L^2}=\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8 cm\end{array}$

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MCQ 101 Mark

Two equal circles of radius r intersect such that each passes through the of the other. The length of the common chord of the circles, is

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

Answer

Correct option: C.

$\sqrt{3} r$

c

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MCQ 111 Mark

The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is

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

Answer

Correct option: B.

$2 \sqrt{5} cm$

b

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MCQ 121 Mark

The greatest chord of a circle is called its

A
radius
B
secant
✓
diameter
D
none of these

Answer

Correct option: C.

diameter

c

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MCQ 131 Mark

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle, is

A
$60^{\circ}$
B
$75^{\circ}$
C
$120^{\circ}$
✓
$150^{\circ}$

Answer

Correct option: D.

$150^{\circ}$

d

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MCQ 141 Mark

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If $\angle Q P R=67^{\circ}$ and $\angle S P R=72^{\circ}$, then $\angle Q R S=$

✓
$41^{\circ}$
B
$23^{\circ}$
C
$67^{\circ}$
D
$18^{\circ}$

Answer

Correct option: A.

$41^{\circ}$

a

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MCQ 151 Mark

One chord of a circle is known to be 10 cm. The radius of this circle must be

A
5 cm
✓
greater than 5 cm
C
greater than or equal to 5 cm
D
less than 5 cm

Answer

Correct option: B.

greater than 5 cm

b

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MCQ 161 Mark

Number of circles that can be drawn through three non-collinear points, is

✓
1
B
$0$
C
2
D
3

Answer

Correct option: A.

1

a

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MCQ 171 Mark

Let C be the mid-point of an arc AB of a circle such that m$\overparen{A B}=183^{\circ}$. If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle lies

✓
in the interior of S
B
in the exterior of S
C
on the segment AB
D
on AB and bisects AB

Answer

Correct option: A.

in the interior of S

a

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MCQ 181 Mark

In Fig. sides AB and AD of quadrilateral ABCD are produced to E and F respectively. If $\angle C B E=100^{\circ}$, then $\angle C D F=$

A
$100^{\circ}$
✓
$80^{\circ}$
C
$130^{\circ}$
D
$90^{\circ}$

Answer

Correct option: B.

$80^{\circ}$

b

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MCQ 191 Mark

In Fig. O is the centre of the circle such that $\angle A O C=130^{\circ}$, then $\angle A B C=$

A
$130^{\circ}$
✓
$115^{\circ}$
C
$65^{\circ}$
D
$165^{\circ}$

Answer

Correct option: B.

$115^{\circ}$

b

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MCQ 201 Mark

In Fig. O is the centre of the circle. If $\angle B A D=75^{\circ}$ and chord BC = chord CD then $\angle B O C=$

A
$80^{\circ}$
B
$65^{\circ}$
✓
$75^{\circ}$
D
$105^{\circ}$

Answer

Correct option: C.

$75^{\circ}$

c

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MCQ 211 Mark

In Fig. O is the centre of the circle and $\angle B D C=42^{\circ}$. The measure of $\angle A C B$ is

A
$42^{\circ}$
✓
$48^{\circ}$
C
$58^{\circ}$
D
$52^{\circ}$

Answer

Correct option: B.

$48^{\circ}$

b

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MCQ 221 Mark

In Fig. if equilateral triangle ABC is inscribed in a circle and ABCD is a quadrilateral, then $\angle B D C=$

A
$90^{\circ}$
B
$60^{\circ}$
✓
$120^{\circ}$
D
$150^{\circ}$

Answer

Correct option: C.

$120^{\circ}$

c

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MCQ 231 Mark

In Fig. if chords AB and CD of the circle intersect each other at right angles, then x + y =

A
$45^{\circ}$
B
$60^{\circ}$
C
$75^{\circ}$
✓
$90^{\circ}$

Answer

Correct option: D.

$90^{\circ}$

d

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MCQ 241 Mark

In Fig. if $\angle A B C=45^{\circ}$, then $\angle A O C=$

A
$45^{\circ}$
B
$60^{\circ}$
C
$75^{\circ}$
✓
$90^{\circ}$

Answer

Correct option: D.

$90^{\circ}$

d

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MCQ 251 Mark

In Fig. chords AD and BC intersect each other at right angles at a point P. If $\angle D A B=35^{\circ}$, then $\angle A D C=$

A
$35^{\circ}$
B
$45^{\circ}$
✓
$55^{\circ}$
D
$65^{\circ}$

Answer

Correct option: C.

$55^{\circ}$

c

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MCQ 261 Mark

In Fig. AOB is a diameter of the circle. If $\angle A D C=120^{\circ}$, then $\angle B A C=$

A
$60^{\circ}$
✓
$30^{\circ}$
C
$45^{\circ}$
D
$20^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

b

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MCQ 271 Mark

In Fig. $\angle O B D=$

A
$30^{\circ}$
✓
$15^{\circ}$
C
$20^{\circ}$
D
$35^{\circ}$

Answer

Correct option: B.

$15^{\circ}$

b

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MCQ 281 Mark

In Fig. $\angle B O D=$

✓
$150^{\circ}$
B
$140^{\circ}$
C
$105^{\circ}$
D
$145^{\circ}$

Answer

Correct option: A.

$150^{\circ}$

a

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MCQ 291 Mark

In Fig. $\angle B C D=$

A
$95^{\circ}$
B
$85^{\circ}$
C
$100^{\circ}$
✓
$105^{\circ}$

Answer

Correct option: D.

$105^{\circ}$

d

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MCQ 301 Mark

In Fig. AB is a diameter of the circle. If $\angle A D C=120^{\circ}$, then $\angle C A B=$

A
$60^{\circ}$
B
$45^{\circ}$
✓
$30^{\circ}$
D
$40^{\circ}$

Answer

Correct option: C.

$30^{\circ}$

c

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MCQ 311 Mark

In Fig. ABCD is a quadrilateral inscribed in a circle. If $\angle D A C=40^{\circ}$ and $\angle B D C=45^{\circ}$, then $\angle B C D=$

A
$85^{\circ}$
✓
$95^{\circ}$
C
$75^{\circ}$
D
$105^{\circ}$

Answer

Correct option: B.

$95^{\circ}$

b

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MCQ 321 Mark

In Fig. ABCD is a cyclic quadrilateral in which AB || DC. If $\angle A=80^{\circ}$, then $\angle B=$

A
$100^{\circ}$
✓
$80^{\circ}$
C
$120^{\circ}$
D
$60^{\circ}$

Answer

Correct option: B.

$80^{\circ}$

b

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MCQ 331 Mark

In a circle with centre O, AB and CD are two diameters perpendicular each other. The length of chord AC, is

A
2AB
B
$\sqrt{2}$
C
$\frac{1}{2} A B$
✓
$\frac{1}{\sqrt{2}} A B$

Answer

Correct option: D.

$\frac{1}{\sqrt{2}} A B$

d

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MCQ 341 Mark

In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are

A
$90^{\circ}$ and $270^{\circ}$
B
$90^{\circ}$ and $90^{\circ}$
✓
$270^{\circ}$ and $90^{\circ}$
D
$60^{\circ}$ and $210^{\circ}$

Answer

Correct option: C.

$270^{\circ}$ and $90^{\circ}$

c

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MCQ 351 Mark

In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

A
34 cm
B
15 cm
C
23 cm
✓
30 cm

Answer

Correct option: D.

30 cm

d

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MCQ 361 Mark

If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a

A
rhombus
B
rectangle
C
parallelogram
✓
square

Answer

Correct option: D.

square

d

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MCQ 371 Mark

If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is

A
15 cm
B
16 cm
✓
17 cm
D
34 cm

Answer

Correct option: C.

17 cm

c

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MCQ 381 Mark

If O is the centre of a circle of radius rand AB is a chord of the circle at a distance r/2 from O, then $\angle B A O=$

A
$60^{\circ}$
B
$45^{\circ}$
✓
$30^{\circ}$
D
$15^{\circ}$

Answer

Correct option: C.

$30^{\circ}$

c

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MCQ 391 Mark

If AB is a chord of a circle, P and Q are the two points on the circle different A and B, then

A
$\angle A P B=\angle A Q B$
✓
$\angle A P B+\angle A Q B=180^{\circ}$ or $\angle A P B=\angle A Q B$
C
$\angle A P B+\angle A Q B=90^{\circ}$
D
$\angle A P B+\angle A Q B=180^{\circ}$

Answer

Correct option: B.

$\angle A P B+\angle A Q B=180^{\circ}$ or $\angle A P B=\angle A Q B$

b

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MCQ 401 Mark

If ABC is an arc of a circle and $\angle A B C=135^{\circ}$, then the ratio of arc $\widehat{A B C}$ to the circumference, is

✓
1:4
B
3:4
C
3:8
D
1:2

Answer

Correct option: A.

1:4

a

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MCQ 411 Mark

If A, B, C are three points on a circle with centre O such that $\angle A O B=90^{\circ}$ and $\angle B O C=120^{\circ}$, then $\angle A B C=$

A
$60^{\circ}$
✓
$75^{\circ}$
C
$90^{\circ}$
D
$135^{\circ}$

Answer

Correct option: B.

$75^{\circ}$

b

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MCQ 421 Mark

If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than $\angle A O B=$

✓
$60^{\circ}$
B
$90^{\circ}$
C
$120^{\circ}$
D
none of these

Answer

Correct option: A.

$60^{\circ}$

a

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MCQ 431 Mark

If A and B are two points on a circle such that $m(\overparen{A B})=260^{\circ}$. A possible value for the angle subtended by arc BA at a point on the circle is

A
$100^{\circ}$
B
$75^{\circ}$
✓
$50^{\circ}$
D
$25^{\circ}$

Answer

Correct option: C.

$50^{\circ}$

c

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MCQ 441 Mark

In Fig. O is the centre of the circle. If $\angle A O C=150^{\circ}$, then $\angle A D C=$

A
$70^{\circ}$
B
$75^{\circ}$
C
$60^{\circ}$
✓
$105^{\circ}$

Answer

Correct option: D.

$105^{\circ}$

(d) $105^{\circ}$
Clearly, arc AC subtends $\angle A O C=105^{\circ}$ at the centre O and $\angle A B C$ at point B on the circumference. Therefore,
$\angle A B C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 150^{\circ}=75^{\circ}$
Since $A B C D$ is a cyclic quadrilateral. Therefore,
$\angle A D C+\angle A B C=180^{\circ} \Rightarrow \angle A D C+75^{\circ}=180^{\circ} \Rightarrow \angle A DC=105^{\circ}$

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MCQ 451 Mark

In Fig. O is the centre of the circle and diameter AB bisects the chord CD of length 16 cm at point E such that $E B=4 cm$. The radius of the circle is

✓
10 cm
B
12 cm
C
6 cm
D
8 cm

Answer

Correct option: A.

10 cm

(a) 10 cm
In right triangle OEC, we obtain
$\begin{array}{ll}& O C^2=O E^2+E C^2 \\
\Rightarrow & O C^2=(O B-E B)^2+8^2 \\
\Rightarrow & O C^2=(O C-4)^2+64 \\
\Rightarrow & 0=-8 O C+16+64 \Rightarrow O C=10 cm\end{array}$

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MCQ 461 Mark

In Fig. if O is the centre of the circle, then $\angle B C D=$

A
$110^{\circ}$
B
$100^{\circ}$
C
$140^{\circ}$
✓
$130^{\circ}$

Answer

Correct option: D.

$130^{\circ}$

(d) $130^{\circ}$
In $\triangle A E D$, we find that
$\angle A E D=90^{\circ} \text { (angle in a semi-circle) and } \angle E A D=60^{\circ} \text {. Therefore, } \angle A DE=30^{\circ} \text {. }$
$A E D C$ is a cyclic quadrilateral such that
$\begin{array}{ll}& \angle E D C=\angle E D A+\angle A D C=30^{\circ}+70^{\circ}=100^{\circ} \\
\therefore \quad & \angle C A E=80^{\circ} \Rightarrow \angle D A C=20^{\circ}\end{array}$
Thus, $A B C D$ is a cyclic quadrilateral, such that $\angle D A B=20^{\circ}+30^{\circ}=50^{\circ}$
$\therefore \quad \angle B C D=180^{\circ}-50^{\circ}=130^{\circ}$

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MCQ 471 Mark

In Fig. if O is the centre of the circle, then $\angle A B C=$

A
$70^{\circ}$
✓
$110^{\circ}$
C
$100^{\circ}$
D
$130^{\circ}$

Answer

Correct option: B.

$110^{\circ}$

(b) $110^{\circ}$
ABCD is a cyclic quadrilateral such that $\angle A D C=70^{\circ}$.
$\therefore \quad \angle A B C+\angle A D C=180^{\circ} \Rightarrow \angle A B C+70^{\circ}=180^{\circ} \Rightarrow\angle A B C=110^{\circ}$

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MCQ 481 Mark

In Fig. if O is the centre of the circle such that $\angle A O C=100^{\circ}$ and side AB of quadrilateral OABC has been produced to D, then $\angle C B D=$

A
$25^{\circ}$
B
$80^{\circ}$
✓
$50^{\circ}$
D
$40^{\circ}$

Answer

Correct option: C.

$50^{\circ}$

(c) $50^{\circ}$
Let $E$ be a point on the circle. We observe that arc ABC makes $\angle A O C=100^{\circ}$ at O and $\angle A E C$ at E.
$\therefore \quad \angle A E C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 100^{\circ}=50^{\circ}$
Using exterior angle property in cyclic quadrilateral $A B C E$, we obtain
$\text { ext } \angle C B D=\angle A E C=50^{\circ}$

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MCQ 491 Mark

In Fig. if $O A=5 cm, A B=8 cm$ and OD is perpendicular to AB, then CD is equal to

✓
2 cm
B
3 cm
C
4 cm
D
5 cm

Answer

Correct option: A.

2 cm

(a) 2 cm
We have, $A B=8 cm$. Therefore, $A C=B C=4 cm$
Applying Pythagoras theorem in $\triangle O C A$, we obtain
$\begin{aligned}& O A^2=O C^2+A C^2 \Rightarrow O C=\sqrt{O A^2-A C^2}=\sqrt{5^2-4^2}=3 cm \\
\therefore & \quad C D=O D-O C=O A-O C=(5-3)=2 cm\end{aligned}$

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MCQ 501 Mark

In Fig. if DC is produced to E and $C F \| A B$, then $\angle B A D=$

A
$95^{\circ}$
B
$85^{\circ}$
✓
$105^{\circ}$
D
$75^{\circ}$

Answer

Correct option: C.

$105^{\circ}$

(c) $105^{\circ}$
Given that $C F \| A B$ and $\angle A B C=85^{\circ}$.
$\therefore \angle B C F=85^{\circ} \Rightarrow \angle B C E=105^{\circ} \Rightarrow \angle B A D=105^{\circ} \quad$ [Exterior angle $=$ Interior opposite angle]

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M.C.Q - Maths STD 9 Questions - Vidyadip