Mathematics Class IX

Question 1 of 35

1. Question
1 point(s)
The center of the circle lies in______ of the circle.
- Interior
- Exterior
- Circumference
- None of the above
Correct

Incorrect
Question 2 of 35

2. Question
1 point(s)
The longest chord of the circle is:
- Radius
- Arc
- Diameter
- Segment
Correct

Incorrect
Question 3 of 35

3. Question
1 point(s)
Equal _____ of the congruent circles subtend equal angles at the centers.
- Segments
- Radii
- Arcs
- Chords
Correct

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Hint

Let $\triangle AOB$ and $\triangle COD$ are two triangles inside the circle.
$OA=OC$ and $OB=OD$ (radii of the circle)
$AB=CD$ (Given)
So, $\triangle AOB\cong\triangle COD$ (SSS congruency)
$\therefore$ By CPCT rule, $\angle AOB=\angle COD$ .
Hence, this prove the statement.
Question 4 of 35

4. Question
1 point(s)
If chords AB and CD of congruent circles subtend equal angles at their centres, then:
- AB = CD
- AB > CD
- AB < AD
- None of the above
Correct

Incorrect

Hint

In triangles $AOB$ and $COD$ ,
$\angle AOB=\angle COD$ (given)
$OA=OC$ and $OB=OD$ (radii of the circle)
So, $\triangle AOB\cong\triangle COD$ . (SAS congruency)
$\therefore AB=CD$ (By CPCT)
Question 5 of 35

5. Question
1 point(s)
If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
- 0
- 1
- 2
- 3
Correct

Incorrect
Question 6 of 35

6. Question
1 point(s)
The angle subtended by the diameter of a semi-circle is:
- 90
- 45
- 180
- 60
Correct

Incorrect

Hint

The semicircle is half of the circle, hence the diameter of the semicircle will be a straight line subtending 180 degrees.
Question 7 of 35

7. Question
1 point(s)
If AB and CD are two chords of a circle intersecting at point E, as per the given figure. Then:
- $\angle BEQ>\angle CEQ $
- $\angle BEQ=\angle CEQ $
- \angle BEQ<\angle CEQ [/latex]
- None of the above
Correct

Incorrect

Hint

$OM=ON$ (Equal chords are always equidistant from the centre)
$OE=OE$ (Common)
$\angle OME=\angle ONE$ (perpendiculars)
So, $\triangle OEM\cong\triangle OEN$ (by RHS similarity criterion)
Hence, $\angle MEO=\angle NEO$ (by CPCT rule)
$\therefore\angle BEQ=\angle CEQ$
Question 8 of 35

8. Question
1 point(s)
If a line intersects two concentric circles with centre O at A, B, C and D, then:
- AB = CD
- AB > CD
- AB < CD
- None of the above
Correct

Incorrect

Hint

From the above fig., $OM\bot AD$ .
Therefore, $AM=MD$ ….(1)
Also, since $OM\bot BC,\ OM$ bisects $BC$ .
Therefore, $BM=MC$ ….(2)
From equation 1 and equation 2.
$AM - BM = MD - MC\\ \\ \therefore AB=CD$
Question 9 of 35

9. Question
1 point(s)
In the below figure, the value of $\angle ADC $ is:
- $ 60^o$
- $ 30^o$
- $ 45^o$
- $ 55^o$
Correct

Incorrect

Hint

$\angle AOC=\angle AOB+\angle BOC$
So, $\angle AOC=60^o+30^o$
$\therefore\angle AOC=90^o$
An angle subtended by an arc at the centre of the circle is twice the angle subtended by that arc at any point on the rest part of the circle.
So, $\angle ADC=\dfrac{1}{2}\angle AOC=\dfrac{1}{2}\times90^o=45^o$
Question 10 of 35

10. Question
1 point(s)
In the given figure, find angle OPR.
- $ 20^o$
- $ 15^o$
- $ 12^o$
- $ 10^o$
Correct

Incorrect

Hint

The angle subtended by major arc PR at the centre of the circle is twice the angle subtended by that arc at point, Q, on the circle.
So, $\angle POR=2\times\angle PQR$ , here $\angle POR$ is the exterior angle
We know the values of angle $PQR$ as $100^o$
So, $\angle POR=2\times100^o=200^o$
$\therefore\angle ROP=360^o-200^o=160^o$
Now, in $\triangle OPR$ ,
OP and OR are the radii of the circle
So, OP = OR
Also, $\angle OPR=\angle ORP$
By angle sum property of triangle, we know:
$\angle ROP+\angle OPR+\angle ORP=180^o\\ \\ \angle OPR+\angle OPR=180^o-160^o$
As, $\angle OPR=\angle ORP\\ \\ 2\angle OPR=20^o$
Thus, $\angle OPR=10^o$
Question 11 of 35

11. Question
1 point(s)
ABCD is a cyclic quadrilateral in which $\angle DBC=55^o $ and $\angle BAC=55^o $. Find $\angle BCA $.
- $60^o $
- $80^o $
- $100^o $
- $120^o $
Correct

Incorrect

Hint

$\angle BCA=180^o-(55^o+55^o)=80^o$
Question 12 of 35

12. Question
1 point(s)
The radius of a circle is 2.5 cm. AB and CD are two parallel chords 2.7 cm apart. If AB = 4.8 cm. Find CD.
- 4.8 cm
- 2.4 cm
- 3 cm
- 4 cm
Correct

Incorrect

Hint

$LM=2.7\ cm\\ \\ OL=x\ cm\\ \\ OM=(2.7-x)\ cm\\ \\ OA^2=OL^2+LA^2\\ \\ (2.5)^2=x^2+(2.4)^2\\ \\ 6.25-5.76=x^2\\ \\ x^2=0.49\Rightarrow x=0.7\ cm\\ \\ OM=2.7-0.7=2\ cm\\ \\ \text{In}\ \triangle OMC\\ \\ OC^2=OM^2+CM^2\\ \\ CM^2=6.25\\ \\ CM^2=2.25\ cm\\ \\ CM=1.5\ cm\\ \\ CD=2\times CM=2\times1.5=3\ cm$
Question 13 of 35

13. Question
1 point(s)
If $ \angle B=125^o$ and $ E$ is a point on the circle. Find $\angle AEC $
- $55^o $
- $125^o $
- $130^o $
- $62.5^o $
Correct

Incorrect
Question 14 of 35

14. Question
1 point(s)
$\triangle ABC$ is an isosceles triangle with $AB = AC$ and $\angle ABC=50^o $. Find the sum of $\angle BDC $ and $\angle BEC $
- $30^o $
- $60^o $
- $50^o $
- $180^o $
Correct

Incorrect

Hint

$AB=AC\\ \\ \Rightarrow\angle BCA=\angle ABC=50^o\\ \\ \Rightarrow\angle BAC=180^o-(50+50)=80^o$
$\angle BDA=\angle BAC=80^o$ (angles in the same segment are equal)
$\angle BEC=180^o-\angle BDC$ (Opposite angles of a cyclic quadrilateral are supplementary)
$\angle BDC+\angle BEC=80^o+100^o=180^o$
Question 15 of 35

15. Question
1 point(s)
O is the centre of the circle. If $\angle POQ=110^o $ and $\angle POR=120^o $. Find $ \angle QPR$
- $65^o $
- $55^o $
- $50^o $
- $60^o $
Correct

Incorrect

Hint

$\angle QDR=360^o-(110^o+120^o)=130^o\\ \\ \angle QPR=\dfrac{1}{2}\times\angle QOR=\dfrac{1}{2}\times130^o=65^o$
Question 16 of 35

16. Question
1 point(s)
Find the value of y
- $135^o $
- $45^o $
- $130^o $
- $115^o $
Correct

Incorrect

Hint

$\angle ADC=\dfrac{1}{2}\times AOC=45^o\\ \\ Y+\angle ADC=180^o\\ \\ Y=180-45=135^o$
Question 17 of 35

17. Question
1 point(s)
The length of side QR is
- 11 cm
- 10 cm
- 12 cm
- 9 cm
Correct

Incorrect

Hint

Length of tangents from a outside point to the circle are equal.
Question 18 of 35

18. Question
1 point(s)
$\triangle ABC $ is inscribed in a circle. The bisector of $\angle BAC $ meets BC at D and the circle at E. If EC is joined and $\angle ECD=30^o $. Find $\angle BAC $.
- $30^o $
- $40^o $
- $50^o $
- $60^o $
Correct

Incorrect

Hint

$\angle BCE=\angle BAE=30^o$ (angles in the same segment are equal)
$\angle BAE=\angle CAE=30^o$ (AD is a bisector)
$\angle BAC=30^o+30^o=60^o$
Question 19 of 35

19. Question
1 point(s)
48 cm long chord of a circle is at a distance of 7 cm from the centre. Find the radius of the circle.
- 5 cm
- 17 cm
- 25 cm
- 16 cm
Correct

Incorrect

Hint

$AB=\sqrt{(24)^2+7^2}\\ \\ =\sqrt{57+49}=\sqrt{625}=25\ cm$
Question 20 of 35

20. Question
1 point(s)
Two circles with radii 6 and 12 are drawn with the same centre. The smaller inner circle is painted Pink and the part outside the smaller circle and inside the larger circle is painted green. What is the ratio of the areas painted green to the area painted pink?
- $ 3:1 $
- $3:2 $
- $ 2:1$
- $ 4:1$
Correct

Incorrect

Hint

$\text{Ratio}\dfrac{\pi\times(R-r)^2}{\pi r^2}\\ \\ =\dfrac{144-36}{36}\\ \\ =\dfrac{108}{36}=3:1$
Question 21 of 35

21. Question
1 point(s)
If $\angle ADB=80^o $ and $\angle APB=125^o $. Find angle $\angle DAC $.
- $40^o $
- $55^o $
- $50^o $
- $45^o $
Correct

Incorrect

Hint

$\angle ADB=\angle ACB\\ \\ \angle APC+\angle ACP+\angle PAC=180^o$
$\angle APC=180^o-125$ (Linear pair)
$\angle DAC=180^o-(80^o+55^o=45^o)$
Question 22 of 35

22. Question
1 point(s)
O is the centre of a circle. If $\angle DAC=54^o $ and $\angle ACB=63^o $. Find $ \angle DAB$.
- $72^o $
- $54^o $
- $81^o $
- $27^o $
Correct

Incorrect

Hint

In $\triangle ABC$
$\angle ABC=90^o$ (Angle in a semicircle is $90^o$ )
$\angle BAC=180^o-(90^o+63^o)=27^o\\ \\ \angle DAB=54^o+27^o=81^o$
Question 23 of 35

23. Question
1 point(s)
If $\angle BAD=100^o $ and $\angle ACB=35^o $ Find angle $\angle ABD $
- $50^o $
- $45^o $
- $40^o $
- $55^o $
Correct

Incorrect

Hint

$\angle ADB=\angle ACB$ (angles inscribed by the same chord AB)
$\angle ADB=35^o$
In $\triangle ABD$
$\angle ABD=180^o-(100^o+35^o)=45^o$
Question 24 of 35

24. Question
1 point(s)
ABCD is a parallelogram. The circle through A, B, C intersect CD at E. Find the value of $\dfrac{AE}{AD} $
- $\dfrac{1}{\sqrt2} $
- $\sqrt2 $
- $\dfrac{1}{\sqrt3} $
- $1 $
Correct

Incorrect

Hint

BC = AD ( $\because$ ABCD is a parallelogram)
AE = BC (through construction)
So, AE = AD
$\mathrm{\dfrac{AE}{AD}}=1:1$
Question 25 of 35

25. Question
1 point(s)
A triangle ABC is inscribed in a circle, the bisector of whose angles meet the circumference at x, y, z. Determine the angles x, y, z
- $90^o-\dfrac{A}{2},90^o-\dfrac{B}{2},90^o-\dfrac{C}{2} $
- $90^o,60^o,30^o $
- $\dfrac{A}{2},\dfrac{B}{2},\dfrac{C}{2} $
- $45^o,60^o,30^o $
Correct

Incorrect

Hint

$\angle BYX=\angle BAX=\dfrac{\angle A}{2}\\ \\ \angle ZYB=\angle ZCB=\dfrac{\angle C}{2}$
( $\because$ angles in the same segment)
$\angle ZYX=\angle ZYB+\angle BYX\\ \\ \Rightarrow\angle ZYX=\dfrac{\angle C}{2}+\dfrac{\angle A}{2}=\dfrac{\angle A+\angle C}{2}\\ \\ =\dfrac{180^o-\angle B}{2}=90^o-\dfrac{\angle B}{2}$
Question 26 of 35

26. Question
1 point(s)
AB is a chord of a circle with centre o and radius 17 cm. If OM $\bot $ AB and OM = 8 cm. Find the length of chord AB.
- 12 cm
- 30 cm
- 15 cm
- 24 cm
Correct

Incorrect

Hint

$MB=\sqrt{17^2-8^2}\\ \\ =\sqrt{289-64}\\ \\ =\sqrt{225}\\ \\ =15\ cm\\ \\ AB=2\times15=30\ cm$
Question 27 of 35

27. Question
1 point(s)
The length of tangent from point p to a circle of radius 5 cm is 12 cm. The distance p from the centre is
- 17 cm
- 14 cm
- 13 cm
- 7 cm
Correct

Incorrect

Hint

$PO=\sqrt{PA^2+OA^2}=\sqrt{5^2+12^2}\\ \\ =\sqrt{25+144}=13\ cm$
Question 28 of 35

28. Question
1 point(s)
If $\angle ADC=125^o $ and $BC=BE $. Find $\angle CBE $
- $120^o $
- $100^o $
- $110^o $
- $115^o $
Correct

Incorrect

Hint

$\angle ADC+\angle CBA=180^o\\ \\ \Rightarrow\angle CBA=180^o-125^o=55^o\\ \\ \angle CBE=2\times55^o=110^o$
Question 29 of 35

29. Question
1 point(s)
Two circles of radius 13 cm and 12 cm intersect at two points and the distance between their centers is 5 cm. Find the length of the common chord.
- 27 cm
- 25 cm
- 23 cm
- 24 cm
Correct

Incorrect

Hint

$AC=AD$ (radius)
$BC=BD$ (radius)
$AB=BA$ (common)
$\therefore\triangle ABC\cong\triangle ABD$ (SSS)
$\angle AOC=\angle AOD$ (cpct)
$AO=OB$ (cpct)
$CO=OD$ (cpct)
$\angle AOC+\angle AOD=180^o$ (Linear pair)
$\angle AOC=90^o$
$CO=\sqrt{13^2-5^2}=\sqrt{169-25}=12\ cm\\ \\ CD=2\times12=24\ cm$
Question 30 of 35

30. Question
1 point(s)
AC is a diameter of the circle and arc AXB = $ \dfrac{1}{3}$ar BYC. Find $\angle BOC $
- $145^o $
- $140^o $
- $125^o $
- $135^o $
Correct

Incorrect

Hint

The ratio of arc length to the circumference $=360^o$
arc $A\times B\dfrac{1}{3}\ ar\ BYC$
$\Rightarrow\angle BOA=\dfrac{1}{3}\times BOC\\ \\ \angle BOA+\angle BOC=180^o\\ \\ \dfrac{1}{3}\angle BOC+\angle BOC=180^o\\ \\ \angle BOC=45^o$
Question 31 of 35

31. Question
1 point(s)
Two circles with centre O and O’ intersect at two points A and B. A line PQ is drawn. Parallel to OO’ through A or B intersecting the circles at P and Q. Find the ratio of OO’ : PQ
- $2:3$
- $ 1:1 $
- $1:2 $
- $2:1 $
Correct

Incorrect

Hint

PA is a chord and OR is a line perpendicular to it.
A perpendicular drawn from the centre of a circle bisects the chord.
PR = RA
$\Rightarrow$ PA = 2RA
Similarly AQ = 2 AS
$\mathrm{OO^1:PQ=1:2}$
Question 32 of 35

32. Question
1 point(s)
A chord of a circle is 12 cm in length and its distance from the centre is 8 cm. Find the length of the chord of the same circle which is at a distance of 6 cm form the centre.
- 30 cm
- 24 cm
- 16 cm
- 18 cm
Correct

Incorrect

Hint

$AB=12\ cm\\ \\ \Rightarrow AM=MB=6\ cm$
In $\triangle OMA$
$OA^2=OM^2+AM^2\\ \\ \Rightarrow OA=\sqrt{100}=10\ cm$
Radius $OA=OC_2=10\ cm$
In $\triangle ONC,OC^2=ON^2+NC^2\\ \\ \Rightarrow NC=\sqrt{64}=8\ cm$
Length of the chord $=2\times8=16\ cm$
Question 33 of 35

33. Question
1 point(s)
Two chords AB and AC of a circle subtend angles equal to $80^o $ and $70^o $ respectively at the centre. Find $\angle BAC $, if AB and AC lie on the opposite sides of the centre.
- $95^o $
- $105^o $
- $110^o $
- $100^o $
Correct

Incorrect

Hint

$\angle AOB=80^o\\ \\ \angle AOC=70^o\\ \\ \angle BAO=ABO\\ \\ \angle BAO+\angle ABO+\angle AOB=180^o\\ \\ \angle BAO=\dfrac{1}{2}\ (180-80)=50^o\\ \\ \angle CAO=\dfrac{1}{2}(180-80)=\dfrac{1}{2}\times110=55^o\\ \\ \angle BAC=50+55=105^o$
Question 34 of 35

34. Question
1 point(s)
If BC is a diameter and $\angle BAO=45^o $ Then the value of $\angle ADC $ is
- $50^o $
- $55^o $
- $45^o $
- $35^o $
Correct

Incorrect

Hint

$OA = OB$ (radius)
$\triangle AOB$ is an isosceles triangle
$\angle ABO=\angle BAO=45^o$
$\angle BAO=\angle ADC=45^o$ (alternate angle)
Question 35 of 35

35. Question
1 point(s)
The chords AB and CD of a circle are perpendicular to each other. If radius of the circle is 14 cm and length of the arc AQD is 24 cm. Find the length of arc BPC
- 22 cm
- 19 cm
- 21 cm
- 20 cm
Correct

Incorrect

Hint

arc RC = arc RD
arc RS = arc RC+ arc CS
$=180^o$
arc RD + arc CS $=180^o$
arc RC+ arc SD $=180^o$
circumference = Length of arc BPC + length of arc AQD $0.5\times88=44\$ cm
Length of arc BPC $= 44-24=20$ cm