Mathematics Class X

Question 1 of 40

1. Question
1 point(s)
In the given figure, O is the centre of a circle. AB is a chord and AT is the tangent at A. If $\angle AOB=100^o $, calculate $ \angle BAT$
- $50^o $
- $40^o $
- $90^o $
- $100^o $
Correct

Incorrect

Hint

$\angle 1=\angle 2\\ \\ (\because OA=OB)\\ \\ \angle1+\angle2+100^o=180^o\\ \\ 2\angle1=80^o\\ \\ \angle 1=90^o\\ \\ \angle 1+\angle BAT=90^o\\ \\ \angle BAT=50^o$
Question 2 of 40

2. Question
1 point(s)
A chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord is
- $10\ cm $
- $10\sqrt2\ cm $
- $15\ cm $
- $5\ cm $
Correct

Incorrect

Hint

$AB^2=10^2+10^2=200\\ \\ AB=\sqrt{200}=10\sqrt2\ cm$
Question 3 of 40

3. Question
1 point(s)
In the given figure, PQ is a tangent at point C. If AB is a diameter and $\angle CAB=30^o $. The value of $\angle PCA $ is
- $60^o $
- $50^o $
- $35^o $
- $55^o $
Correct

Incorrect

Hint

$\angle ACB =90^o$ (angle in the semicircle)
In $\triangle ABC$
$\angle CAB+\angle ACB+\angle ABC=180^o\\ \\ 30^o+90+\angle ABC=180^o\\ \\ \angle ABC=60^o\\ \\ \angle PCA=\angle CBA=60^o$ (Angle in the alternate segment)
Question 4 of 40

4. Question
1 point(s)
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm, BC = 4 cm, the value of AP is
- 7.5 cm
- 15 cm
- 6 cm
- 4 cm
Correct

Incorrect

Hint

2 AP = Perimeter of triangle
$=5+6+4=15\ cm$
AP $=\dfrac{15}{2}=7.5\ cm$
Question 5 of 40

5. Question
1 point(s)
$PQ $ and $PR $ are two tangents to a circle with centre $O $. If $\angle QPR=46^o $ then the value of $\angle QOR $ is
- $100^o $
- $134^o $
- $135^o $
- $165^o $
Correct

Incorrect

Hint

$\angle OQP=90^o\\ \\ \angle ORP=90^o\\ \\ \angle OQP+\angle QPR+\angle QRP+\angle QOR=360^o\\ \\ 90+46+90+\angle QOR=360^o\\ \\ \angle QOR=360^o-(90+90+46)=134$
Question 6 of 40

6. Question
1 point(s)
The sides AB, BC, CA of a triangle ABC touch a circle at P,Q,R respectively. If PA = 4 cm, BP = 3 cm, AC = 11 cm. The length of BC is
- 10 cm
- 15 cm
- 12 cm
- 5 cm
Correct

Incorrect

Hint

$BC=BQ+QC\\ \\ AP=AR=4\ cm\\ \\ BQ=BP=3\ cm\\ \\ RC=QC=7\ cm\\ \\ BC=3+7=10\ cm$
Question 7 of 40

7. Question
1 point(s)
The perimeter of a square circumscribing a circle of radius $a $ cm is
- $4a $
- $2a $
- $8a $
- $5a $
Correct

Incorrect

Hint

$AB=a+a=2a$
Perimeter $=4\times AB$
$=4\times2a=8a\ cm$
Question 8 of 40

8. Question
1 point(s)
$AB $ is the diameter of a circle with centre $O $ and $AT $ is a tangent. If $\angle AOQ=58^o $, the value of $\angle ATQ $ is
- $65^o $
- $61^o $
- $58^o $
- $55^o $
Correct

Incorrect

Hint

$\angle ABQ=\dfrac{1}{2}\angle AOQ=\dfrac{58}{2}=29$
$\angle BAT=90^o$ (Tangent is perpendicular to the radius)
$\angle ATQ=180^o-(\angle ABQ+\angle BAT)\\ \\ =180-(29+90)=61^o$
Question 9 of 40

9. Question
1 point(s)
Two concentric circles are of radius $7\ cm$ and $r\ cm $ respectively where $r > 7$. A chord of the larger circle of length $48\ cm $ touches the smaller circle. The value of $r $ is
- 24 cm
- 25 cm
- 26 cm
- 30 cm
Correct

Incorrect

Hint

$\angle OCA=90^o$ ( $\because$ Tangent is perpendicular to the radius)
$AC=\dfrac{AB}{2}=\dfrac{48}{2}=24\ cm$
In $\triangle OCA$
$OA^2=OC^2+AC^2=(7)^2+(24)^2\\ \\ r^2=49+576=625\\ \\ r=25\ cm$
Question 10 of 40

10. Question
1 point(s)
A right triangle ABC circumscribes a circle of radius $r $. If AB and AC are of length 8 cm and 6 cm respectively. The value of $r $ is
- 2 cm
- 1 cm
- 3 cm
- 1.5 cm
Correct

Incorrect

Hint

Area of $\triangle ABC$
$=\dfrac{1}{2}\times6\times8=24\ cm^2$
Also
Area of $\triangle ABC=ar(\triangle AOB)+ar(\triangle BOC)+ar(\triangle AOC)$
$=\dfrac{1}{2}\times r\times AB+\dfrac{1}{2}\times r\times BC+\dfrac{1}{2}\times r\times AC\\ \\ =\dfrac{1}{2}\times r[AB+BC+AC]\\ \\ =\dfrac{1}{2}\times r\times24=12r\ cm^2\\ \\ 12r=24\Rightarrow r=2\ cm$
Question 11 of 40

11. Question
1 point(s)
From a point Q, the length of the tangent to a circle is 24 cm and distance of Q from the centre is 25 cm. The radius of the circle is
- 10 cm
- 8 cm
- 7 cm
- 9 cm
Correct

Incorrect

Hint

$OP=\sqrt{25^2-24^2}\\ \\ =\sqrt{625-576}=\sqrt{49}=7\ cm$
Question 12 of 40

12. Question
1 point(s)
Two concentric circles are of radius $5\ cm$ and $3\ cm $. Find the length of the chord of the larger circle which touches the smaller circle
- 8 cm
- 4 cm
- 6 cm
- 3 cm
Correct

Incorrect

Hint

$OA^2=AP^2+OP^2\\ \\ \Rightarrow 5^2=AP^2+3^2\\ \\ \Rightarrow AP^2=16\Rightarrow AP=4\\ \\ AB=2\times AP=8\ cm$
Question 13 of 40

13. Question
1 point(s)
From an external point P, tangents PA and PB are drawn to a circle with centre O. If $\angle PAB=50^o $ then the value of $\angle AOB $ is
- $80^o $
- $100^o $
- $50^o $
- $40^o $
Correct

Incorrect

Hint

$\angle PAB+\angle OAB=90^o$ ( $\because$ angle between radius and tangent is $90^o$ )
$\angle OAB=40^o\\ \\ PA=PB\\ \\ \Rightarrow \angle PBA=\angle PAB\\ \\ \Rightarrow \angle PBA=50^o\\ \\ \angle PBA+\angle OBA=90^o\\ \\ \angle OBA=90-50=40$
In $\triangle AOB$
$\angle AOB+40+40=180\\ \\ \Rightarrow \angle AOB=100^o$
Question 14 of 40

14. Question
1 point(s)
$PQ $ is a tangent at a point $C $ to a circle with centre $ O$. If $AB $ is a diameter and $\angle CAB=30^o $, the value of $ \angle PCA$ is
- $60^o $
- $ 50^o$
- $40^o $
- $20^o $
Correct

Incorrect

Hint

In $\triangle AOC$
$OA=OC$ (radius)
$\Rightarrow\angle CAO=\angle OCA$ ( $\because$ Angle opposite to equal sides are equal)
$\angle CAB=\angle OCA\\ \\ \angle CAB=\angle OCA=30^o$
$\angle PCO=90^o$ (Tangent is perpendicular to radius at point of contact)
$\angle OCA+\angle PCA=90^o\\ \\ \angle PCA=90-30=60^o$
Question 15 of 40

15. Question
1 point(s)
From an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If PO = 2r. The value of $\angle $OTS is
- $45^o $
- $30^o $
- $15^o $
- $90^o $
Correct

Incorrect

Hint

Let $\angle TOP=\theta$
In $\triangle OTP$ ,
$\cos\theta=\dfrac{OT}{OP}=\dfrac{r}{2r}=\dfrac{1}{2}\\ \\ \Rightarrow \theta=60^o\\ \\ \angle TOS=120^o$
In $\triangle OTS$
$OT=OS$ (radius)
$\angle OTS=\angle OST$
In $\triangle OTS$
$\angle OTS+\angle OST+\angle TOS=180^o\\ \\ 2\angle OST=60^o\\ \\ \angle OST=\angle OTS=30^o$
Question 16 of 40

16. Question
1 point(s)
Two equal circles with centre O and O’ touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O at the point C. O’D is perpendicular to AC. The value of $\mathrm{\dfrac{DO’}{CO}} $ is
- $1:1 $
- $1:2 $
- $2:1 $
- $1:3 $
Correct

Incorrect

Hint

$\angle ADO'=\angle ACO$ (each $90^o$ )
$\angle OAO=\angle CAO$ (Common)
$\triangle ADO'\sim\triangle ACO$ (AA centre)
$\dfrac{AO'}{AO}=\dfrac{DO'}{CO}\\ \\ AO=AO'+O'X+XO\\ \\ =r+r+r=3r\\ \\ \dfrac{DO'}{CO}=\dfrac{r}{3r}=\dfrac{1}{3}$
Question 17 of 40

17. Question
1 point(s)
$PA $ and $PB $ are tangents to the circle with centre O such that $APB=50^o $. The value of $\angle OAB $ is
- $30^o $
- $20^o $
- $25^o $
- $50^o $
Correct

Incorrect

Hint

$\angle OAP=\angle OBP=90^o$ ( $\because$ tangent is perpendicular to radius)
$\angle OAP+\angle APB+\angle OBP+AOB=360^o\\ \\ \angle APB+\angle AOB=180^o\\ \\ 50^o+\angle AOB=180^o\\ \\ \angle AOB=130^o$
In $\triangle AOB$
$OA=OB$ (radius)
$\Rightarrow \angle A=\angle B\\ \\ \Rightarrow\angle A+\angle B+\angle AOB=180^o\\ \\ \Rightarrow \angle A+\angle B=50^o\\ \\ \Rightarrow 2\angle A=50\\ \\ \Rightarrow \angle A=25^o$
Question 18 of 40

18. Question
1 point(s)
Two concentric circle of radius $a $ and $b(b>a) $ are given. The length of the chord of the larger circle which touches the smaller circle is
- $a^2+b^2 $
- $2\sqrt{a^2-b^2} $
- $a^2+b^2-2 $
- $b^2-a^2 $
Correct

Incorrect

Hint

$OC\bot AB\\ \\ \angle OCB=90^o\\ \\ AC=BC\\ \\ \Rightarrow AB=2AC$
In $\triangle OCA$ ,
$AO^2=OC^2+AC^2\\ \\ a^2=b^2+AC^2\\ \\ AC=\sqrt{a^2-b^2}\\ \\ AB=2AC=2\sqrt{a^2-b^2}$
Question 19 of 40

19. Question
1 point(s)
O is the centre of a circle PT and PQ are tangent to the circle from an external point P. If $\angle TPQ=70^o $. Find $\angle TRQ $
- $55^o $
- $60^o $
- $45^o $
- $90^o $
Correct

Incorrect

Hint

$\angle OTP=\angle OQP=90^o$ (Tangent is perpendicular to radius)
$\angle QOT+\angle OTP+\angle TPQ+\angle OQP=360^o\\ \\ \angle TOQ+\angle TPQ=180^o\\ \\ \angle TOQ=110^o$
$\angle TOQ=2\angle TRQ$ (Central angle is twice of inscribed angle)
$\angle TRQ=\dfrac{110}{2}-55$
Question 20 of 40

20. Question
1 point(s)
Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radius 8 cm and 5 cm respectively. If AP = 15 cm the length of BP is
- $17\ cm $
- $19\ cm $
- $65\ cm $
- $2\sqrt{66}\ cm $
Correct

Incorrect

Hint

In $\triangle PAO$
$PA^2+OA^2=OP^2\\ \\ 15^2+8^2=OP^2\\ \\ OP^2=289\Rightarrow OP=17\ cm$
In $\triangle PBO$
$PB^2+5^2=17^2\\ \\ PB^2=289-25=264\Rightarrow PB=\sqrt{264}=2\sqrt{66}\ cm$
Question 21 of 40

21. Question
1 point(s)
CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm, BR = 4 cm. The length of BC is
- 11 cm
- 4 cm
- 7 cm
- 16 cm
Correct

Incorrect

Hint

$CP=CQ$ (Tangents drawn from an external point are equal)
$CP=CQ=11\ cm$
$BR=BQ$ (Tangent drawn from an external point are equal)
$BR=BQ=4\ cm\\ \\ BC=CQ-BQ\\ \\ =11-4=7\ cm$
Question 22 of 40

22. Question
1 point(s)
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 13 cm. The length of PQ is
- 12 cm
- 13 cm
- 11 cm
- 5 cm
Correct

Incorrect

Hint

$OQ^2=OP^2+PQ^2\\ \\ 169=25+PQ^2\\ \\ PQ=\sqrt{169-25}=\sqrt{144}=12\ cm$
Question 23 of 40

23. Question
1 point(s)
ABCD is a cyclic quadrilateral. If $\angle BAC=50^o $ and $\angle DBC=60^o $. The value of $\angle BCD $ is
- $50^o $
- $60^o $
- $70^o $
- $ 90^o$
Correct

Incorrect

Hint

$\angle BDC=\angle BAC=50^o$ (Angles in same segment)
In $\triangle BCD$
$\angle BCD=180^o-(50+60)=70$
Question 24 of 40

24. Question
1 point(s)
The quadrilateral ABCD circumscribes a circle with centre O. If $\angle AOB=115^o $ the value of $\angle COD $ is
- $75^o $
- $90^o $
- $120^o $
- $115^o $
Correct

Incorrect

Hint

$\angle AOB=\angle COD$ (V.O.A)
So, $\angle COD=115^o$
Question 25 of 40

25. Question
1 point(s)
O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of $50^o $ with PQ. The value of $\angle $POQ is
- $100^o $
- $50^o $
- $75^o $
- $105^o $
Correct

Incorrect

Hint

$\angle OPQ=90-50=40$
$OP=OQ$ (radius)
$\angle OPQ=\angle OQP=40^o\\ \\ \angle POQ=180^o-(\angle OPQ+\angle OQP)\\ \\ =180-(40+40)=100^o$
Question 26 of 40

26. Question
1 point(s)
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of a circle such that $\angle POR=120^o $. The value of $\angle OPQ $ is
- $30^o $
- $45^o $
- $60^o $
- $15^o $
Correct

Incorrect

Hint

$\angle OQP=90^o\\ \\ \angle QOP=180^o-120^o=60^o\\ \\ \angle OPQ=180^o-(\angle OQP+\angle QOP)\\ \\ =180-(90+60)\\ \\ =30^o$
Question 27 of 40

27. Question
1 point(s)
In the given figure, AB and AC are tangents to the circle with centre O such that $\angle $BAC = $ 40^o$. Then calculate $\angle $BOC
- $90^o $
- $140^o $
- $80^o $
- $70^o $
Correct

Incorrect

Hint

$\angle ABO=\angle ACO=90^o\\ \\ \angle ABO+\angle ACO+\angle BAC+\angle BOC=360^o\\ \\ 90+90+40+\angle BOC=360^o\\ \\ \angle BOC=140^o$
Question 28 of 40

28. Question
1 point(s)
If angle between two tangents drawn from a point P to a circle of radius ‘a’ and centre O is $90^o $, the value of OP is
- $a $
- $a\sqrt2 $
- $3a $
- $a^2 $
Correct

Incorrect

Hint

$\angle P=90^o$ So, $\angle QOR=90^o$
$OR=OQ=a$
So, $PQOR$ is a square
$OP=\sqrt{a^2+a^2}\\ \\ =\sqrt{2a^2}=a\sqrt2$
Question 29 of 40

29. Question
1 point(s)
Two concentric circle of radius 6 cm and 4 cm with centre O are drawn. If AP is a tangent to the larger circle and BP the smaller circle and length of AP is 8 cm. The length of BP is
- $14\ cm $
- $2\sqrt{21}\ cm $
- $\sqrt{21}\ cm $
- $28\ cm $
Correct

Incorrect

Hint

$OP^2=OA^2+AP^2\\ \\ =36+64=100\\ \\ OP=10\ cm\\ \\ BP^2=OP^2-OB^2=100-16=84\\ \\ BP=\sqrt{84}=2\sqrt{21}\ cm$
Question 30 of 40

30. Question
1 point(s)
$PQ $ is a tangent from an external point $P $ to a circle with centre $O $ and $OP $ with the circle at $T $ and $QOR$ is a diameter. If $\angle POR=130^o $ and $S $ is a point on the circle. The value of $\angle 1+\angle 2 $ is
- $95^o $
- $135^o $
- $105^o $
- $75^o $
Correct

Incorrect

Hint

$\angle POR=130^o=\angle ROT\\ \\ \angle 2=\dfrac{1}{2}\times\angle ROT=65^o$ ( $\because$ Central angle is twice of inscribed angle)
$\angle POQ=180^o-130^o=50^o$ (linear pair)
$\angle Q=90^o\\ \\ \angle 1=180-(50+90)=40^o\\ \\ \angle 2+\angle 1=65+40=105^o$
Question 31 of 40

31. Question
1 point(s)
If TP and TQ are two tangents to a circle with centre O, so that $\angle $POQ = $110^o $, the value of $\angle $PTQ is
- $70^o $
- $40^o $
- $95^o $
- $105^o $
Correct

Incorrect

Hint

$OP\bot PT$ and $OQ\bot QT$
$\Rightarrow \angle OPT=90^o$ and $\angle OQT=90^o$
In quadrilateral $OPTQ$ we have
$\angle POQ+\angle OPT+\angle PTQ+\angle TQO=360^o\\ \\ 110^o+90^o+\angle PTQ+90^o=360^o\\ \\ \angle PTQ=360^o-290=70^o$
Question 32 of 40

32. Question
1 point(s)
PA and PB are tangents from a point P to a circle with centre O. The quadrilateral OAPB is a
- Square
- Rhombus
- Parallelogram
- Cyclic quadrilateral
Correct

Incorrect

Hint

$\angle OAP=\angle OBP=90^o$ (Tangents and radius are perpendicular)
$\angle OAP+\angle OBP=180^o$
Which are opposite angles of quadrilateral $OAPB$
So, the sum of remaining two angles is also $180^o$ .
Question 33 of 40

33. Question
1 point(s)
If $PQR $ is a tangent to a circle at $Q $ whose centre is $O $, $AB $ is a chord parallel to $PR $ and $\angle BQR=70^o $. The value of $\angle AQB $ is
- $40^o $
- $15^o $
- $75^o $
- $70^o $
Correct

Incorrect

Hint

$\angle ABQ=\angle BQR=70$ (alternate angles)
In $\triangle QDA$ and $\triangle QDB$
$\angle QDA=\angle QDB$ (each $90^o$ )
$QD=DQ$ ( Common)
$AD=BD\\ \\ \triangle QDA\cong\triangle QDB$
So, $\angle QAD=\angle QBD$ (CPCT)
$\angle A+\angle B+\angle Q=180^o\\ \\ \angle Q=180^o-(70+70)=40^o$
Question 34 of 40

34. Question
1 point(s)
$PT $ touches the circle at $R $ whose centre is $O $. Diameter $SQ $ when produced meets $PT $ at $ P$. If $\angle SPR=x^o $ and $\angle QRP=y^o $. Then
- $x+2y=90^o $
- $x+2y=180^o $
- $x+y=120^o $
- $4x+5y=180^o $
Correct

Incorrect

Hint

$\angle QRP=\angle QSR$ (Same chord QR)
$\angle QSR=y\\ \\ \Rightarrow \angle PSR=y$
$\angle QRS=90^o$ (angle in a semicircle is $90^o$ )
$\angle PRS=\angle QRP+\angle QRS=y+90^o$
In $\triangle PRS$
$\angle SPR+\angle PRS+\angle PSR=180^o\\ \\ x+y+90+y=180^o\\ \\ x+2y=90^o$
Question 35 of 40

35. Question
1 point(s)
AB and PQ intersect at M. If A and B are centres of circles then
- $PM=MQ $
- $PQ\bot AB $
- Both (a) and (b)
- $PQ=AB $
Correct

Incorrect

Hint

PQ is the common chord
$\Rightarrow$ AB bisects PQ at M so, PM = MQ
PQ is perpendicular to AB
Question 36 of 40

36. Question
1 point(s)
PQ is tangent at a point R of the circle with centre O. If $\angle $ TRQ = $30^o $, the value of $\angle $PRS is
- $60^o $
- $30^o $
- $45^o $
- $75^o $
Correct

Incorrect

Hint

$OR\bot RQ$
So, $\angle ORQ=90^o$
$\angle TRQ+\angle ORT=90^o\\ \\ \angle ORT=60^o$
$\angle SRT=90^o$ (angles in a semicircle is $90^o$ )
$\angle ORT+\angle SRO=90^o\\ \\ \angle SRO+60^o=90^o\\ \\ \angle SRO=30^o\\ \\ \angle PRS=90-30=60$
Question 37 of 40

37. Question
1 point(s)
A circle has $\underline{\qquad} $ number of tangents
- 0
- 4
- 1
- Infinite
Correct

Incorrect

Hint

A circle has infinitely many tangents
Question 38 of 40

38. Question
1 point(s)
If the angle between the two radius of a circle is $120^o $, then the angle between the tangents at the end of the radius is
- $60^o $
- $70^o $
- $120^o $
- $180^o $
Correct

Incorrect

Hint

Angle between the tangents $=180-120=60^o$
Question 39 of 40

39. Question
1 point(s)
If a parallelogram circumscribes a circle then it is a
- Rhombus
- Square
- rectangle
- none of these
Correct

Incorrect

Hint

In $||^{gm}\ ABCD$
$AB=CD,AD=BC$
Hence $AP=AS,BP=BQ,CR=CQ,DR=DS$
$AP+BP+CR+DR=AS+BQ+CQ+DS\\ \\ AB+CD=AD+AD\\ \\ 2AB=2AD\Rightarrow AB=AD$
So, $AB=AD$ and $AB=CD$ and $AD=BC$
So, $AB=BC=CD-AD$
So, $ABCD$ is a rhombus as parallelogram with equal sides is a rhombus
Question 40 of 40

40. Question
1 point(s)
If $d_1,d_2\ (d_2>d_1) $ be the diameters of two concentric circles and C be the length of a chord of a circle which is tangent to the other circle then
- $d_2^2=d^2_1+c^2 $
- $d^2_2=c^2-d^2_1 $
- $2d_1^2=c^2+d^2_2 $
- $d^2_1=c^2-d^2_2 $
Correct

Incorrect

Hint

$OC^2+CB^2=OB^2\\ \\ \left(\dfrac{1}{2}d_1\right)^2+\left(\dfrac{1}{2}c\right)^2=\left(\dfrac{1}{2}d_2\right)^2\\ \\ d_1^2+c^2=d^2_2$