Maths Class X Question Bank

Question 1 of 45

1. Question
1 point(s)
In the figure, find the length of PR.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$OP=\sqrt{8^2+6^2}\\ \\ =\sqrt{64+36}=\sqrt{100}=10\ cm\\ \\ PR=OR+OP=6+10=16\ cm$
Question 2 of 45

2. Question
1 point(s)
In the figure, if BC = 4.5 cm, find the length of AB.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BC=4.5\ cm\\ \\ PC=BC=4.5\ cm\\ \\ AC=PC=4.5\ cm\\ \\ AB=AC+BC\\ \\ =4.5+4.5\\ \\ =9\ cm$
Question 3 of 45

3. Question
1 point(s)
A pair of tangents $PA $ and $ PB$ are drawn from an external point $P $ to a circle with centre $O $. If $\angle APB=90^o $ and $PA=6\ cm $, find the radius of tehcircle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle OAP$
$\tan45=\dfrac{AP}{OA}\\ \\ \Rightarrow 1=\dfrac{6}{OA}\\ \\ \Rightarrow OP=\ \Rightarrow OA=6\ cm$
Question 4 of 45

4. Question
1 point(s)
In the figure, BOA is a diameter of a circle and the tangent at point P meets BA produced at T. If $\angle PBO=30^o $, find $\angle $PTA
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle BPA=90^o\\ \\ \angle OPT=90^o\\ \\ \angle PAB=180^o-(30+\angle BPA)\\ \\ =180-30-90=60$
$OP=OA$ (radius)
$\angle OAP=\angle OPA=60^o\\ \\ \angle APT=\angle OPT-\angle OPA=90-60=30^o\\ \\ \angle PAT=180-60=120$
In $\triangle PAT$ ,
$\angle PAT+\angle PTA+\angle APT=180\\ \\ 120+\angle PTA+30=180\\ \\ \angle PTA=30^o$
Question 5 of 45

5. Question
1 point(s)
The length of tangent from an external point on a circle is always greater than the radius of the circle. Is it true?
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

False
It may be or may not be greater than the radius.
Question 6 of 45

6. Question
1 point(s)
If the angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then find OP.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\sin45=\dfrac{OT}{OP}\\ \\ \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{a}{OP}\\ \\ \Rightarrow 2a=\sqrt3\\ \\ \Rightarrow OP=\dfrac{2a}{\sqrt3}$
Question 7 of 45

7. Question
1 point(s)
In the figure, circles C(O, r) and C’ (O’, r/2) touch ubternally at a point A and AB is a chord of the circle C (O, r), intersecting C’ (O’,r) at C. Prove that AC = CB.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OCA=90^o$ (angle in semicircle)
In $\triangle OCA$ and $\triangle OCB$
$OA=OB=r$
$\angle OCA=\angle OCB=90$ and $OC=OC$
$\triangle OCA\cong\triangle OCB$ (RHS)
$AC=BC$ (cpct) (Proved)
Question 8 of 45

8. Question
1 point(s)
P is the mid-point of an arc QPR of a circle. Show that the tangent at P is parallel to the chord QR.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$QP=PR\\ \\ OP\bot MN$
Also by construction $OP\bot QR$
So $OP$ must be parallel to $MN$
Question 9 of 45

9. Question
1 point(s)
In the figure, PQL and PRM are tangents to the circle with centre O at the points Q and R respectively and S is a point on the circle such that $\angle SOL=50^o $ and $\angle SRM=60^o $. Find $ \angle QSR$.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OSQ=\angle OQS=90-50=40\\ \\ \angle RSO=\angle SRO=90-60=30\\ \\ \angle QSR=40+30=70$
Question 10 of 45

10. Question
1 point(s)
In the figure, if $\angle CBA=140^o $, find $\angle ADB $.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle ABE=180-\angle ABC\\ \\ =180-140=40\\ \\ \angle ADB=\angle ABE\ (alt\ \angle S)=40^o$
Question 11 of 45

11. Question
1 point(s)
Show that the tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parrallel to BC.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$AB=AC$ (given)
$\Rightarrow \angle ACB=\angle ABC$ …(1)
$DAE$ is the tangent and $AB$ is the chord
$\angle DAB=\angle ACB$ …(2)
From (1) and (2), $\angle ABC=\angle DAB$
But they form alternate angles so $DE||BC$
Question 12 of 45

12. Question
1 point(s)
$PQ $ and $PR $ are tangent segment to a circle with centre O. If $\angle QPR=80^o $, find $\angle QOR $.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OQP=\angle ORP=90^o\\ \\ \angle OQP+\angle ORP+\angle PRO+\angle ROQ=360\\ \\ 90+46+90+\angle ROQ=360\\ \\ \angle ROQ=360-226=134^o$
Question 13 of 45

13. Question
1 point(s)
$AB $ is a diameter of a circle and $AC $ is its chord such that $\angle BAC=30^o $. If the tangent at $C $ intersects $AB $ at $D $, then prove that $ BC=BD$.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle ACB=90^o$
In $\triangle ACB$
$\angle A+\angle ACB+\angle CBA=180^o\\ \\ \angle CBA=180-(90+30)=60$ …(1)
In $\triangle OCB$
$OC=OB\\ \\ \Rightarrow \angle OCB=\angle OBC=60^o$
Now
$\angle OCB+\angle BCD=90\\ \\ \angle BCD=90-60=30$ …(2)
$\angle CBO=\angle BCO+\angle CDB\\ \\ 60=30+\angle CDB\\ \\ \angle CDB=30$ …(3)
From (2) and (3)
$BC=BD$
Question 14 of 45

14. Question
1 point(s)
In the figure, $\angle ADC=90^o\ BC=38\ cm,CD=28\ cm $ and $BP=25\ cm $. Find the radius of the circle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BP=BQ=25\ cm\\ \\ CQ=38-25=12 \ cm\\ \\ CR=CQ=12\\ \\ \Rightarrow DR=28-12=16\ cm$
Let $OR=r=OS$
$\Rightarrow ORDS$ is a square
$\Rightarrow r=16\ cm$
Question 15 of 45

15. Question
1 point(s)
If a number of circles pass through the end points P and Q of a line segment PQ, then prove that their centres lie on the perpendicular bisector of PQ.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

PQ is either a chord to a circle or the diameter.
If PQ is diameter than clearly centre lies on the left of diameter i.e. on the perpendicular bisector of PQ.
But if PQ is a chord then we know than perpendicular from the centre to the chord of circle bisect the chord implies that centre lies on the perpendicular bisector of the chord
So for each circle centre lies on the perpendicular bisector of PQ.
Question 16 of 45

16. Question
1 point(s)
Two tangent segments $ BC$ and $BD $ are drawn to a circle with centre $O $ such that $\angle CBD=120^o $. Show that $OB=2BC $.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $\triangle BOC$ and $\triangle BOD$
$\angle BCO=\angle BDO=90$
$OB=OB$ (Common)
$OC=OD$ (radius)
$\triangle CBO\cong\triangle BOD$ (RHS)
$\Rightarrow \angle CBO=\angle DBO$ (cpct)
$\angle CBD=120^o\\ \\ \angle CBO=\dfrac{120}{2}=60^o\\ \\ \cos60=\dfrac{BC}{BO}\Rightarrow \dfrac{1}{2}=\dfrac{BC}{BO}\\ \\ =BO=2BC$
Question 17 of 45

17. Question
1 point(s)
In the figure, $\angle ADC=90^o\ BC=38\ cm,\ CD=28\ cm $ and $BP=25\ cm $. Find the radius of the circle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BP=BQ=25\ cm\\ \\ CQ=38-25=12 \ cm\\ \\ CR=CQ=12\\ \\ \Rightarrow DR=28-12=16\ cm$
Let $OR=r=OS$
$\Rightarrow ORDS$ is a square
$\Rightarrow r=16\ cm$
Question 18 of 45

18. Question
1 point(s)
Out of the two concentric circles, the radius of the outer is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$OA^2=AD^2+DO^2\\ \\ \Rightarrow DO^2=5^2-4^2=25-16=9\\ \\ DO=3\ cm$
Question 19 of 45

19. Question
1 point(s)
In the isosceles triangle ABC shown below, AB = AC, show that BF = FC
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Since tangents from an exterior point to a cirlcle are equal
$AE=AG\\ \\ BE=BF\\ \\ CG=CF$
$AB=AC$ (given)
$AB-AE=AC-AE\\ \\ AB-AE=AC-AG\ (\because AE=AG)\\ \\ BE=CG\\ \\ BF=CF$ (Proved)
Question 20 of 45

20. Question
1 point(s)
In the figure, a circle is inscribed in a $\triangle $ABC with sides AB = 12 cm, BC = 8 cm and AC = 10 cm. Find the lengths of AD, BE and CF.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Let $AD=AF=x\ cm$
$BD=BE=y\ cm\\ \\ CE=CF=z \ cm\\ \\ 2(x+y+z)=AB+BC+AC=AD+DB+BE+EC+AF+FC=30\ cm\\ \\ \Rightarrow x+y+z=15\ cm\\ \\ AB=AD+DB=x+y=12\ cm\\ \\ Z=CF=15-12=3\ cm\\ \\ AC=AF+FC=x+z=10\ cm\\ \\ y=BE=15-10=5\ cm\\ \\ x=AD=x+y+z-2-y=15-3-5=7\ cm$
Question 21 of 45

21. Question
1 point(s)
Prove that opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of circle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$2(1+2+3+4)=360\\ \\ 1+2+3+4=180^o$
In $\triangle AOB$
$\angle BOA=180-(1+2)$
In $\triangle COD,\ \angle COD=180-(3+4)$
$\angle BOA+\angle COD=360-(1+2+3+4)\\ \\ =360-180=180$
This opposite sides of a quadrilateral circumscribing a circle subtend supplementry angles at the centre of the circle
Question 22 of 45

22. Question
1 point(s)
If all sides of a parallelogram touch a circle, then prove that it is a rhombus.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$AP=AS$ …(1)
$BP=BQ$ …(2)
$CR=CQ$ …(3)
$DR=DS$ …(4)
Adding 1, 2, 3, 4 we get
$AP+BP+CR+DR=AS+BQ+CQ+DS\\ \\ \Rightarrow (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)\\ \\ \Rightarrow AB+CD=AD+BC\\ \\ \Rightarrow AB+AB=AD+AD$
( $\because$ opposite sides are equal
$2AB=2AD\\ \\ \Rightarrow AB=AD$
But $AB=CD$ and $AD=BC$
So $AB=CD=BC=DA$
So it is a rhombus
Question 23 of 45

23. Question
1 point(s)
Two tangents PA and PB are drawn from an external point P to a circle with centre. O. Prove that AOBP is a cyclic quadrialteral.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OAP=90^o$ and $\angle OBP=90^o$
$\Rightarrow \angle OAP+\angle OBP=90+90=180$ ..(1)
In quadrilateral $OAPB$
$\angle OAP+\angle APB+\angle AOB+\angle OBP=360\\ \\ \Rightarrow \angle APB+\angle AOB=360-180=180$ …(2)
From (1) and (2), the quadrilateral $AOBP$ is cyclic.
Question 24 of 45

24. Question
1 point(s)
Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $DCMN$
$CM=CN$ (radius)
$\therefore\ \angle CMN=\angle CNM$
Now $\angle OMC=\angle ONC$
$\Rightarrow OMC-\angle OMC-\angle CMN=\angle ONC-\angle CNM\\ \\ \Rightarrow \angle OML=\angle ONL$
This tangents make equal angle with the chord
Question 25 of 45

25. Question
1 point(s)
In the figure, is the centre of two concentric circles of radii 6 cm and 4 cm. PQ and PR are tangents to the two circles from an external point P. If PQ = 10 cm, find the length of PR (upto one decimal place).
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$OP=\sqrt{OA^2+PA^2}=\sqrt{6^2+10^2}=\sqrt{136}$
In $\triangle OAP$ ,
$OP^2=OB^2+PB^2\\ \\ PB=\sqrt{OP^2-OB^2}=\sqrt{136-16}\\ \\ =\sqrt{120}\ cm=10.9\ cm$
Question 26 of 45

26. Question
1 point(s)
Prove that the tangents drawn at the end-points of a diameter of a circle are parallel.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OAR=90\\ \\ \angle OAS=90\\ \\ \angle OBP=90\\ \\ \angle OBQ=90$
But $\angle OAR=\angle OBQ$ (alt interior angles)
$\angle OAS=\angle OBP$ (alt interior angles)
Since alt $\angle S$ are equal line $PQ$ and $RS$ will be parallel.
Question 27 of 45

27. Question
1 point(s)
In the figure, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$AB=AP+PB\\ \\ CD=CR+DR$
But $AP=AS$
$PB=BQ\\ \\ DR=DS\\ \\ CR=CQ\\ \\ AB+CD=AP+PB+CR+DR\\ \\ =AS+BQ+DS+CQ\\ \\ =(AS+DS)+(BQ+CQ)\\ \\ AB+CD=AD+BC$ (Proved)
Question 28 of 45

28. Question
1 point(s)
Prove that the line segment joining the points of contact of two parallel tangents to a circle is a diameter of the circle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OPA=90^o$
Since $OR||BA$ (by construction)
$\angle OPA+\angle POR=180\\ \\ \angle POR=90^o$
Similarly $\angle QOR=90^o$
$\angle POR+\angle QOR=90+90=180$
$PQ$ is a straight line so it is a diameter.
Question 29 of 45

29. Question
1 point(s)
In the figure, $O $ is the centre of a circle and $BCD $ is tangent to it at $C $. Prove that $\angle BAC+\angle ACD=90^o $.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$OA=OC$ (radius)
$\Rightarrow \angle OAC=\angle OCA$ …(1)
$\angle OCD=90^o\\ \\ \Rightarrow \angle ACD+\angle OAC=90^o$
$\Rightarrow \angle ACD+\angle BAC=90^o$
Question 30 of 45

30. Question
1 point(s)
In the figure, two circles touch each other externally at C. Prove that the common tangent at C bisects the other two common tangents.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

We know that tangent are equal in length from an external point to circle.
$EP=EC$ and $EQ=EC\Rightarrow EP=EQ$ and $GH=FC,\ FH=FC\Rightarrow GH=FH$
Hence $ECF$ bisects $PQ$ and $ECF$ bisects $GH$
Question 31 of 45

31. Question
1 point(s)
In the figure, a circle touches the side BC of triangle ABC at P and touches AB and AC produced at Q and R respectively. Show that AQ=$\dfrac{1}{2} $ (Perimeter of $\triangle $ABC )
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$AB=AP,BQ=BP,BP=CP\\ \\ AC+AB+BC=2(AB)+BP+PC\\ \\ =2(AB)+2(B2)\\ \\ AC+AB+BC=2(A2)\\ \\ \Rightarrow AQ=\dfrac{AB+BC+AC}{2}\\ \\ AQ=\dfrac{Perimeter\ of\ \triangle ABC}{2}$
Question 32 of 45

32. Question
1 point(s)
In the figure, all three sides of a triangle touch the circle. Find the value of x.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BP=BQ=10\ cm\\ \\ CP=CR=6\ cm\\ \\ AQ=AR=18-10=8\ cm\\ \\ x=AR+RC=8+6=14\ cm$
Question 33 of 45

33. Question
1 point(s)
Two tangents PA and PB are drawn to a circle with centre O, such that $\angle APB=120^o $. Prove that OP = 2 AP.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OAP=\angle OBP=90\\ \\ OA=OB\\ \\ \triangle OAP\cong\triangle OBP$
So $\angle OPA=\angle OPB=\dfrac{120}{2}=60^o$
In $\triangle OAP$
$\cos\angle OPA=\cos60=\dfrac{AP}{OP}\\ \\ \Rightarrow \dfrac{1}{2}=\dfrac{AP}{OP}\Rightarrow OP=2\ AP$
Question 34 of 45

34. Question
1 point(s)
In the figure, O is the centre of the circle. PA and PB are tangents to the circle from the point P prove that AOBP is a cyclic quadrilateral.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle OBP=\angle OAP=90^o$
In quadrilateral $AOBP$
$\angle APB+\angle AOB+\angle OBP+\angle OAP=360\\ \\ \Rightarrow \angle APB+\angle AOB+90+90=360\\ \\ \Rightarrow \angle APB+\angle AOB=180$
Since the sum of the opposite angles of quadrilateral is $180^o$
Hence $AOBP$ is a cyclic quadrilateral
Question 35 of 45

35. Question
1 point(s)
In the figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11 cm, and BC = 7 cm, then find the length of BR.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$CP=CQ=11\ cm\\ \\ BC=7\ cm\\ \\ BQ=11-7=4\ cm\\ \\ BQ=BR=4\ cm$
Question 36 of 45

36. Question
1 point(s)
In the figure, $\triangle $ABC is circumscribing a circle. Find the length of BC.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BP=BR=3\ cm\\ \\ AR=AQ=4\ cm$
So $QC=11-4=7\ cm$
$QC=PC=7\ cm\\ \\ BC=BP+PC=3+7=10\ cm$
Question 37 of 45

37. Question
1 point(s)
A tangent PT is drawn parallel to a chord AB as shown in figure. Prove that APB is an isosceles triangle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$\angle ADP=90^o$
$AD=DB$ ( $\because PD$ is a bisector of $AB$ )
In $\triangle ADP$ and $\triangle BDP$
$AB=DB$
$\angle ADP=\angle BDP$ (each 90)
$PD=DP$ (common)
$\triangle ADP\cong\triangle BDP$ (SAS)
$\angle PAD=\angle PBD$ (cpct)
$\Rightarrow AP=PB$
$\Rightarrow APB$ is an isosceles $\triangle$
Question 38 of 45

38. Question
1 point(s)
Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Since, $OA\bot AP$ and $OB\bot BP$
$\angle PAO+\angle PBO=90^o+90^o=180^o\\ \\ \Rightarrow \angle APB+\angle AOB=180^o$
$\Rightarrow AOBP$ is a cyclic quadrilateral.
Question 39 of 45

39. Question
1 point(s)
In the figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BA=PC\\ \\ PB=PD\\ \\ PA-PB=PC-PD\\ \\ \Rightarrow AB=CD$
Question 40 of 45

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, prove that $d^2_2=c_2+d^2_1 $.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$OC^2+BC^2=OB^2\\ \\ \left(\dfrac{1}{2}d_1^2\right) +\left(\dfrac{1}{2}C\right)^2=\left(\dfrac{1}{2}d_2\right)^2$
(as C bisects AB)
$\Rightarrow d^2_2=C^2+d_1^2$
Question 41 of 45

41. Question
1 point(s)
ABC is a right triangle, right angled at A. A circle is inscribed in it. The lengths of the sides containing the right angle are 8 cm and 6 cm. Find the radius of the incircle.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$BC=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\ cm$
Area of $\triangle ABC$ = Area of $OAB$ + Area of $\triangle OBC$ + Area of $\triangle OCA$
$\Rightarrow \dfrac{1}{2}\times AB\times AC=\dfrac{1}{2}\times r\times AB+\dfrac{1}{2}\times r\times BC+\dfrac{1}{2}\times r\times CA\\ \\ \Rightarrow \dfrac{1}{2}\times 6\times 8=\dfrac{1}{2}\times r\times 6+\dfrac{1}{2}\times r\times 10+\dfrac{1}{2}\times r\times 8\\ \\ \Rightarrow 12r=24\Rightarrow 24\Rightarrow r=2\ cm$
Question 42 of 45

42. Question
1 point(s)
a, b, c are the sides of a right triangle, where c is the hypotenune. Prove that the radius r of the circle which touches the sides of the triangle is given by $r=\dfrac{a+b-c}{2} $
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

$AE=AF,\ BD=BF\\ \\ CE=CD=r\\ \\ AF=b-r\\ \\ BF=a-r\\ \\ AB=C=AF+BF\\ \\ \Rightarrow C=b-r+a-r\\ \\ \Rightarrow C=b+a-2r\\ \\ \Rightarrow \dfrac{a+b-c}{2}=r$
Question 43 of 45

43. Question
1 point(s)
AB is a diameter of a circle APB. AH and BK are perpendicular from A and B respectively to the tangent at P. Prove that AB + AH = BK.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Let $AH=x,BK=y$
$OP=r,BM=z$
$\triangle MKB\sim MHA$ (AA)
$\Rightarrow \dfrac{BK}{AH}=\dfrac{MB}{MA}\Rightarrow \dfrac{y}{z}=\dfrac{z}{2r+z}\\ \\ \Rightarrow z=\dfrac{2ry}{x-y}$ …(1)
Also $\triangle MKB\sim\triangle MPO$ (AA)
$\Rightarrow \dfrac{BK}{OP}=\dfrac{BM}{OM}\\ \\ \Rightarrow \dfrac{y}{r}=\dfrac{z}{r+z}\Rightarrow z=\dfrac{ry}{r-y}$ ..(2)
From (1) and (2)
$\dfrac{ry}{r-y}=\dfrac{2ry}{x-y}\Rightarrow 2r=x+y\\ \\ \Rightarrow AB=AH+BK$ (Proved)
Question 44 of 45

44. Question
1 point(s)
Show that the tangents drawn at the ends of chord of a circle make equals angles with the chord.
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

In $DCMN$
$CM=CN$ (radius)
$\angle CMN=\angle CNM$
Now, $\angle OMC=\angle ONC$
$\angle OMC-\angle CMN=\angle ONC-\angle CNM\\ \\ \Rightarrow \angle OML=\angle ONL$
Tangents make equal angles with the chord.
Question 45 of 45

45. Question
1 point(s)
Let A be the one point of intersection of two intersecting circles with centre O and Q. The tangents at A to the two circles meet the circles again at B and C respectively. Let the point P be located such that AOPQ is a parallelogram. Prove that P is the circumcentre of $\triangle $ABC .
In $\triangle ABC $
$ PA=PB$
Similarly $PA=PC $
Hence $PA=PB=PC $
Hence $P $ is equdratant from the 3 vertices of $\triangle ABC $.
So $P $ is the circumcentre of $ \triangle ABC$
- Upload your answer to this question.
  
  This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Grading can be reviewed and adjusted.

Grading can be reviewed and adjusted.

Hint

Perpendicular from the centre of circle to a chord bisects the chord
$AM=MB$
This $OM$ and $OP$ is the perpendicular bisector of $AB$ . Similarly $PQ$ is the perpendicular bisector of $AC$ .