Maths Class X Question Bank

Question 1 of 44

1. Question
1 point(s)
A circle with centre O and radius 5 cm has been inscribed in an equilateral triangle ABC. Find the perimeter of $\triangle $ABC .
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Hint

$\begin{matrix}BF=BD=x \\FC=CE=x\\AE=AD=x \end{matrix}\ \because$ it is an equilateral triangle
Area of equilateral $\triangle=\dfrac{\sqrt3 a^2}{4}$
$=\dfrac{\sqrt 3}{4}(2x)^2 =x^2\sqrt3$
Area of $\triangle ABC$ = Area of $\triangle AOB$ + Area of $\triangle AOC$ + Area of $\triangle BOC$
$\Rightarrow x^2\sqrt3=\dfrac{1}{2}\times5\times 2x+\dfrac{1}{2}\times5\times2x+\dfrac{1}{2}\times 5\times 2x\\ \\ \Rightarrow x^2\sqrt3=5x+5x+5x\\ \\ \Rightarrow x=5\sqrt3\Rightarrow 2x=10\sqrt3$
Perimeter $=3\times10\sqrt3=30\sqrt3\ cm$
Question 2 of 44

2. Question
1 point(s)
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.
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Hint

$OR\bot m\\ \\ PQ|| m\\ \\ \Rightarrow \angle OMP=\angle OMQ=90$
In $\triangle OPM$ and $\triangle OQM$
$OP=OQ$ (radius)
$OM=OM$ (common)
$\angle OMP=\angle OMQ\\ \\ \triangle OPM\cong\triangle OQM$ (RHS)
$\Rightarrow \angle POM=\angle QOM$ (cpct)
$ar\ (PR)=arc\ QR$
This $R$ bisects $arc\ PRQ$
Question 3 of 44

3. Question
1 point(s)
In the figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.
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Hint

$AE=CE$ (E is the external point)
$BE=DE$ (E is the external point)
Adding eqn (1) are (2)
$AE+BE=CE+DE$
$AB=CD$ (Proved)
Question 4 of 44

4. Question
1 point(s)
Prove that the centre of a circle touching two intersecting lines lies on the angles bisector of the lines.
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Hint

$OM=ON$ (radius)
$\angle OMP=\angle ONP$ (radius is perpendicular to its tangent)
$OP=OP$ (common)
$\triangle OPM\cong\triangle OPN$ (SAS)
$\angle MPO=\angle NPO$ (cpct)
$\therefore$ bisects $\angle MPN$
$\therefore O$ lies on the bisector of the angle between $l_1$ and $l_2$
Question 5 of 44

5. Question
1 point(s)
Prove that there is one and only one tangent at any point on the circumference of a circle.
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Hint

Let point P on the circle
If posible let PT’ be the tangent instead of PT
$OP\bot PT$ and $OP \bot PT'\\ \\ \angle OPT=\angle OPT'=90$
This is only possible if PT’ concide PT
Therefore there is only one tangent to a point on circle.
Question 6 of 44

6. Question
1 point(s)
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
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Hint

$OA=OC$ (radii)
$OB=OC+CB\\ \\ OB>OC\\ \\ \Rightarrow AO<OB$
$B$ is an arbitrary point on the tangent
Thus $OA$ is shorter than any other line segment joining O to any point on tangent. Shortest distance of a point from a given line is the perpendicular distance from that line
Hence the tangent at any point of circle is perpendicular to the radius
Question 7 of 44

7. Question
1 point(s)
In the figure, I and m are two parallel tangents at A and B. The tangent at C makes an intercept DE between l and m. Prove that DE subtends a right angle at the centre of the circle.
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Hint

In $\triangle ODA$ and $\triangle ODC$
$OA=OC$ (radii)
$AD=DC$ (length of tangent )
$DO=OD$ (common)
$\triangle DOA\cong\triangle ODC$ (SSS)
$\Rightarrow \angle DOA=\angle COD$
Similarly $\triangle OEB\cong\triangle OEC$
$\Rightarrow \angle EOB=\angle COE$
$AOB$ is a diameter of the circle
Hence it is a straight line
$\angle DOA+\angle COD+\angle COE+\angle EOB=180\\ \\ 2\angle COD+2\angle COE=180\Rightarrow \angle DOE=90$
Question 8 of 44

8. Question
1 point(s)
In the figure, triangle $ABC $ is a right angled triangle with $AB = 6\ cm, AC = 8\ cm\ \angle A=90^o$ . A circle with centre $O $ is inscribed inside the triangle. Find the radius r.
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Hint

$BC=\sqrt{8^2+6^2}\\ \\ =\sqrt{100}=10\ cm$
Area of $(\triangle OAC+\triangle OAB+\triangle DOC)$ = area of $\triangle ABC$
$\Rightarrow \dfrac{1}{2}\times8\times r+\dfrac{1}{2}\times6\times r+\dfrac{1}{2}\times 10\times r=\dfrac{1}{2}\times 6\times 8\\ \\ \Rightarrow 12r=24\Rightarrow r=2\ cm$
Question 9 of 44

9. Question
1 point(s)
If the an external point B of the circle with centre O, two tangents BC and BD are drawn such that $\angle DBC=120^o $, prove that BO = 2BC.
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Hint

$\angle BCO=\angle BDO=90^o$
In $\triangle OBC$ and $\triangle OBD$
$\angle BCO=\angle BDO=90^o$
$OB=OB$ (common)
$OC=OD$ (radius)
$\triangle OBC\cong\triangle OBD$ (RHS)
$\Rightarrow \angle OBC=\angle OBD$ (cpct)
$\therefore\angle OBC=\angle OBD=60^o\\ \\ \cos60=\dfrac{BC}{OB}\Rightarrow \dfrac{1}{2}=\dfrac{BC}{OD}\\ \\ \Rightarrow OB=2BC$
Question 10 of 44

10. Question
1 point(s)
In the figure, from an external point P, PA and PB are tangent to the circle with centre O. If CD is another tangent at point E to the circle and PA = 12 cm, find the perimeter of $\triangle $PCD .
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Hint

$PA=PB=10\ cm\\ \\ CE=CA\\ \\ DE=DB$
Perimeter of $\triangle PCD=PC+PD+CD$
$=PC+PD+CE+ED\\ \\ =(PC+CA)+(PD+DB)\\ \\ =PA+PB=10+10=20\ cm$
Question 11 of 44

11. Question
1 point(s)
In the figure, OP is equal to the diameter of the circle. Prove that ABP is an equilateral triangle
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Hint

$OP=2r\\ \\ \angle OAP=90$
In $\triangle OAP$
$\sin Q=\dfrac{OA}{OP}=\dfrac{r}{2r}=\dfrac{1}{2}\Rightarrow \angle OPA=30^o$
Similarly $\angle OPB=30^o$
$\angle APB=30+30=60$
Since $PA=PB$ (length of tangent from an external point are equal)
$\Rightarrow \angle PAB=\angle PBA$
In $\triangle PBA$
$\angle APB+\angle PAB+\angle PBA=180\\ \\ \Rightarrow 60+2\angle PAB=180\\ \\ \Rightarrow \angle PAB=60=\angle PBA$
Since all angles are $60^o,\ \triangle ABP$ is equilateral $\triangle$
Question 12 of 44

12. Question
1 point(s)
In the figure, a circle touches all the four sides of a quadrilateral ABCD with sides AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
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Hint

$AP=AS\\ \\ BP=BQ\\ \\ DR=DS\\ \\ CR=CQ$
Adding all the equation we get
$(AP+BP)+(CR+RD)=(BQ+QC)+(DS+AS)\\ \\ \Rightarrow AB+CD=BC+DA\\ \\ \Rightarrow 6+4=7+DA\\ \\ \Rightarrow AD=10-7=x \ m$
Question 13 of 44

13. Question
1 point(s)
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact are of lengths 8 cm and 6 cm respectively.
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Hint

$S=\dfrac{x+6+x+8+6+8}{2}\\ \\ =\dfrac{2x+28}{2}=x+14$
Area of $\triangle ABC=\sqrt{S(S-a)(S-b)(S-c)}$
$=\sqrt{(x+14)\times x\times 8\times 6}\\ \\ =\sqrt{48x(x+14)}$ …(1)
Area of $\triangle ABC$ = area of $(\triangle OBC+\triangle OCA+OAB)\\ \\ =\dfrac{1}{2}\times4\times a+\dfrac{1}{2}\times4\times b+\dfrac{1}{2}\times 4\times c\\ \\ =2\times 2S=4S=4(x+14)$ …(2)
From eqn (1) and (2)
$\sqrt{48x(x+14)}=4(x+14)\\ \\ \Rightarrow 3x=x+14\Rightarrow x=7\ cm\\ \\ AB=x+8=7+8=15\ cm\\ \\ AC=x+6=7+6=13\ cm$
Question 14 of 44

14. Question
1 point(s)
A circle touches the side BC of $\triangle $ABC at P and sides AB and AC produced at Q and R respectively. Prove that $AQ=\dfrac{1}{2} $ (Perimeter of $\triangle $ABC)
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Hint

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 )
Question 15 of 44

15. Question
1 point(s)
In the figure, triangle ABC is isosceles in which AB = AC, circumscribed about a circle. Prove that base is bisected by the point of contact.
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Hint

$AP=AQ\\ \\ BP=BR\\ \\ CQ=CR\\ \\ AB=AC\ (\because\triangle \ ABC)$ is an isosceles $\triangle$ )
$\Rightarrow AB-AP=AC-AP\\ \\ \Rightarrow AB-AP=AC-AQ\\ \\ \Rightarrow BR=CQ\Rightarrow BR=CR$
So $BC$ is bisected at the point of contact
Question 16 of 44

16. Question
1 point(s)
Prove that the angle between the two tangents to a circle drawn from an external point, is supplementary to the angle subtended by the line segment joining the points of contact at the centre
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Hint

$\angle OBP=90\\ \\ \angle OAP=90\\ \\ \angle OAP+\angle OBP+\angle BOA+\angle APB=360\\ \\ 90+90+\angle BOA+\angle APB=360\\ \\ \angle BOA+\angle APB=180$
If proves that angle between the two tangents drawn from an external point to a circle supplementry to the angle subtended by the line segment.
Question 17 of 44

17. Question
1 point(s)
Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that $\angle APB=2\ \angle OAB $.
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Hint

$\angle OAP=90^o\\ \\ \angle OBP=90^o\\ \\ \angle APB+\angle PBO+\angle BOA+\angle OAP=360\\ \\ \angle APB+\angle BOA=360-180\\ \\ \angle APB+\angle BOA=180$ ..(1)
$\triangle AOB$ is an isosceles triangle since
$OA=OB\\ \\ \angle OAB=\angle ABO$ ….(2)
$\angle OAB+\angle ABO+\angle AOB=180\\ \\ \Rightarrow 2\angle OAB+180-\angle APB=180\\ \\ \Rightarrow 2\angle AOB=\angle APB$
Question 18 of 44

18. Question
1 point(s)
In the figure, XP and XQ are tangents from an external point X to the circle with centre O. R is a point on the circle where another tangent ARB is drawn to the circle. Prove that XA + AR = XB + BR.
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Hint

$XP=XQ$ (from X) …(1)
$AP=AR$ (from A) ….(2)
$BQ=BR$ (from B) ….(3)
$XP=XQ \\ \\ \Rightarrow XA+AP=XB+BQ\\ \\ \Rightarrow XA+AR=XB+BR$ (using eqn (2) and (3))
Question 19 of 44

19. Question
1 point(s)
In the figure, PO $\bot $ QO. The tangents to the circle with centre O at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.
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Hint

$OP=OQ$ (radius)
$\angle POQ=90^o$ …(1)
$TP=TQ$ …(2)
$\angle TPO=\angle TQO=90^o$ …(3)
From (1), (2), (3)
$OPTQ$ form a square and bisector of square bisector each other at 90.
Question 20 of 44

20. Question
1 point(s)
In the figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at point P and Q intersect at point T. Find the length of tangent TP.
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Hint

Let $TR=y$
Since $OT$ is the perpendicular bisector of $PQ$
$PR=QR=4\ cm\\ \\ TP^2=TR^2+PR^2$
Also $OT^2=TP^2+OP^2\\ \\ OT^2=TR^2+PR^2+OP^2\\ \\ (y+3)^2=y^2+16+25\\ \\ \Rightarrow 6y=32\Rightarrow y=\dfrac{16}{3}\\ \\ TP^2=TR^2+PR^2\\ \\ \Rightarrow TP^2=\left(\dfrac{16}{3}\right)^2+4^2=\dfrac{400}{9}\\ \\ \Rightarrow TP=\dfrac{20}{3}\ cm$
Question 21 of 44

21. Question
1 point(s)
In the figure, ABC is a right triangle, right angled at A, with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the radius of the inscribed circle.
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Hint

$BC=\sqrt{8^2+6^2}\\ \\ =\sqrt{100}=10\ cm$
Area of $(\triangle OAC+\triangle OAB+\triangle DOC)$ = area of $\triangle ABC$
$\Rightarrow \dfrac{1}{2}\times8\times r+\dfrac{1}{2}\times6\times r+\dfrac{1}{2}\times 10\times r=\dfrac{1}{2}\times 6\times 8\\ \\ \Rightarrow 12r=24\Rightarrow r=2\ cm$
Question 22 of 44

22. Question
1 point(s)
In the figure, $PT $ and $PS $ are tangents to a circle from a point $P $ such that $PT=5\ cm $ and $\angle TPS=60^o $. Find the length of chord $TS $.
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Hint

$PT=PS\\ \\ \Rightarrow \angle PTS=\angle PST\\ \\ \angle PTS+\angle PST+60=180\\ \\ 2\angle PST=120\Rightarrow PST=\angle PTS=60\\ \\ \therefore PTS\ \text{is an equilateral}\ \triangle \\ \\ TS=5\ cm$
Question 23 of 44

23. Question
1 point(s)
$PAQ $ is a tangent to the circle with centre $O $ at a point $A $ as shown in the figure. If $\angle OBA=35^o $, find the value of $\angle BAQ $ and $\angle ACB $.
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Hint

$OA=OB\\ \\ \angle OAB=35=\angle OBA$
But $\angle OBA+\angle AOB+\angle OAB=180\\ \\ \angle AOB=180-70=110\\ \\ \angle ACB=\dfrac{1}{2}\times \angle AOB=55^o$
$\angle BAQ=\angle ACB=55^o$ (Angles in the same segment)
Question 24 of 44

24. Question
1 point(s)
In the figure, a circle is inscribed in a quadrilateral $ABCD $ in which $\angle B=90^o $. If $AD=23\ cm,\ AB=29\ cm $ and $DS=5\ cm $, find the radius (r) of the circle
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Hint

$DS=DR\\ \\ DS=5\ cm$
So $DR=5\ cm$
$AD=23\ cm\\ \\ AR=AD-DR=23-5=18\ cm$
Also $AR=AQ$
$\therefore AQ=18\ cm\\ \\ BQ=AB-AQ=29-18=11\ cm$
In quadrilatral $BQOP$
$BQ=BP\\ \\ OQ=OP\\ \\ \angle QBP=\angle QOP=90$ (given)
$\angle OQB=\angle OPB=90^o$ (angle between the radius and tangent at point of contact)
$\therefore BQOP$ is a square
So radius = 11 cm
Question 25 of 44

25. Question
1 point(s)
In the figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle.
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Hint

Let $OA=r$
$OP=2r$
In $\triangle OPA$
$\sin(\angle OPA)=\dfrac{OA}{OP}=\dfrac{r}{2r}=\dfrac{1}{2}\\ \\ \Rightarrow \angle OPA=30^o$
Similarly $\angle OPB=30^o$
$\angle APB=30+30=60$
Since $PA=PB$
$\Rightarrow \angle PAB=\angle PBA$
In $\triangle APB$ ,
$\angle APB+\angle PAB+\angle PBA=180\\ \\ 60+2\angle PAB=180\\ \\ \angle PBA=\angle PAB=60$
Since all angle are $60^o$ , So $\triangle ABP$ is an equilateral $\triangle$
Question 26 of 44

26. Question
1 point(s)
Two circles with centres O and O’ of radii 3 cm and 4 cm respectively intersect at two points P and Q such that OP and O’ P are tangents to the two circles. Find the length of the common chord PQ.
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Hint

$OO'=\sqrt{4^2+3^2}=5\ cm$
Let $O'L=x$ , then $OL=5-x$
$PL^2=4^2-x^2=3^2-(5-x)^2\\ \\ \Rightarrow 16-x^2=9-(25+x^2-10x)\\ \\ \Rightarrow 16=-16+10x\\ \\ \Rightarrow 10x=32\Rightarrow x=3.2\ cm\\ \\ PL=\sqrt{4^2-(3.2)^2}=2.4\\ \\ PQ=2\times 2.4=4.8\ cm$
Question 27 of 44

27. Question
1 point(s)
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
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Hint

$OP\bot PT\\ \\ \Rightarrow \angle OPT=90^o$
Since P is the mid point of arc APB
$\Rightarrow arc\ AP=arc\ BP\\ \\ \Rightarrow \angle AOP=\angle BOP\\ \\ \Rightarrow \angle AOM=\angle BOM$
In $\triangle AOM$ and $\triangle BOM$
$OM=OM$ (common)
$\angle AOM=\angle BOM$ (Proved above)
$\triangle AOM\cong\triangle BOM$ (SAS)
$\angle AMO=\angle BMO$ (cpct)
$\angle AMO+\angle BMO=180^o\\ \\ \Rightarrow \angle AMO=\angle BMO=90^o\\ \\ \Rightarrow \angle BMO=\angle OPT=90^o$
But they are corresponds angle
So $AB||PT$
Question 28 of 44

28. Question
1 point(s)
In a right triangle ABC, in which $\angle B=90^o $, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.
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Hint

$\angle ABC=90^o\\ \\ \Rightarrow \angle 1+\angle 5=90$
$\angle 3=\angle 1$ (angle between tangent and the chord equals angle made by the chord in alternate segment)
$\angle 3+\angle 5=90^o$ …(1)
$\angle APB=90^o$ (angle in semicircle)
$\angle3+\angle4=90^o$ …(2)
From (1) and (2)
$\angle3+\angle5=\angle3+\angle4\\ \\ \Rightarrow \angle5=\angle4\\ \\ \Rightarrow PQ=QC$
Also $QP=QB$ (tangent drawn from an internal point are equal )
$\Rightarrow QB=QC$ (Proved)
Question 29 of 44

29. Question
1 point(s)
From a point P, two tangents PA and PB are drawn to a circle C (O, r). If OP = 2r, show that $\triangle $APB is equilateral.
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Hint

$OA\bot AP\\ \\ \Rightarrow \angle OAP=90^o$
In $\triangle OAP$
$\sin\angle OPA=\dfrac{OA}{OP}=\dfrac{r}{2r}=\dfrac{1}{2}\\ \\ \Rightarrow \angle =30^o$
Similarly it can be proced that $\angle OPB=30^o$
$\angle APB=\angle OPA+\angle OPB=30+30=60^o$
In $\triangle PAB$ ,
$PA=PB\\ \\ \Rightarrow \angle PAB=\angle PBA\\ \\ \angle PAB+\angle PBA+\angle APB=180^o\\ \\ \Rightarrow \angle PAB+\angle PBA=180-60=120\\ \\ \Rightarrow 2\angle PAB=60^o\\ \\ \angle PAB=\angle PBA=\angle APB=60^o$
$\therefore \triangle PAB$ is an equilateral $\triangle$
Question 30 of 44

30. Question
1 point(s)
O is the centre of a circle, PA and PB are tangents to the circle from a point P. Prove that (i) PAOB is a cyclic quadrilateral (ii) PO is the bisector of $\angle $APB. (iii) $\angle $OAB = $\angle $OPA
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Hint

(i) $\angle OBP=\angle OAP=90^o$
$\angle APB+\angle AOB+\angle OBP+\angle OAP=360^o\\ \\ \angle APB+\angle AOB=180^o$
Since the sum of the opposite angles of the quadrilateral is $180^o$
Hence $AOBP$ is a cyclic quadrilateral
(ii) In $\triangle APO$ and $\triangle BPO$
$\angle PAO=\angle PBO$ (each 90)
$OP=OP$ (common)
$OA=OB$ (radius)
$\triangle APO\cong\triangle BPO$ (RHS)
$\Rightarrow \angle APO=\angle BPO$ (cpct)
$OP$ is the bisector of $\angle APB$
Question 31 of 44

31. Question
1 point(s)
If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R respectively, prove that $AQ=\dfrac{1}{2}(AB+BC+CA) $
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Hint

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 )
Question 32 of 44

32. Question
1 point(s)
If an isosceles triangle ABC, in which AB = AC = 6 cm is inscribed in a circle of radius 9 cm, find the area of the triangle.
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Hint

Let $Ax=x\ cm$
$AB^2=AX^2+BX^2\\ \\ \Rightarrow 6^2=AX^2+BX^2\\ \\ \Rightarrow BX^2=36-x^2\\ \\ OA-9\ cm,\ Ax=x\ cm\\ \\ Ox=(9-x)\ cm\\ \\ OB^2=Ox^2+Bx^2\\ \\ \Rightarrow 9^2=(9-x)^2+(36-x)^2\\ \\ \Rightarrow 18x=36\Rightarrow x=2\\ \\ Bx=\sqrt{36-x^2}=\sqrt{36-4}=\sqrt{32}=4\sqrt2\ cm$
$\triangle ABC$ is an isosceles $\triangle$
$BX=CX\\ \\ \Rightarrow BC=2Bx=2\times 4\sqrt2=8\sqrt2$
Area of $\triangle ABC=\dfrac{1}{2}\times BC\times Ax$
$=\dfrac{1}{2}\times 8\sqrt2\times2=8\sqrt2\ cm^2$
Question 33 of 44

33. Question
1 point(s)
In the figure, from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular to the chord AB. Prove that :
(a) $PA. PB = PN^2 – AN^2 $
(b) $OP^2 – OT^2 = PN^2 – AN^2 $
(c) $ PA. PB = PT^2.$
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Hint

(i) $PA\cdot PB=(PN-AN)\ (PN+BN)$
$(PN-AN)\ (PN+AN)\\ \\ (\because AN=BN)\\ \\ PA\cdot PB=PN^2-AN^2$
(ii) $PN^2-AN^2=(OP^2-ON^2)-AN^2\\ \\ (\because In \ \triangle ONA,OA^2=ON^2+AN^2)\\ \\ PN^2-AN^2=OP^2-OT^2$
From parts (1) and (2)
$PA\cdot PB=OP^2-OT^2=PT^2\\ \\ (\because PT^2=OP^2-OT^2)\\ \\ \Rightarrow PA.PB=PT^2$
Question 34 of 44

34. Question
1 point(s)
If a hexagon ABCDEF circumscribes a circle, prove that AB + CD + EF = BC + DE + FA.
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Hint

$AM=AR$ ….(1)
$BM=BN$ …(2)
$CN=CO$ …(3)
$DO=DP$ …(4)
$EP=EQ$ …(5)
$FQ=FR$ …(6)
Adding (1) and (2)
$AM+BM=AR+BN\\ \\ \Rightarrow AB=AR+BN$
Adding (3) and (4)
$CO+DO=CN+DP\\ \\ \Rightarrow CD=CN+DP$
Adding (5) and (6)
$EQ+FQ=EP+ER \\ \\ \Rightarrow EF=EP+FR$
Adding we obtain
$AB+CD+EF=AR+(BN+CN)+(DP+EP)+FR\\ \\ \Rightarrow AB+CD+EF=BC+DE+FA$
Question 35 of 44

35. Question
1 point(s)
In the figure, O is the centre of the circle. If OR = 5 cm and OA = 13 cm, find the perimeter of $\triangle $ABC
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Hint

$AP=AQ\\ \\ BP=PR\\ \\ CR=CQ$
Perimeter of $\triangle ABC=AB+BR+RC+CA$
$=AB+BP+CQ+CA=2\ AP\\ \\ AO^2=AP^2+PO^2\\ \\ \Rightarrow 13^2=AP^2+5^2\Rightarrow AP=12$
Perimeter of $\triangle ABC=2\times12=24\ cm$
Question 36 of 44

36. Question
1 point(s)
The transverse common tangents AB and CD of two circles with centre O and O’ intersect at E. Prove that the points O, E and O’ are collinear.
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Hint

In $\triangle OAE$ and $\triangle OCE$
$OA=OC\\ \\ OE=OE$
$\angle OAE=\angle OCE\ (each\ 90)\\ \\ \triangle OAE\cong\triangle OCE\ (RHS)\\ \\ \angle AEO=\angle CEO\ (cpct)$
Similarly $\angle BEO'-\angle DEO'$
$\angle AEC=\angle DEB\\ \\ \dfrac{1}{2}\angle AEC=\dfrac{1}{2}\angle DEB\\ \\ \Rightarrow \angle AEO=\angle CEO=\angle BEO'=\angle DEO'$
Since there angles are equal and are bisected by OE and O’E.
O, E, O’ are collinear
Question 37 of 44

37. Question
1 point(s)
Let s denote the semi perimeter of a $\triangle $ABC in which BC = a, CA = b and AB = c. If a circle touches BC, CA and AB at D, E and F respectively, prove that BD = s – b.
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Hint

Let $AC=b$
$BD=BF=x\\ \\ CD=CE=z\\ \\ AF=AE=y$
Perimeter of $\triangle ABC=2(x+y+z)=25$
$x+y+z=s\\ \\ x=s-y-z\\ \\ =s-(y+z)\\ \\ =s-AC\\ \\ x=s-b\ (\because AC=b)$
Question 38 of 44

38. Question
1 point(s)
In the figure, tangents $PQ $ and $PR $ are drawn to a circle such that $\angle RPQ=30^o $. A chord $RS $ is drawn parallel to the tangents $PQ $. Find $ \angle RQS$.
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Hint

$\angle PRQ=30^o\\ \\ PR=PQ\\ \\ \Rightarrow \angle PRQ=\angle PQR$
In $\triangle PQR$
$\angle RQP+\angle QRP+\angle RPQ=180^o\\ \\ 2\angle RQP=150\\ \\ \angle RQP=75\\ \\ \angle RQP=\angle RSQ=75^o$
(Alternate segment theorem)
$\angle RQP=\angle SRQ=75^o$ (alt $\angle S$ )
$\Rightarrow \angle RSQ=\angle SRQ=75\\ \\ \angle RSQ+\angle SRQ+\angle RQS=180\\ \\ 75+75+\angle RQS=180\\ \\ \angle RQS=30^o$
Question 39 of 44

39. Question
1 point(s)
In the figure, the tangent at a point $C $ of a circle and a diameter $AB $ when extended intersect at $ P$. If $\angle PCA=110^o $, find $\angle CBA $.
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Hint

$\angle BCA=90^o$ (Angle in a semicircle)
$\angle OCP=90^o\\ \\ \Rightarrow \angle PCA=\angle PCO+\angle OCA\\ \\ \Rightarrow 110=90+\angle OCA\Rightarrow \angle OCA=20$
In $\triangle ABC$
$\angle CAB+\angle CBA+\angle BCA=180\\ \\ \angle CBA=180-(20+90)=70^o$
Question 40 of 44

40. Question
1 point(s)
OABC is a rhombus whose three vertices A,B and C lie on a circle with centre O. If the radius of the circle is 10 cm, find the area of the rhombus.
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Hint

$OABC$ is a rhombus
$\Rightarrow AB=BC=OA=OC=10\ cm\\ \\ OA=OB$
$\triangle OAB$ is an isosceles $\triangle$
$AB^2=AD^2+DB^2\\ \\ \Rightarrow AD^2=10^2-5^2=75\\ \\ AD=5\sqrt3\ cm\\ \\ AC=2\times AD=10\sqrt3\ cm$
Area of rhombus $=\dfrac{1}{2}\times OB\times AC$
$\Rightarrow \dfrac{1}{2}\times 10\sqrt3\times10=50\sqrt3\ cm^2$
Question 41 of 44

41. Question
1 point(s)
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
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Hint

$OA\bot PA,OB\bot PB$
In $\triangle OPA$ and $\triangle OPB$
$\angle OAP=\angle OBP$
$OA=OB$ (radius)
$OP=OP$ (common)
$\triangle OAP\cong\triangle OPB$ (RHS)
$PA=PB$ (cpct)
Thus it is proved that the length of the two tangents drawn from an external point to a circle are equal.
Question 42 of 44

42. Question
1 point(s)
Prove that the angle between the two tangents drawn from any external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
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Hint

AS radius of the circle is perpendicular to the tangent $OA\bot PA$
Similarly $OB\bot PB$
$\angle OBP=90^o,\ \angle OAP=90^o\\ \\ \angle OAP+\angle OBP+\angle BOA+\angle APB=360^o\\ \\ \Rightarrow \angle BOA+\angle APB=180^o$
It proves that the angle between the two tangents drawn from an external point to a circle supplementry to the angle subtended by the line segment.
Question 43 of 44

43. Question
1 point(s)
A circle touches the sides of a quadrilateral ABCD at P, Q R, S respectively. Show that angle subtended at the centre by pairs of opposite sides are supplementary.Using the above, find $\angle $PTQ in the figure. If TP and TQ are the two tangents to a circle with centre O so that $\angle POQ=110^o $.
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Hint

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre
$\angle 1=\angle 2,\angle 3=\angle 4,\angle 5=\angle 6,\angle 7=\angle 8\\ \\ \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle8=360^o\\ \\ 2(\angle2+\angle3+\angle6+\angle7)=360^o$ and $2(\angle 1+\angle8+\angle4+\angle5)=360^o\\ \\ \Rightarrow \angle2+\angle3+\angle6+\angle7=180$ and $\angle1+\angle8+\angle4+\angle5=180^o\\ \\ \angle AOB+\\angle COD=180^o\ (\because \angle AOB=\angle2+\angle3)\\ \\ \angle AOD+\angle BOC=180^o\\ \\ \angle POQ+\angle PTQ=180^o\\ \\ \angle PTQ=180-110=70^o$
Question 44 of 44

44. Question
1 point(s)
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using the above , do the following :In figure, O is the centre of the two concentric circles. AB is a chord of the larger circle touching the smaller circle at C. Prove that AC = BC.
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Hint

$OA=OC$ (radii)
$OB=OC+BC$
$OB>OC$ (radius is $OC$ )
$OAThis OA is shorter than any other line segment joining O to any point on tangent. Hence the tangent at any point of circle is perpendicular to the radius [latex]AS\ OC\bot AB$
So $AC=BC$ (as perpendicular from the centre bisects the chord)