Maths Class X Question Bank

Question 1 of 27

1. Question
1 point(s)
The area of a square inscribed in a circle of radius 8 cm is
- $64\ cm^2 $
- $100\ cm^2 $
- $125\ cm^2 $
- $128\ cm^2 $
Correct

Incorrect

Hint

Let ABCD be the square inscribed by a circle
$\therefore OA=OB=OC=OD$
$AB=8+8=16$
$AC^2=AB^2+BC^2=2AB^2\\ \\ AB=8\sqrt2$
Area of square $=\left(\dfrac{8}{2}\right)^2$
$=128\ cm^2$
Question 2 of 27

2. Question
1 point(s)
A line that intersects a circle in two distinct points is called :
- a diameter
- a secant
- a tangent
- a radius
Correct

Incorrect

Hint

Secant is a line that intersects a circle in two distinct points.
Question 3 of 27

3. Question
1 point(s)
Two circles touch each externally at C and AB is a common tangent to the circles, then $\angle $ACB is:
- 60°
- 45°
- 30°
- 90°
Correct

Incorrect

Hint

Let $\angle PAC=\alpha$ ,
$\angle PBC=\beta$
$\angle CAB+\angle CBA+\angle ACB=180^o\\ \\ \alpha+\beta+(\alpha+\beta)=180^o\\ \\ 2(\alpha+\beta)=180\\ \\ \alpha+\beta=90^o\\ \\ \alpha ACB=90^o$
Question 4 of 27

4. Question
1 point(s)
The angle between two tangents drawn from an external point to a circle is $110^o $. The angle subtended at the centre by the segments joining the points of contact to the centre of circle is :
- 70°
- 110°
- 90°
- 55°
Correct

Incorrect

Hint

$\angle PAO=\angle PBO=90^o\\ \\ 110+90+90+\angle AOB=360\\ \\ \angle AOB=70$
Question 5 of 27

5. Question
1 point(s)
In the figure, O is centre of a circle. MN is a chord and the tangent ML at M makes an angle of $70^o $ with MN. $\angle $MON is equal to:
- 120°
- 90°
- 140°
- 70°
Correct

Incorrect

Hint

$\angle OMN=90-70=20^o\\ \\ \angle OMN=\angle ONM=20^o\ (\because\ OM=ON)\\ \\ \angle MON=180-(20+20)\\ \\ =180-40=140$
Question 6 of 27

6. Question
1 point(s)
The distance between two parallel tangents of a circle of radius 5 cm is
- 5 cm
- 10 cm
- 15 cm
- 2.5 cm
Correct

Incorrect

Hint

Distance between two parallel tangents $=5+5=10\ cm$
Question 7 of 27

7. Question
1 point(s)
A quadrilateral ABCD is drawn to circumscribe a circle. If AB = 12 cm, BC = 15 cm and CD = 14 cm, then AD is
- 10 cm
- 11 cm
- 12 cm
- 14 cm
Correct

Incorrect

Hint

$a+b=12$ …(1)
$a+c=15$ …(2)
$c+d=14$ …(3)
$a+b=12\\c+d=14\\ a+b+c+d=26$
$15+(b+d)=26\\ \\ (b+d)=11\\ \\ AD=11\ cm$
Question 8 of 27

8. Question
1 point(s)
If tangents PA and PB from an external point P to a circle with centre O are inclined to each other at an angle of $ 80^o$, then $\angle $POA is equal to
- 50°
- 60°
- 70°
- 80°
Correct

Incorrect

Hint

$\angle OPA=\dfrac{1}{2}\times80=40$
In $\triangle OPA$
$\angle POA+\angle OPA+\angle OAP=180^o\\ \\ \angle POA=180-130=50$
Question 9 of 27

9. Question
1 point(s)
If two tangents inclined at an angle of $60^o $ are drawn to a circle of radius 3 cm, then the length of each tangent is equal to :
- $\dfrac{3\sqrt3}{2}\ cm $
- $2\sqrt3\ cm $
- $3\sqrt3\ cm $
- $6\ cm $
Correct

Incorrect

Hint

$\dfrac{OB}{AB}=\tan30\\ \\ \Rightarrow AB=\dfrac{3}{1/\sqrt3}=3\sqrt3$
Question 10 of 27

10. Question
1 point(s)
In the figure, AB, AC, PQ are tangents. If AB = 5 cm, then perimeter of $ \triangle$APQ is:
- 10 cm
- 15 cm
- 12.5 cm
- 20 cm
Correct

Incorrect

Hint

$AB=AC=5\ cm\\ \\ PX=PB\\ \\ QX=QC$
Perimeter $=AP+AQ+(PA+QC)$
$=(AP+PB)+(AQ+QC)\\ \\ =AB+AL=5+5=10\ cm$
Question 11 of 27

11. Question
1 point(s)
In the figure, $PQ $ and $PR $ are the tangnets to the circle with centre O such that $\angle QPR=50^o $. Then $\angle OQR $ is equal to:
- 25°
- 30°
- 40°
- 50°
Correct

Incorrect

Hint

$\angle QPR=50\\ \\ \angle QOR+\angle QPR=180^o\\ \\ \angle QOR=180-50=130\\ \\ \angle OQR=\angle ORQ=\dfrac{180-130}{2}=\dfrac{50}{2}=25^o$
Question 12 of 27

12. Question
1 point(s)
In two concentric circles, if chords are drawn in the outer circle which touch the inner circle, then
- all chords are of different lengths
- all chords are of same length
- only parallel chords are of same length
- only perpendicular chords are of same length
Correct

Incorrect

Hint

All chords are of the same length as their distances from the centre are equal which is radius of the inner circle.
Question 13 of 27

13. Question
1 point(s)
A tangent $PQ $ at the point $P $ of a circle meets a line through the centre $ O$ at a point $Q $, so that $OQ=12\ cm $ and $PQ=\sqrt{119}\ cm $, the diameter of circle is
- 13 cm
- 26 cm
- 10 cm
- 5 cm
Correct

Incorrect

Hint

$OP=\sqrt{(12)^2-(\sqrt{119})^2}\\ \\ =\sqrt{144-119}\\ \\ =\sqrt{25}=5$
Diameter $=2\times5=10\ cm$
Question 14 of 27

14. Question
1 point(s)
In the figure, a quadrilateral ABCD is drawn to circumscrible a circle. Then
- AD + BC = AB + CD
- AB + BC = AD + CD
- BC + CD = AD + AB
- AB + BC + CD + AD = AC + BD
Correct

Incorrect

Hint

$AG=AF\\ \\ BG=BH\\ \\ CE=CH\\ \\ DF=DE\\ \\ AG+BG+CE+DE=AF+BH+CH+DF\\ \\ AB+CD=AD+BC$
Question 15 of 27

15. Question
1 point(s)
In the figure, if the semiperimeter of $\triangle $ABC is 23 cm, then AF + BD + CE is
- 46 cm
- 11.5 cm
- 23 cm
- 34.5 cm
Correct

Incorrect

Hint

$AB+BC+AC=2\times23=46\ cm\\ \\ AF+BF+BD+DC+AE+CE=46\\ \\ 2(AF+BD+CE)=46\\ \\ AF+BD+CE=23\ cm$
Question 16 of 27

16. Question
1 point(s)
In the figure, AP = 2 cm, BQ = 3 cm and RC = 4 cm, then the perimeter of $\triangle $ABC (in cm) is :
- 16
- 18
- 20
- 21
Correct

Incorrect

Hint

$AP=AR=2\ cm\\ \\ BQ=BP=3\ cm\\ \\ RC=CQ=4\ cm\\ \\ AB=2+3=5\ cm\\ \\ BC=3+4=7\ cm\\ \\ AC=2+4=6\ cm\\ \\ AB+BC+CA=5+7+6=18\ cm$
Question 17 of 27

17. Question
1 point(s)
In the figure, AQ, AR and BC are tangents to circle with centre O. If AB = 7 cm, BC = 5 cm, AC =5 cm, then length of the tangent AQ is :
- 5 cm
- 7 cm
- 8.5 cm
- 17 cm
Correct

Incorrect

Hint

$AR=AQ\\ \\ BQ=BS\\ \\ CR=CS\\ \\ AB+BC+CA=7+5+5=17\ cm\\ \\ AB+BS+SC+AC=17\ cm\\ \\ AB+BQ+CR+AC=17\ cm\\ \\ AQ+AR=17\ cm\\ \\ 2AQ=17\ cm\\ \\ AQ=\dfrac{17}{2}=8.5\ cm$
Question 18 of 27

18. Question
1 point(s)
In the figure, triangle ABC is circumscribing a circle. Then the length of BC is :
- 7 cm
- 8 cm
- 9 cm
- 10 cm
Correct

Incorrect

Hint

$BL=BN=4\ cm\\ \\ LC=CM=6\ cm\\ \\ BC=BL+LC=6+4=10\ cm$
Question 19 of 27

19. Question
1 point(s)
In the figure, angle $OBC=30^o $, then value of x is
- $100^o $
- $110^o $
- $30^o $
- $150^o $
Correct

Incorrect

Hint

$\angle OBC=30^o\\ \\ \angle OBD=90^o\\ \\ \angle CBD=60^o\\ \\ \angle OBC=\angle OCB$
In $\triangle ABC$
$\angle BAC+\angle ACB=\angle CBD\\ \\ x+30=60\\ \\ x=30^o$
Question 20 of 27

20. Question
1 point(s)
From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilaterla PQOR is:
- $60\ cm^2 $
- $65\ cm^2 $
- $30\ cm^2 $
- $32.5\ cm^2 $
Correct

Incorrect

Hint

$PQ=\sqrt{13^2-5^2}=12\ cm$
Area of $\triangle POQ=\dfrac{OQ\times PQ}{2}=30\ cm^2$
Area of quadrilateral $PQOR=60\ cm^2$
Question 21 of 27

21. Question
1 point(s)
In the figure, $AT $ is the tangent to the circle with centre $O $ such that $OT=4\ cm $ and $\angle OTA=30^o $. Then $AT $ is equal to:
- $4 \ cm $
- $2\ cm $
- $2\sqrt3\ cm $
- $8\ cm $
Correct

Incorrect

Hint

$\cos30=\dfrac{AT}{OT}=\dfrac{AT}{4}\\ \\ \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{AT}{4}\Rightarrow AT=\dfrac{4\sqrt3}{2}=2\sqrt3$
Question 22 of 27

22. Question
1 point(s)
In the figure, $ O$ is the centre of the circle with $\angle TQM=35^o $, then angle $ ATQ$ would be equal to
- 55°
- 56°
- 35°
- 54°
Correct

Incorrect

Hint

$OQ=OT\\ \\ \angle OQT=\angle OTQ=35$
But $\angle OQT+\angle ATQ=90^o$
$\angle ATQ=90-35=55$
Question 23 of 27

23. Question
1 point(s)
In the figure, $AB $ is a chord of cirlce and $AOC $ is diameter such that $\angle ACB=55^o $. If $ AT$ is a tangent to the circle at point $A $, then angle $BAT $ is :
- 65°
- 40°
- 50°
- 55°
Correct

Incorrect

Hint

$\angle ABC=90^o\\ \\ 55+90+\angle BAC=180\\ \\ \angle BAC=35^o\\ \\ \angle BAT+\angle CAB=90\\ \\ \angle BAT=90-35=55^o$
Question 24 of 27

24. Question
1 point(s)
A circle touches all the four sides of quadrilaterial ABCD whose sides are AB = 6 cm, BC = 7 cm, CD = 4 cm. The length of side AD is :
- 5 cm
- 7 cm
- 6.5 cm
- 3 cm
Correct

Incorrect

Hint

$AP=AS\\ \\ BQ=BP\\ \\ DR=DS\\ \\ RC=CQ\\ \\ (AP+BP)+(DR+RC)\\ \\ =AS+BQ+OS+CQ\\ \\ \Rightarrow (AP+BP)+(DR+RC)=(AS+DS)+(BQ+CQ)\\ \\ \Rightarrow AB+DC=AD+BC\\ \\ \Rightarrow 6+4=AD+7\\ \\ \Rightarrow AD=3\ cm$
Question 25 of 27

25. Question
1 point(s)
PQ is a tangent to a circle with centre O at point P. If $\triangle $OPQ is an isosceles triangle, then $\angle $OQP is equal to:
- 15°
- 30°
- 45°
- 60°
Correct

Incorrect

Hint

$OP=OQ\\ \\ \Rightarrow \angle OQP=\angle QOP\Rightarrow \angle OQP+\angle QOP=2\angle OQP\\ \\ \angle OPQ+2\angle OQP=180\\ \\ \Rightarrow 2\angle OQP=90\Rightarrow OQP=45^o$
Question 26 of 27

26. Question
1 point(s)
In the figure, PQR is the tangent to a circle at Q whose centre is O. AB is a chord parallel to PR and $\angle BQR=70^o $, then angle $\angle AQB $ is equal to
- 20°
- 40°
- 35°
- 45°
Correct

Incorrect

Hint

$\angle OQR=\angle DQR=90^o\\ \\ \angle DQR=\angle QBD+\angle BQR\\ \\ \Rightarrow 90=\angle QBD+70\Rightarrow \angle QBD=20^o\\ \\ \angle BDQ+\angle DQR=180\\ \\ \angle BDQ=180-90=90\\ \\ \angle ADQ+\angle BDQ=180\\ \\ \angle ADQ=90$
In $\triangle AQD$ and $\triangle BQD$
$OD=QD$ (common)
$\angle ADQ=\angle BDQ$
$AD=BD$ ( Perpendicular from the centre bisect two chord)
$\triangle AQD\cong\triangle BQD$ (SAS)
$\angle AQD=\angle BQD$ (cpct)
$\angle AQD+\angle BQD=\angle AQB\\ \\ \angle AQB=2\angle BQD=2\times20=40$
Question 27 of 27

27. Question
1 point(s)
In the figure, the pair of tangents AP and AQ, drawn from an external point A to a circle with centre O, are perpendicular to each other and length of each tangent is 4 cm, then the radius of the circle is :
- 10 cm
- 7.5 cm
- 2.5 cm
- 4 cm
Correct

Incorrect

Hint

$\angle PAO=\angle QAO=45\\ \\ \angle OPA=90^o\\ \\ \angle OAP=\angle AOP=45\\ \\ OP=AP=4\ cm$