Math Class 9 Question Bank

Question 1 of 43

1. Question
1 point(s)
Two consecutive angles of a parallelogram are in the ratio 1 : 3. Then the smaller angle is :
- 50°
- 90°
- 60°
- 45°
Correct

Incorrect

Hint

$2(1x+3x)=360^o\\ \\ 4x=180^o\Rightarrow x=\dfrac{180}{4}=45^o$
The smaller angle is $45^o$
Question 2 of 43

2. Question
1 point(s)
ABCD is a rhombus. Diagonal AC is equal to one of its sides. Then $\triangle $ ABC
- a right angled triangle
- an equilateral triangle
- an isosceles triangle
- none of these
Correct

Incorrect

Hint

$AB=AC=BC$
Question 3 of 43

3. Question
1 point(s)
In the given figure, ABCD is a rhombus. If $\angle A=80^o $, then $\angle CDB $ is equal to
- 80°
- 90°
- 50°
- 100°
Correct

Incorrect

Hint

$\angle C=80^o$
$\angle b=\angle d=100^o$ (opposite angles are equal)
$\angle A=80^o\\ \\ \angle CDB=50^o$
Question 4 of 43

4. Question
1 point(s)
The sum of three angles of a quadrilateral is 3 right angles. Then the fourth angle is a/an :
- Right angle
- Obtuse angle
- Acute angle
- Reflex angle
Correct

Incorrect

Hint

$3\times90+$ fourth angle $=360^o$
fourth angle $=360-270=90$
Question 5 of 43

5. Question
1 point(s)
In the figure, D, E and F are the mid-points of the sides AB, BC and CA respectively. If AC = 8.2 cm, then value of DE is :
- 8.2 cm
- 4.1 cm
- 2.05 cm
- None of these
Correct

Incorrect

Hint

$DE=\dfrac{1}{2}AC=\dfrac{1}{2}\times8.2=4.1$
Question 6 of 43

6. Question
1 point(s)
In a rectangle ABCD, diagonals AC and BD intersect at O. If AO = 3 cm, then the length of the diagonal BD is equal to :
- 3 cm
- 9 cm
- 6 cm
- 12 cm
Correct

Incorrect

Hint

$AO=3\ cm\\ \\ AC=6\ cm$
$AC=BD=6\ cm$ (Diagonals of a rectangle are equal)
Question 7 of 43

7. Question
1 point(s)
If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form :
- A square
- A rhombus
- A rectangle
- Any other parallelogram
Correct

Incorrect

Hint

$\angle APQ=\angle PQD\ (\because\ APB||CQD)\\ \\ 2\angle MPQ=2\angle NQP\\ \\ \angle MPQ=\angle NQP\Rightarrow PM||QN$
Similarly $\angle BPQ=\angle CQP\Rightarrow PN||QM$
So, $PNQM$ is a parallelogram
Now $\angle CQP+\angle PQD=180^o$
$2\angle MQP+2\angle NQP=180^o\\ \\ \angle MQP+\angle NQP=90^o\\ \\ \angle MQN=90^o$
Hence it is a rectangle
Question 8 of 43

8. Question
1 point(s)
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is :
- A rhombus
- A rectangle
- A square
- Any parallelogram
Correct

Incorrect

Hint

It will form rectangle.
Question 9 of 43

9. Question
1 point(s)
The sides BA and DC of a quadrilateral ABCD are produced as shown in the figure. Then which of the following relations is true?
- $x+y=\angle1+\angle2+\angle3+\angle4 $
- $x-y=\angle1+\angle2+\angle3+\angle4 $
- $x+y=2(\angle1+\angle2+\angle3+\angle4) $
- none of these
Correct

Incorrect

Hint

$x=\angle1+\angle2\\ \\ y=\angle3+\angle4\\ \\ x+y=\angle1+\angle2+\angle3+\angle4$
Question 10 of 43

10. Question
1 point(s)
In the figure, if ABCD is a square, then value of x is :
- 50°
- 55°
- 80°
- 60°
Correct

Incorrect

Hint

$\angle COP=180-100=80^o$ (Linear pair)
$\angle OCP=\dfrac{90}{2}=45^o\\ \\ x=180-(80+45)=55$
Question 11 of 43

11. Question
1 point(s)
Lengths of two adjacent sides of a parallelogram are in the ratio 2 : 7. If its perimeter is 180 cm, then the adjacent sides of the parallelogram are
- 10 cm, 20 cm
- 20 cm, 70 cm
- 41 cm, 140 cm
- None of these
Correct

Incorrect

Hint

$2(2x+7x)=180\ cm\\ \\ 9x=90\\ \\ x=10$
Adjacent sides are $20\ cm,\ 10\ cm$
Question 12 of 43

12. Question
1 point(s)
The triangle formed by joining the mid points of the sides of a right angled triangle is :
- An acute angled triangle
- An obtuse angled triangle
- A right angled triangle
- None of these
Correct

Incorrect

Hint

The triangle formed by joining the midpoint of the sides of a right triangle also have are right angle.
Question 13 of 43

13. Question
1 point(s)
The diagonals of a rectangle PQRS intersect at $O.\ \angle ROQ=60^o $, then $\angle OSP $ is equal to
- 90°
- 120°
- 60°
- None of these
Correct

Incorrect

Hint

$\angle SOP=60^o$ (VOA)
$\angle OPS=\dfrac{90}{2}=45^o\\ \\ \angle OSP=180-(60+45)\\ \\ =180-105=75$
Question 14 of 43

14. Question
1 point(s)
In the $\triangle $ABC, $\angle $B is a right angle, D and E are the mid-points of the sides AB and AC respectively. If AB = 6 cm and AC = 10 cm, then the length of DE is :
- 3 cm
- 5 cm
- 4 cm
- 6 cm
Correct

Incorrect

Hint

$DE=\dfrac{1}{2}BC\\ \\ BC=\sqrt{100-36}=\sqrt{64}=8\\ \\ DE=4\ cm$
Question 15 of 43

15. Question
1 point(s)
In a $\triangle $ABC, D, E and F are respectively the mid-points of BC, CA and AB as shown in the figure. The perimeter of $\triangle $DEF is :
- $\dfrac{1}{2}(AB+BC+CA) $
- $AB+BC+CA $
- $2\ (AB+BC+CA) $
- none of these
Correct

Incorrect

Hint

Perimeter of $\triangle DEF=DE+EF+DF$
$=\dfrac{1}{2}AB+\dfrac{1}{2}BC+\dfrac{1}{2}CA\\ \\ =\dfrac{1}{2}(AB+BC+CA)$
Question 16 of 43

16. Question
1 point(s)
In a parallelogram ABCD $\angle A=(3x+15^o) $ and $\angle B=(5x-35^o) $. The measure of $\angle D $ is
- $125^o $
- $90^o $
- $180^o $
- cannot be determined
Correct

Incorrect

Hint

$3x+15+5x-35=180$ (Linear pair)
$8x-20=180\\ \\ 8x=200\\ \\ x=\dfrac{200}{8}=25$
$\angle A=\angle D$ (opposite angles of parallelogram)
$\angle A=3\times25+15=75+15=90^o$
Question 17 of 43

17. Question
1 point(s)
In the given, if ABCD is a parallelogram, then the value of $2\angle ABC-\angle ADC $ is:
- 40°
- 220°
- 70°
- 75°
Correct

Incorrect

Hint

$\angle C=180-70=110^o\\ \\ \angle B=180-110=70\\ \\ \angle A=\angle C=110^o\\ \\ \angle B=\angle D=70^o\\ \\ 2\times70-70=140-70=70$
Question 18 of 43

18. Question
1 point(s)
If one angle of a parallelogram is 56° more than three times of its adjacent angle, then measures of all the angles are :
- 31°, 149°, 31°, 149°
- 59°, 121°, 59°, 121°
- 37°, 143°, 37°, 143°
- None of these
Correct

Incorrect

Hint

$x+3x+56=180\\ \\ 4x=124\Rightarrow x=31$
other angles are $3x+56$
$=3\times31+56\\ \\ =93+56=149$
Question 19 of 43

19. Question
1 point(s)
The quadrilateral formed by joining the mid– points of the sides of a quadrilateral PQRS, taken in order is the rectangle, if :
- PQRS is a rectangle
- PQRS is a parallelogram
- diagonals of PQRS are perpendicular
- diagonals of PQRS are equal
Correct

Incorrect

Hint

If diagonals of PQRS are perpendicular to each other than ABCD is a rectangle.
Question 20 of 43

20. Question
1 point(s)
The quadrilateral formed by joining the mid– points of the sides of a quadrilateral PQRS, taken in order is a rhombus, if :
- PQRS is a rhombus
- PQRS is a parallelogram
- diagonals of PQRS are perpendicular
- diagonals of PQRS are equal
Correct

Incorrect

Hint

$ABCD$ is a rhombus
So, $AB=BC=CD=AD$
In $\triangle PQS$
$DC=\dfrac{1}{2}QS$
In $\triangle PSR$
$BC=\dfrac{1}{2}PR\\ \\ \dfrac{1}{2}QS=\dfrac{1}{2}PR\ (\because\ BC=DC)\\ \\ QS=PR$
Question 21 of 43

21. Question
1 point(s)
If the angles of a quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a :
- rhombus
- kite
- Parallelogram
- trapezium
Correct

Incorrect

Hint

$3x+7x+6x+4x=360\\ \\ x=18^o\\ \\ \angle A=3\times18=54\qquad \angle C=6\times18=108\\ \\ \angle B=7\times18=126\qquad \angle D=4\times18=72\\ \\ \angle A+\angle B=180^o\\ \\ \angle C+\angle D=180^o$
So, $ABCD$ is a trapezium
Question 22 of 43

22. Question
1 point(s)
Two adjacent angles of a rhombus are $3x-40^o $ and $2x+20^o $. The measurement of the greater angle is :
- $160^o $
- $100^o $
- $80^o $
- $120^o $
Correct

Incorrect

Hint

$3x-40+2x+20=180$ (Linear pair)
$5x-20=180\Rightarrow 5x=200\\ \\ \Rightarrow x=40\\ \\ 3x-40=3\times40-40=120-40=80\\ \\ 2x+20=2\times40+20=100$
Question 23 of 43

23. Question
1 point(s)
$ABCD $ is a parallelog ram in which $\angle DAC=40^o.\ \angle BAC=30^o;\ \angle DOC=105^o $ then $\angle CDO $ equals:
- $75^o $
- $70^o $
- $45^o $
- $85^o $
Correct

Incorrect

Hint

$\angle OCD=30^o$ (alt $\angle S$ )
$\angle CDO=180-(30+105)\\ \\ =180-135=45$
Question 24 of 43

24. Question
1 point(s)
Angles of a quadrilateral are in the ratio : 3 : 6 : 8 : 13. The largest angle is :
- $178^o $
- $90^o $
- $156^o $
- $36^o $
Correct

Incorrect

Hint

$3x+6x+8x+13x=360\\ \\ 30x=360\Rightarrow x=12\\ \\ 13x=13\times12=156$
Question 25 of 43

25. Question
1 point(s)
ABCD is a rhombus such that one of its diagonals is equal to its side. Then the angles of rhombus ABCD are
- $45^o,135^o,45^o,135^o $
- $100^o,80^o,100^o,80^o $
- $120^o,60^o,120^o,60^o $
- $60^o,60^o,60^o,60^o $
Correct

Incorrect

Hint

The diagonal will divide the rhombus in two equilateral $\triangle S$
$x+x+x=180\Rightarrow x=60$
Two angle as $60^o$ other two angle $=180-60$
$=120$ (Linear pair)
Question 26 of 43

26. Question
1 point(s)
D, E, F are midpoints of sides BC, CA and AB of $\triangle $ABC . If perimeter of $\triangle $ABC is 12.8 cm, then perimeter of $\triangle $DEF is
- 17 cm
- 38.4 cm
- 25.6 cm
- 6.4 cm
Correct

Incorrect

Hint

$DE=\dfrac{1}{2}AB\\ \\ DF=\dfrac{1}{2}AC\\ \\ EF=\dfrac{1}{2}BC\\ \\ DE+DF+EF=\dfrac{1}{2}\ (AB+AC+BC)\\ \\ =\dfrac{1}{2}\times12.8=6.4\ cm$
Question 27 of 43

27. Question
1 point(s)
Two adjacent angles of a parallelogram are $(2x+30)^o $ and $(3x+30)^o $. The value of $x $ is:
- $30^o $
- $60^o $
- $24^o $
- $36^o $
Correct

Incorrect

Hint

$2x+30+3x+30=180$ (Linear pair)
$5x+60=180\\ \\ 5x=120\Rightarrow x=24$
Question 28 of 43

28. Question
1 point(s)
In the figure, $ABCD $ is a parallelogram. If $\angle DAB=60^o $ and $\angle DBC=80^o $, then $\angle CDB $ is:
- $40^o $
- $80^o $
- $60^o $
- $20^o $
Correct

Incorrect

Hint

$\angle C=\angle A=60^o$ (opposite angles of a parallelogram)
In $\triangle DCB$
$\angle CDB=180-(80+60)\\ \\ =180-140=40$
Question 29 of 43

29. Question
1 point(s)
In the figure. ABCD is a parallelogram. If $\angle B=100^o $. then $(\angle A+\angle C) $ is equal to:
- $360^o $
- $200^o $
- $180^o $
- $160^o $
Correct

Incorrect

Hint

$\angle B=\angle D=100\\ \\ \angle B+\angle D=100+100=200\\ \\ \angle A+\angle B+\angle C+\angle D=360\\ \\ \angle A+\angle C=360-200=160$
Question 30 of 43

30. Question
1 point(s)
All the angles of a convex quadrilateral are congruent. However, not all its sides are congruent. What type of quadrilateral is it ?
- parallelogram
- square
- rectangle
- trapezium
Correct

Incorrect

Hint

Rectangle as all angles of a rectangle measures $90^o$ but all sides are not equal.
Question 31 of 43

31. Question
1 point(s)
ABCD is a quadrilateral and AP and DP are bisectors of $\angle $A and $\angle $D .The value of x is :
- $60^o $
- $85^o $
- $95^o $
- $100^o $
Correct

Incorrect

Hint

$\angle A+\angle B=180\\ \\ \angle A=50\\ \\ \angle C+\angle D=180\\ \\ \angle D=120\\ \\ \dfrac{1}{2}\angle A+\dfrac{1}{2}\angle D+x=180\\ \\ 25+60+x=180\Rightarrow x=95^o$
Question 32 of 43

32. Question
1 point(s)
If $\angle C=\angle D=50^o $, then four points $A,B,C,D $:
- are concyclic
- do not lie on same circle
- are collinear
- A, B, D and A, B, C lie on different circles
Correct

Incorrect

Hint

Since $\angle ADB=\angle ACB$ and both are subtended by the same line segment $AB$ at different points $C$ and $D$ , these angles must lie in the same segment of a circle which passes through $A,B,C,D$
So, all the four points are concyclic
Question 33 of 43

33. Question
1 point(s)
If $PQRS $ is a parallelogram, then $\angle Q-\angle S $ is equal to
- $90^o $
- $120^o $
- $180^o $
- $0^o $
Correct

Incorrect

Hint

$\angle Q=\angle S$ (opposite angles are equal)
$\angle Q-\angle S=0$
Question 34 of 43

34. Question
1 point(s)
Which of the following is not true for a parallogram ?
- opposite sides are equal
- opposite angles are equal
- opposite angles are bisected by diagonals
- diagonals bisect each other
Correct

Incorrect

Hint

In a parallelogram, the opposite angles are not bisected by the diagonals.
Question 35 of 43

35. Question
1 point(s)
If APB and CQD are parallel lines and a transversal PQ cut then at P and Q, then the bisectors of angle APQ. BPQ. CQP and PQD from a
- rectangle
- rhombus
- square
- any other parallelogram
Correct

Incorrect

Hint

$\angle APQ=\angle PQD\ (\because\ APB||CQD)\\ \\ 2\angle MPQ=2\angle NQP\\ \\ \angle MPQ=\angle NQP\Rightarrow PM||QN$
Similarly $\angle BPQ=\angle CQP\Rightarrow PN||QM$
So, $PNQM$ is a parallelogram
Now $\angle CQP+\angle PQD=180^o$
$2\angle MQP+2\angle NQP=180^o\\ \\ \angle MQP+\angle NQP=90^o\\ \\ \angle MQN=90^o$
Hence it is a rectangle
Question 36 of 43

36. Question
1 point(s)
$ABCD $ is a rhombus such that $\angle ACB=40^o $, then $\angle ADB $
- $40^o $
- $45^o $
- $50^o $
- $60^o $
Correct

Incorrect

Hint

In $\triangle OBC$
$\angle OBC=180-(40+90)=50$
In $\triangle BOC$ and $\triangle AOD$
$\angle BOC=\angle DOA$ (VOA)
$OC=OA$
$BO=OD$ (diagonals bisect each other)
$\triangle BOC\cong\triangle DOA$ (SAS)
$\angle OBC=\angle ADB=50^o$
Question 37 of 43

37. Question
1 point(s)
If the diagonals AC and BD of a quadrilateral ABCD bisect each other, then ABCD is a :
- parallelogram
- rectangle
- rhombus
- trepezium
Correct

Incorrect

Hint

If diagonals of a quadrilateral bisect each other then it is a parallelogram.
Question 38 of 43

38. Question
1 point(s)
The diagonals of a parallelogram $PQRs$ intersect at $ Q$. If $\angle QOR=90^o $ and $\angle QSR=50^o $, then $\angle ORS $ is
- $90^o $
- $40^o $
- $70^o $
- $50^o $
Correct

Incorrect

Hint

$\angle SOR=180-90=90$ (Linear pair)
In $\triangle SOR$
$\angle ORS=180-(90+50)=180-140=40^o$
Question 39 of 43

39. Question
1 point(s)
A quadrilatral whose diagonals are equal and bisect each other at right angles is a :
- rhombus
- square
- trapezium
- rectangle
Correct

Incorrect

Hint

A quadrilateral whose diagonals are equal and bisect each other at $90^o$ is a square.
Question 40 of 43

40. Question
1 point(s)
In the given figure, ABCD is a rhombus in which diagonals AC and BD intersect at O. Then $\angle $AOB is:
- $60^o $
- $80^o $
- $90^o $
- $45^o $
Correct

Incorrect

Hint

The diagonals of a rhombus intersect each other at right-angles.
So, $\angle AOD=90^o$ .
Question 41 of 43

41. Question
1 point(s)
In an equilateral triangle ABC, D and E are the mid points of sides AB and AC respectively. Then length of DE is :
- not possible to find
- $3\ \text{cm} $
- $\dfrac{1}{2}\ \text{BC} $
- $\dfrac{3}{2}\ \text{BC} $
Correct

Incorrect

Hint

$D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$ .
$DE=\dfrac{1}{2}BC$ (by midpoint theorem)
Question 42 of 43

42. Question
1 point(s)
In the figure, $ PQRS$ is a rectangle. If $\angle RPQ=30^o $, then the value of $ (x+y)$ is:
- $90^o $
- $120^o $
- $150^o $
- $180^o $
Correct

Incorrect

Hint

$\angle RPQ=30^o\\ \\ \angle SPR=60^o\\ \\ \angle SPR=\angle SQR$
$\triangle QOR$ is an equlateral $\triangle$
$\angle QOR=60^o,\ \angle SOR=120^o\\ \\ \angle SPR+\angle SOR=60+20=180$
Question 43 of 43

43. Question
1 point(s)
In a $\triangle $ABC . E and F are the mid–points of AB and AC respectively. The altitude AP intersects EF at Q. The correct relation between AQ and QP is :
- AQ > QP
- AQ = QP
- AQ < QP
- None of these
Correct

Incorrect

Hint

$EF=\dfrac{1}{2}BC$
Since $E$ is the mid point of $AB,EQ||DP$ by converse of mid point theorem $Q$ is the mid point of $AP$
$AQ=QP$