Mathematics Class IX

Question 1 of 30

1. Question
1 point(s)
If two adjacent angles of a parallelogram are in the ratio 2:4 then the measure of the angles are.
- $60^o,300^o $
- $60^o,120^o $
- $60^o,129^o $
- $67^o,116^o $
Correct

Incorrect

Hint

$2x+4x=180^o$ (adjacent angles are supplementary)
$6x=180^o\\ \\ x=30^o\\ \\ \text{Angles are}\ 2x=2\times30^o=60^o\\ \\ 4x=4\times30^o=120^o$
Question 2 of 30

2. Question
1 point(s)
In a rhombus ABCD, the diagonals bisect at o. what kind of a triangle AOB is?
- An isosceles right angled triangle.
- Neither isosceles nor a right angles triangle.
- an isosceles but not right angles triangle.
- a right angled but not an isosceles triangle.
Correct

Incorrect

Hint

The triangle AOB is a right angled triangle but not an isosceles triangle
Question 3 of 30

3. Question
1 point(s)
In the quadrilateral ABCD, AO and DO are the bisectors of $\angle A $ and $\angle D $ respectively.
If $\angle B=90^o $ and $\angle C=99^o $ then $\angle AOD $ is
- $94.5^o $
- $107.5^o $
- $189^o $
- $79.5^o $
Correct

Incorrect

Hint

$\angle B=90^o\\ \\ \angle C=99^o\\ \\ \angle B=180^o-90^o=90^o\\ \\ \angle D=180^o-90^o=81^o\\ \\ \text{In}\ \triangle AOD\\ \\ \dfrac{90}{2}+\dfrac{81}{2}+\angle AOD=180^o\\ \\ 90+81+2\angle AOD=36000\\ \\ 2\angle AOD=189^o\\ \\ \angle AOD=94.5$
Question 4 of 30

4. Question
1 point(s)
In a parallelogram ABCD find $\angle CDB $ if $\angle DAB=64^o $ and $ \angle DBC=50^o$
- $66^o $
- $50^o $
- $26^o $
- $40^o $
Correct

Incorrect

Hint

$\angle A=\angle C=64^o$ (opposite angles)
$50^o+64^o+\angle CDB=180^o\\ \\ \angle CDB=66^o$
Question 5 of 30

5. Question
1 point(s)
In a parallelogram ABCD, $\angle A-\angle C $ is
- $360^o $
- $0^o $
- $90^o $
- cannot be determined
Correct

Incorrect

Hint

$\angle A=\angle C$ (Opposite angles of a parallelogram are equal)
$\angle A-\angle C=0$
Question 6 of 30

6. Question
1 point(s)
In a parallelogram ABCD if $AB= 2x+5,\ CD=y+1,\ AD=y+5,\ BC=3x-4 $
What is the ratio of AB and BC?
- $71:21 $
- $ 12:11 $
- $ 31:35$
- $7:7 $
Correct

Incorrect

Hint

$AB=CD\\ \\ 2x+5=y+1\\ \\ \Rightarrow 2x-y=\ -4\ …..(1)\\ \\ BC=AD\\ \\ 3x-4=y+5\\ \\ 3x-y=9\ ….(2)$
Subtracting eqn (1) from (2) we get
$2x-y-3x+y=-13\\ \\ -x=-13\\ \\ x=13\\ \\ 2\times13-y=-4\\ \\ 26-y=-4\\ \\ -y=-4-26\\ \\ -y=-30\\ \\ y=30\\ \\ AB= 2x+5=2\times13+5=26+5=31\\ \\ BC= 3x - 4 = 3\times13 - 4 = 39- 4=35\\ \\ AB:BC = 31:35$
Question 7 of 30

7. Question
1 point(s)
In quadrilateral ABCD if diagonals AC and BD are equal and perpendicular to each other what is ABCD?
- a square
- a parallelogram
- a rhombus
- a trapezium
Correct

Incorrect

Hint

In square diagonals are equal and perpendicular to each other
Question 8 of 30

8. Question
1 point(s)
D and E are the midpoint of AB and AC respectively of $\triangle $ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
- $\angle DAE=\angle EFC $
- $AE = EF $
- $DE=EF $
- $\angle ADE=\angle ECF $
Correct

Incorrect

Hint

$AE=CE$ (given)
$DE=EF$
$\angle AED=\angle FEC$ (V.O.A)
$\triangle ADE\cong\triangle CFE$ (SAS)
$AD=CF$ (Cpct)
So, the additional information is $DE=EF$
Question 9 of 30

9. Question
1 point(s)
The diagonals AC and BD of a parallelogram ABCD intersect each other at O. If $\angle DAC=32^o $ and $\angle AOB=70^o $ then $\angle DBC $ is equal to
- $24^o $
- $86^o $
- $38^o $
- $32^o $
Correct

Incorrect

Hint

$\angle ACB=32^o$ (alternate angle)
$\angle BOC=180^o-70^o$ (Co-interior angle)
$=110^o\\ \\ \angle DBC=180^o-(32^o+110^o)=38^o$
Question 10 of 30

10. Question
1 point(s)
The figure obtained by joining the midpoints of the sides of a quadrilateral ABCD taken in order is a square only if
- ABCD is a rhombus
- Diagonals of ABCD are equal
- Diagonals of ABCD are equal and perpendicular.
- Diagonals of ABCD are perpendicular.
Correct

Incorrect

Hint

In square diagonals are equal and perpendicular.
Question 11 of 30

11. Question
1 point(s)
Three angles of a quadrilateral are $ 75^o\ 90^o,75$. The fourth angle is
- $90^o $
- $95^o $
- $105^o $
- $120^o $
Correct

Incorrect

Hint

Let the fourth angle be x.
$75+90+75+x=360^o\\ \\ x=360-240=120^o$
Question 12 of 30

12. Question
1 point(s)
A diagonal of a rectangle is inclined to one side of the rectangle at $ 25^o$. The acute angle between the diagonals is
- $55^o $
- $ 50^o$
- $40^o $
- $25^o $
Correct

Incorrect

Hint

$\angle OAD=25^o\\ \\ \angle BAO=90^o-25^o=65^o\\ \\ \angle ABO=\angle BAO\\ \\ \angle BOA=180^o-(65^o+65^o)=50^o$
Question 13 of 30

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

Incorrect

Hint

Diagonal bisect each other at right angles $\angle BOC=90^o$
$\angle DBC=180^o-(90^o+40^o)=180^o-130^o=50^o$
$\angle ADB=\angle DBC=50^o$ (Alternate angle)
Question 14 of 30

14. Question
1 point(s)
If angles A, B, C, D of quadrilateral ABCD taken in order are in the ratio $ 3:7:6:4 $ then ABCD is a
- Rhombus
- parallelogram
- trapezium
- kite
Correct

Incorrect

Hint

$3x+7x+6x+4x=360^o\\ \\ 20x=360^o\\ \\ x=18^o$
angles are $54^o,126^o,108^o,72^o$
$54+126=180^o\\ \\ 108+72=180^o$
ABCD is a trapezium.
Question 15 of 30

15. Question
1 point(s)
If PQ and RS are two perpendicular diameters of a circle what is PQRS?
- rectangle
- trapezium
- square
- rhombus
Correct

Incorrect

Hint

As in square diagonals are equal and bisect at right angles.
Question 16 of 30

16. Question
1 point(s)
If an angle of a parallelogram is four-fifth of its adjacent angle. Find the angles of a parallelogram.
- $60^o,120^o\ 60^o,120^o $
- $90^o,90^o,90^o,90^o $
- $80^o,100^o,80^o,100^o $
- $30^o,150^o,30^o,100^o $
Correct

Incorrect

Hint

Let one angle be x
Other angle $=\dfrac{4}{5}x$
$x+\dfrac{4}{5}x=180^o\\ \\ x=100^o\\ \\ \dfrac{4x}{5}=\dfrac{4\times100}{5}=80$
Question 17 of 30

17. Question
1 point(s)
Two opposite angles of a parallelogram are (3p – 4) and (48 – p) what is the value of p?
- 13
- 15
- 17
- 20
Correct

Incorrect

Hint

Opposite angles of a parallelogram are equal.
$3p-4=48-p\\ \\ \Rightarrow 4p=52\\ \\ \Rightarrow P=13$
Question 18 of 30

18. Question
1 point(s)
$\triangle $ABC and $\triangle $DEF are such that AB = DE, BC = EF, AB || DE, and BC || EF. What is ADFC?
- square
- rectangle
- parallelogram
- rhombus
Correct

Incorrect

Hint

AB = DE and AB || DE
$\Rightarrow$ ABED is a parallelogram
$\therefore$ BE||AD and BE = AD ….(1)
Given BC= EF and BC || EF
$\Rightarrow$ BEFC is a parallelogram
$\therefore$ BE || CF and BE = CF …..(2)
From (1) and (2) we get
AD || CF and AD = CF
$\therefore$ So, ADFC is a parallelogram.
Question 19 of 30

19. Question
1 point(s)
In the parallelogram PQRS find the measurement of $\angle $PSQ and $\angle $PQS when $\angle $QPS is $55^o $ and $\angle $QSR is $70^o $
- $55^o,35^o $
- $55^o,70^o $
- $20^o,70^o $
- $70^o,55^o $
Correct

Incorrect

Hint

$\angle QRS=\angle QPS=55^o$ (Opposite angles are equal)
$\angle RQS=180^o-(55+70)\\ \\ =180-125=55\\ \\ \angle PSQ=\angle RQS=55^o\ \text{ (Alternate angle)}\\ \\ \angle PQS=180-(55+55)=70^o$
Question 20 of 30

20. Question
1 point(s)
If angles P, Q, R, S of the quadrilateral PQRS taken in order are in the ratio 9:11 : 12:8 then what kind of slope does PQRS represent?
- Square
- trapezium
- rectangle
- parallelogram
Correct

Incorrect

Hint

$9x+11x+12x+8x=360^o\\ \\ 40x=360^o\\ \\ x=9\\ \\ 9x=9\times9=81\\ \\ 11x=11\times9=99\\ \\ 12x=12\times9=108^o\\ \\ 8x=8\times9=72$
Clearly adjacent angles are supplementary but opposite angles are not equal so, it is a trapezium.
Question 21 of 30

21. Question
1 point(s)
ABCD is a square A line segment DX cuts the sides BC at x and diagonal AC at O such that $\angle \text{COD}=110^o $ and $\angle \text{OXC}=x^o$. Find the value of x
- $65^o $
- $63^o $
- $60^o $
- $66^o $
Correct

Incorrect

Hint

$\angle COX=180-110=70^o\\ \\ \angle XCO=45^o\\ \\ X=180-(70+45)=65^o$
Question 22 of 30

22. Question
1 point(s)
The diagonals of a rhombus ABCD intersect at o. if AB=10 cm, BD= 16 cm find AC
- 6 cm
- 13 cm
- 12 cm
- 11 cm
Correct

Incorrect

Hint

$AB=10\ cm\\ \\ BD=16\ cm\\ \\ OB= 8\ cm\\ \\ \text{In}\ \triangle ABO\\ \\ AB^2=OA^2+OB^2\\ \\ 10^2=OA^2+8^2\\ \\ 100-64=OA^2\\ \\ 36=OA^2\\ \\ OA=6\ cm\\ \\ AC=2\times OA=2\times6=12\ cm$
Question 23 of 30

23. Question
1 point(s)
ABCD is a parallelogram. The angle bisectors of $\angle $A and $\angle $D meet at a O. What is the measure of $\angle $AOD?
- sum of $\angle $A and $\angle $D
- $0^o $
- $90^o $
- $45^o $
Correct

Incorrect

Hint

$\angle A+\angle D=180^o\\ \\ \dfrac{1}{2}\angle A+\dfrac{1}{2}\angle D=\dfrac{180}{2}\\ \\ \angle OAD+\angle ODA=90^o\\ \\ \text{So,}\ \angle AOD=90^o$
Question 24 of 30

24. Question
1 point(s)
The length and breadth of a rectangle is 5 cm and 12 cm. Find the radius of circumcircle of rectangle
- 3.25 cm
- 19.5 cm
- 6.5 cm
- 13 cm
Correct

Incorrect

Hint

$AC=\sqrt{12^2+5^2}\\ \\ AC=\sqrt{144+25}=13\ cm\\ \\ OA=\dfrac{13}{2}=6.5\ cm$
Question 25 of 30

25. Question
1 point(s)
ABCD is a rectangle. PA = 3 cm, PB = 5 cm, PC = 3 cm. Find the value of PD.
- $3 $ cm
- $3\sqrt3 $ cm
- $5 $ cm
- $ 1$ cm
Correct

Incorrect

Hint

Diagonals are equal.
AC = AP + PC = 3 + 3 = 6 cm
PD = 6 – 5 = 1 cm
Question 26 of 30

26. Question
1 point(s)
In a square ABCD, E, F, G, H are the midpoint of four side. What kind of shape is represented by EFGH.
- Rectangle
- square
- trapezium
- parallelogram
Correct

Incorrect

Hint

AB = BC = CD = DA
( $\therefore$ ABCD is a square)
AE = EB
BF= FC
CG = GD
DG = HA
Question 27 of 30

27. Question
1 point(s)
ABCD is a parallelogram in which p is the midpoint of CD and PA bisect $\angle $DAB taken
- $ PC=AB $
- $DC=BC $
- $PC^2=PB^2+BC^2 $
- $PA^2=PB^2+AB^2 $
Correct

Incorrect

Hint

$\angle PAB=\angle APD$ (alt. angle)
$APD=\angle PAD\\ \\ \Rightarrow AD=PD\\ \\ \Rightarrow PC=BC$
Question 28 of 30

28. Question
1 point(s)
Find the measurement of angles of a parallelogram if one of its angles is $60^o $ less than the twice of the smallest angle.
- $ 80^o, 100^o, 80^o, 100^o $
- $70^o, 110^o, 70^o, 110^o $
- $75^o, 105^o, 75^o, 105^o $
- $ 85^o, 95^o, 85^o, 95^o $
Correct

Incorrect

Hint

Let the smallest angle be x
$x+(2x-60) = 180^o\\ \\ 3x=240^o\Rightarrow x=80^o$
Another angle $=100^o$
Question 29 of 30

29. Question
1 point(s)
PQ is the largest side and RS is the shortest side then:-
- $\angle $R is less than $\angle $P
- $\angle $R is equal to $\angle $P
- $\angle $R is greater than $\angle $P
- $\angle $S is less than $\angle $Q
Correct

Incorrect

Hint

$\text{In}\ \triangle PQR\\ \\ PQ\rangle QR\\ \\ \Rightarrow \angle PRQ\rangle\angle QPR\ ….(1) \\ \\ \text{In}\ \triangle PRS\\ \\ PS\rangle SR\\ \\ \Rightarrow\angle PRS\rangle\angle SPR\ ….(2)$
Adding…. (1) and (2) we get
$\angle R\rangle\angle P$
Question 30 of 30

30. Question
1 point(s)
In quadrilateral PQRS, $\angle P+\angle R=140^o $ and $\angle Q:\angle S=5:6 $. Find $\angle Q $
- $100^o $
- $120^o $
- $105^o $
- $ 35^o$
Correct

Incorrect

Hint

$\angle P+\angle R=140^o\\ \\ \angle Q:\angle S=5:6\\ \\ \angle P+\angle Q+\angle R+\angle S=380\\ \\ 140^o+11x=360^o\\ \\ 11x=220^o\\ \\ x=20^o\\ \\ \angle Q=5x=5\times20^o=100^o$
Another angle $=100^o$