Mathematics Class IX Questions Bank

Question 1 of 27

1. Question
1 point(s)
How many triangle can be drawn having its angles as 50°, 67° and 63° ? Give reasons for your answer.
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Hint

Sum of angles $=50+67+63=180$
Infinitely many triangles can be drawn
Question 2 of 27

2. Question
1 point(s)
Let OA, OB, OC and OD are rays in the anti-clockwise direction such that $\angle $AOB = $\angle $COD $ =100^o$, $\angle $BOC = $82^o $ Is it true to say that AOC and BOD are lines ?
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Hint

$\angle AOC=100+82=182\\ \\ \angle BOD=100+82=182$
No they are not lines
Question 3 of 27

3. Question
1 point(s)
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
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Hint

Let the first angle be $a$ and second angle be $b$
Third angle $=a+b$
$a+b+a+b=180\\ \\ 2(a+b)=180^o\\ \\ (a+b)=90^o$
Question 4 of 27

4. Question
1 point(s)
In the figure, find the value of x, if AB ||CD and BC||ED.
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Hint

$\angle DCB=180-77=103^o$ (Linear pair)
$\angle DCB=x=103^o$ (Alternate angles)
Question 5 of 27

5. Question
1 point(s)
In the figure $PQ||SR,\ \angle SQR=25^o,\ \angle QRT=65^o, $ find $x $ and $ y$:
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Hint

$\angle PQR=\angle QRT$ (Alternate angles)
$x+25=65\\ \\ x=65-25=40$
In $\triangle SPQ$
$90^o+x+y=180^o\\ \\ 90^o+40+y=180^o\\ \\ y=50^o$
Question 6 of 27

6. Question
1 point(s)
If $x + y = s + w$, prove that AOB is a straight line.
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Hint

$x+y=s+w$
We know that
$x+y+s+w=360^o\\ \\ x+y+x+y=360^o\\ \\ x+y=180^o$
Since $x+y=s+w$
So $s+w=180^o$
Question 7 of 27

7. Question
1 point(s)
In the figure, OQ bisects $\angle $AOB. OP is a ray opposite to ray OQ. Prove that $\angle $POA = $\angle $POB
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Hint

Let $\angle AOQ=\angle QOB=x$
$\angle POA=180-x$ (Linear pair)
$\angle POB=180-x$ (Linear pair)
So $\angle POA=\angle POB$
Question 8 of 27

8. Question
1 point(s)
In the figure, if $ x+ y=w+z$, then prove that AOB is a line.
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Hint

We know that
$x+w+z+y=360^o\\ \\ (x+y)+(w+z)=360^o\\ \\ 2(x+y)=360^o\\ \\ x+y=180^o$
Similarly $w+z=180^o$
Question 9 of 27

9. Question
1 point(s)
In the figure, AB||CD, find the value of x.
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Hint

$\angle 1=180-40=140$ (Co-interior angles)
$\angle2=180-35=145$ (Co-interior angles)
$x=\angle1+\angle2=140+145=285$
Question 10 of 27

10. Question
1 point(s)
In the figure, prove that AB||EF.
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Hint

$\angle ABC=\angle BCD\\ \\ 50=30+20=50$
But they form alternate angle
So $AB||CD\ ….(1)\\ \\ \angle FEC+\angle ECD=180^o$
$150+30=180^o$ (Interior angles on the same side)
$CD||EF\ ...(2)$
From (1) and (2)
$AB||EF$
Question 11 of 27

11. Question
1 point(s)
AB and CD are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (shown in the figure). Show that AB||CD.
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Hint

$\angle EAC=\angle ACH$ (alt interior angles)
$\dfrac{1}{2}\angle EAC=\dfrac{1}{2}\angle ACH\\ \\ \angle BAC=\angle ACD$ ( $\because$ AB and CD are bisectors)
But they form alternate angles
So $AB || CD$
Question 12 of 27

12. Question
1 point(s)
In the figure, DE ||QR and AP and BP are bisectors of $\angle $EAB and $\angle $RBA respectively. Find $ \angle$APB.
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Hint

$\angle EAB+\angle PBA=180^o$ (Co-interior angles)
$\dfrac{1}{2}(\angle EAB+\angle RBA)=90^o\\ \\ \angle PAB+\angle PBA=90^o$
In $\triangle APB$
$\angle PAB+\angle PBA+\angle APB=180^o\\ \\ \angle APB=180-90=90^o$
Question 13 of 27

13. Question
1 point(s)
Prove that if two lines intersect, the vertically opposite angles are equal.
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Hint

To Prove
$\angle AOD=\angle BOC\\ \\ \angle POC=\angle BOD\\ \\ \angle AOD+\angle BOD=180^o\\ \\ \angle BOC+\angle BOD=180^o\\ \\ \angle AOD+\angle BOD=\angle BOC+\angle BOD\\ \\ \angle AOD=\angle BOC$ (Proved)
On line $AB$
$\angle AOC+\angle BOC=180^o$
On line $CD$
$\angle BOC+\angle BOD=180^o\\ \\ \angle AOC+\angle BOC=\angle BOC+\angle BOD\\ \\ \angle AOC=\angle BOD$ (Proved)
Question 14 of 27

14. Question
1 point(s)
In the figure, if PQ||RS and $\angle $PXM = $50^o $ and $\angle $MYS = $120^o $, find the value of $ x$.
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Hint

$\angle1+50=180^o\\ \\ \angle 1=130\\ \\ \angle2=120^o$ (alternate angle)
$\angle 1+\angle 2=x-20\\ \\ 130+120=x-20\\ \\ 250+20=x\\ \\ x=270^o$
Question 15 of 27

15. Question
1 point(s)
In the figure, PQ||RS and T is any point as shown in the figure, then show that $\angle $PQT + $\angle $QTS + $ \angle $RST = $360^o $.
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Hint

$\angle PQT+\angle QTU=180^o$ (Co-interior angles)
$\angle OTS+\angle RST=180^o$ (Co-interior angles)
$\angle PQT+(\angle QTU+\angle UTS)+\angle RST=360^o\\ \\ \angle PQT+\angle QTS+\angle RST=360^o$
Question 16 of 27

16. Question
1 point(s)
In the figure. $POQ $ is a line, ray $OR $ is perpendicular in line $PQ $. $OS $ is another ray lying between rays $OP $ and $OR $. Prove that $\angle ROS=\dfrac{1}{2}(\angle QOS-\angle POS) $
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Hint

$\angle POR=\angle QOR=90\\ \\ \angle POS+\angle ROS=90=\angle QOR\\ \\ \angle POS+\angle ROS=\angle QOR$
Adding $\angle ROS$ on both the sides
$\angle POS+\angle ROS+\angle ROS=\angle QOR+\angle ROS\\ \\ \angle POS+2\angle ROS=\angle QOS\\ \\ \angle ROS=\dfrac{1}{2}(\angle QOS-\angle POS)$
Question 17 of 27

17. Question
1 point(s)
In the figure, $l_1||l_2 $ and $ a_1||a_2$. Find the value of $x $.
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Hint

$\angle 1=4x-15$ (Cooresponding angles)
$\angle1=2x$ (Corresponding angles)
$2x=4x-15\\ \\ 2x=15\\ \\ x=\dfrac{15}{2}=7.5$
Question 18 of 27

18. Question
1 point(s)
In the figure, find the value of x.
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Hint

$\angle ADC=\angle AEC+\angle DCE\\ \\ x=\angle AEC+50\ ….(1)$
In $\triangle AEB$
$\angle AEC=\angle ABC+\angle BAE\\ \\ =45+35=80\\ \\ x=80+50=130^o$
Question 19 of 27

19. Question
1 point(s)
Prove that the sum of the angles of a triangle is $180^o $
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Hint

$\angle 2=\angle 4$ (Alternate angles)
$\angle3=\angle5$ (Alternate angles)
For line $xy$
$\angle1+\angle4+\angle5=180^o$ (Linear pair)
$\angle1+\angle2+\angle3=180^o\ (\because\ \angle2=\angle4, \ \angle3=\angle5)$
Question 20 of 27

20. Question
1 point(s)
In the figure, AB and BC are two plane mirrors perpendicular to each other. Prove that the incident ray PQ is parallel to ray RS.
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Hint

$\angle QOR=40$
So $\angle3+2=90^o\ …(1)$ (Angle sum property)
$\angle1=\angle2,\angle3=\angle4$ (Angle of incidence = Angle of reflection)
$\angle1+\angle4=90^o\ ….(2)$
Adding equation (1) and (2)
$\angle1+\angle2+\angle3+\angle4=180^o\\ \\ \angle PQR+\angle SRQ=180^o\\ \\ PQ||RS$
Question 21 of 27

21. Question
1 point(s)
Prove that a triangle must have atleast two acute angles.
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Hint

Let us assume that only one angle have to be acute, other two angles will be right or obtuse
Let assume that both one of $90^o$
Then third angle (x) which is one acute angle will be O which is not possible.
Question 22 of 27

22. Question
1 point(s)
In the figure, $\angle Q>\angle R $ and $M $ is a point on $QR $ such that $PM $ is the bisector of $\angle QPR $. If the perpendicular from $P $ on $QR $ meets $QR $ at $ N$, prove that $\angle MPN=\dfrac{1}{2}(\angle Q-\angle R) $
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Hint

In $\triangle NPM$
$\angle PNM=90\\ \\ \angle NPM=90-\angle PMN$
In $\triangle PMR,\ \angle PMN=\angle MPR+\angle PRM$ (exterior angle property)
But $\angle MPR=\angle QPM=\dfrac{\angle P}{2}$ (MP is the bisector)
So $\angle PMN=\dfrac{\angle P}{Q}+\angle PRM$
In $\triangle PQR,\angle P+\angle Q+\angle R=180$
$\dfrac{\angle P}{2}+\dfrac{\angle Q}{2}+\dfrac{\angle R}{2}=90\\ \\ \dfrac{\angle P}{2}=90-\left(\dfrac{\angle Q}{2}+\dfrac{\angle R}{2}\right)\\ \\ \angle PMN=90^o-\left(\dfrac{\angle Q}{2}+\dfrac{\angle R}{2}\right)+\angle R\\ \\ =90-\dfrac{\angle Q}{2}-\dfrac{\angle R}{2}+\angle R\\ \\ =90-\dfrac{\angle Q}{2}+\dfrac{\angle R}{2}\\ \\ \angle NPM=90-\left[90-\dfrac{\angle Q}{2}+\dfrac{\angle R}{2}\right]\\ \\ =\dfrac{\angle Q}{2}-\dfrac{\angle R}{2}=\dfrac{1}{2}\ (\angle Q-\angle R)\\ \\ \angle MPN=\dfrac{1}{2}\ (\angle Q-\angle R)$ (Proved)
Question 23 of 27

23. Question
1 point(s)
In the figure, AB||CD and CD||EF, Also EA $\bot $ AB. If $\angle $BEF = $40^o $, then find x, y, z
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Hint

$x=y$ (Corresponding angles)
$y+40=180^o$ (Co-interior angle)
$y=140^o\\ \\ x=y=140^o$
$\angle BAE+\angle FEA=180^o$ (Co-interior angles)
$90+(z+40)=180^o\\ \\ 130+z=180\Rightarrow z=50^o$
Question 24 of 27

24. Question
1 point(s)
In the figure, if AB||CD. EF $ \bot$ CD and $\angle $GED = $126^o $ find $\angle $AGE, $\angle $GEF and $\angle $FGE.
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Hint

$\angle AGE=\angle GED$ (alt. interior $\angle S$ )
$=126^o$
$\angle GED=\angle GEF+\angle FED=126^o\\ \\ \angle GEF+90=126^o\\ \\ \angle GEF=126-90=36^o\\ \\ \angle CEG+\angle GED=180^o\\ \\ \angle CEG+126^o=180^o\ \Rightarrow \ \angle CEG=54^o\\ \\ \angle FGE=\angle CEG=54^o$
Question 25 of 27

25. Question
1 point(s)
In the figure, OP bisects $\angle $AOC, OQ bisects $\angle $BOC and OP $\bot $ OQ. Show that point A,O and B are collinear
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Hint

$\angle AOC=2\angle POC\ ….(1)\\ \\ \angle COB=2\angle COQ\ …(2)\\ \\ \angle AOC+\angle COD=2(\angle POC+\angle COQ)\\ \\ \angle AOC+\angle COB=2\angle POQ\\ \\ \angle AOC+\angle COB=2\times90=180^o\\ \\ \angle AOB=180^o$
Question 26 of 27

26. Question
1 point(s)
P is a point equidistant from two lines l and m intersecting at a point A. Show that AP bisects the angle between them.
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Hint

$PB=PC$ (Given)
$\angle PBA=\angle PCA=90^o$
$PA=PA$ (Common)
$\triangle PAB\cong\triangle PAC\ \ (RHS)\\ \\ \angle PAB=\angle PAC\ \ (cpct)$
Question 27 of 27

27. Question
1 point(s)
In the figure, $l || m$, show that $\angle1+\angle2-\angle3=180^o $
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Hint

$\angle 1+x=180^o\\ \\ x=180^o-\angle 1\\ \\ \angle2+y=180^o\\ \\ \angle y=180^o-\angle2\\ \\ \angle3+\angle x+\angle y=180^o\\ \\ \angle 3+180^o-\angle1+180^o-\angle2=180^o\\ \\ 180^o=\angle1+\angle2-\angle3$