Mathematics Class IX

Question 1 of 30

1. Question
1 point(s)
In $\triangle $LMN, LP is the bisector of MN and QR is the bisector of LP. Points P and R are the points on linesegment MN and LN respectively. LR=
- $\dfrac{2}{3}\text{LN} $
- $\dfrac{1}{3}\text{LN} $
- $\dfrac{2}{3}\text{LM} $
- $\dfrac{1}{3}\text{LM} $
Correct

Incorrect

Hint

RS = SN ….(1)
LR = RS …..(2)
From (1) and (2)
LR = RS = SN
$\Rightarrow \mathrm{LR=\dfrac{1}{3}LN}$
Question 2 of 30

2. Question
1 point(s)
In $\triangle ABC,\ AD\ \bot\ BC $ and $\angle B=\angle C $ By which property is $\triangle ADB \cong\triangle ADC $?
- S A S
- S S S
- R H S
- A S A
Correct

Incorrect

Hint

$\mathrm{\angle B=\angle C \ (given)\\ \\ \Rightarrow AB=AC\\ \\ \angle ADB=\angle ADC\ and\ BD=DC\ (\because AD\bot BC)\\ \\ \triangle ADB\cong\triangle\ ADC\ (ASA)}$
Question 3 of 30

3. Question
1 point(s)
An exterior angle of a triangle is $(4x-20)^o$ and the measure of two remote interior angles are $25^o$ and $(x+15^o)$. The value of x is
- $25^o $
- $20^o $
- $24^o $
- $29^o $
Correct

Incorrect

Hint

$(4x-20)^o=25^o+x+15^o$
( $\because$ exterior angle is equal to sum of interior opposite angle)
$3x=60\\ \\ x=20^o$
Question 4 of 30

4. Question
1 point(s)
An exterior angle of a triangle is $135^o$ and one of the interior opposite angle is $40^o$. Then the other two angles are.
- $95^o,45^o $
- $85^o,35^o $
- $60^o,30^o $
- $95^o,35^o $
Correct

Incorrect

Hint

$\angle R=135^o-40^o$ (exterior angle is equal to the sum of interior opposite angle)
$=95^o\\ \\ \angle RTS=180^o-(40^o+95^o)\\ \\ =180^o-135^o=45^o$
Question 5 of 30

5. Question
1 point(s)
Which of the following is not a criteria for congruence of triangles?
- SAS
- ASA
- SSA
- SSS
Correct

Incorrect

Hint

SSA specifies two sides and a non-included angle
Question 6 of 30

6. Question
1 point(s)
If AB = QR, BC = PR, CA = PQ then
- $\triangle ABC\cong\triangle PQR $
- $\triangle CBA\cong\triangle PRQ $
- $\triangle BAC\cong\triangle RPQ $
- $\triangle PQR\cong\triangle BCA $
Correct

Incorrect

Hint

$AB=QR\\ \\ BC=PR\\ \\ CA=PQ\\ \\ \triangle CBA\cong\triangle PRQ$
Question 7 of 30

7. Question
1 point(s)
In $\triangle ABC,\ AB=AC $ and $\angle B=50^o,\angle C $ is equal to
- $40^o $
- $50^o $
- $90^o $
- $80^o $
Correct

Incorrect

Hint

$AB = AC\\ \\ \Rightarrow\angle C=\angle B\\ \\ \angle C=\angle B=50^o$
Question 8 of 30

8. Question
1 point(s)
In $\triangle PQR\ \angle R=P $ and $QR=4 $ cm and $ PR=5$ cm. The length of PQ is
- 4 cm
- 5 cm
- 2 cm
- 2.5 cm
Correct

Incorrect

Hint

$\angle R=\angle P\\ \\ \Rightarrow PQ=RQ\\ \\ PQ=QR=4\ cm$
Question 9 of 30

9. Question
1 point(s)
Two sides of a triangle are of length 5 cm and 1.5 cm. The length of the third side cannot be.
- 3.6 cm
- 4.1 cm
- 3.8 cm
- 3.4 cm
Correct

Incorrect

Hint

$5-1.5 = 3.5\ cm$
The length of the third side cannot be less than 3.5 cm.
Question 10 of 30

10. Question
1 point(s)
Which of the following statements is true?
- Two line segments having the same length are congruent.
- Two squares having the same side length are congruent.
- Two circles having the same radius are congruent.
- All of the above.
Correct

Incorrect

Hint

All the aboue points are correct as line segments having same length, square having same side circle having same radius are congruent
Question 11 of 30

11. Question
1 point(s)
AB = AC. E is midpoint of AB and F is midpoint of AC. What is the length of BF?
- AE
- CF
- BE
- CE
Correct

Incorrect

Hint

In $\triangle ABC,\ AB=AC$ (given)
$AE=\dfrac{1}{2}AB$ and $AF=\dfrac{1}{2}AC\\ \\ \Rightarrow AE=AF$
In $\triangle ABF$ and $\triangle ACE$
$AB=AC$ (given)
$\angle BAF=\angle CAE\\ \\ AE=AF\\ \\ \triangle ABF\cong\triangle ACE\ \text{(SAS)}\\ \\ BF=CE\ (CPCT)$
Question 12 of 30

12. Question
1 point(s)
If the length of the largest side of a triangle is 12 cm. Find the possible values of its other two sides.
- 4.8, 8.2
- 3.2, 6.5
- 2.9, 7.2
- 4.1, 3.9
Correct

Incorrect

Hint

The other two sides must be greater than 12 cm.
So, 4.8 and 8.2 cm are the possible values.
Question 13 of 30

13. Question
1 point(s)
$DP=BQ $ and $\angle ADP=\angle CBQ $. To which triangle is $\angle ADP $ congruent to?
- $\triangle CBQ $
- $\triangle PDQ $
- $\triangle QBC $
- $\triangle PQB $
Correct

Incorrect

Hint

$\mathrm{DP=BQ\ (Given)\\ \\ \angle ADP=\angle CBQ\ (Given)\\ \\ \angle DPB=\angle BQD\ (Given)\\ \\ 180^o-\angle DPB=180^o-\angle BQD\\ \\ \angle APD=\angle BQC\\ \\ So,\ \triangle ADP\cong\triangle CBQ}$
Question 14 of 30

14. Question
1 point(s)
In $\triangle $ABC, if $AD $ is the perpendicular bisector of $BC $. Which of the following is true?
- $AB\rangle AC $
- $AD=AB $
- $AB=AC $
- $AD=AC $
Correct

Incorrect

Hint

$\mathrm{AD\bot BC\ (Given)\\ \\ \Rightarrow BD=DC\\ \\ In\ \triangle ABD\ and \ \triangle ACD\\ \\ BD=DC\ (Given)\\ \\ AD=AD\ (Common)\\ \\ \angle ADB=\angle ADC\ (each\ 90^o)\\ \\ \triangle ABD\cong\triangle ADC\ (SAS)\\ \\ AB=AC\ (cpct)}$
Question 15 of 30

15. Question
1 point(s)
In the given figure $GH||EF $ and $\dfrac{DG}{GE}=\dfrac{3}{5} $ and $ DF=5.6$ cm. The value of $DE $ is
- 2.1 cm
- 3.36 cm
- 9.46 cm
- 2.26 cm
Correct

Incorrect

Hint

$\dfrac{DG}{GE}=\dfrac{DH}{HF}$ (Basic proportionality theorem)
$\dfrac{3}{5}=\dfrac{x}{5.6-x}\\ \\ 5x=16.8-3x\\ \\ 8x=16.8\\ \\ x=2.1\ cm$
Question 16 of 30

16. Question
1 point(s)
In an isosceles $\triangle PQR,PQ=QR $ and $QR^2=2PR^2 $ then
- $\triangle PQR $, is an acute angled triangle.
- $\triangle $, is an obtuse angled triangle.
- $\triangle PQR $, is a right angled triangle.
- $PQ\rangle QR $
Correct

Incorrect

Hint

$QR^2=2PR^2\\ \\ \Rightarrow QR^2=PR^2+ PR^2\\ \\ \Rightarrow QR^2=PR^2+ PQ^2$
$\triangle\ PQR$ is a right angled triangle (by the converse of Pythagoras theorem)
Question 17 of 30

17. Question
1 point(s)
If $\triangle ABC\cong\triangle PQR $ and $ \triangle ABC$ is not congruent to $\triangle RPQ $ then which of the following is not true?
- $AC=PR $
- $ QR=BC$
- $BC=PQ $
- $ AB=PQ$
Correct

Incorrect

Hint

As BC = QR
Question 18 of 30

18. Question
1 point(s)
In a triangle other than an equilateral triangle angle opposite the largest side is greater than ………… of a right angle.
- $\dfrac{1}{2} $
- $\dfrac{2}{3} $
- $\dfrac{1}{3} $
- $\dfrac{1}{4} $
Correct

Incorrect

Hint

Let $AB=AD=DB$
$\angle ABD=\angle ADB=\angle BAD=60^o$
Question 19 of 30

19. Question
1 point(s)
Two sides of a triangle are of length 6 cm and 2 cm. The length of the third side cannot be
- 4.3 cm
- 3.9 cm
- 5.1 cm
- 6 cm
Correct

Incorrect

Hint

The third side cannot be less than (6 – 2) cm = 4 cm
Question 20 of 30

20. Question
1 point(s)
In $\triangle PQR\ \angle P=90^o $ and $\angle R=30^o $. Which side is largest?
- PR
- PQ
- QR
- all of are equal
Correct

Incorrect

Hint

$\angle P=90^o,\angle R=30^o\ So, \ \angle Q=180^o-(90+30)\\ \\ So,\ \angle Q=180^o-(90+30)=60^o$
QR will be the largest side as $\angle P$ is greater among $\angle Q$ and $\angle R$ .
Question 21 of 30

21. Question
1 point(s)
XA is a median and AB || xy . Which of the following statement is true?
- $XA = ZB $
- $XA=YB $
- $XB\rangle ZB $
- None of these
Correct

Incorrect

Hint

XA is a median
$\Rightarrow YA=AZ\\ \\ AB||XY$
By mid point theorem
$\dfrac{YA}{AZ}=\dfrac{XB}{BZ}\\ \\ as\ YA=AZ\\ \\ So,\ XB=BZ$
Question 22 of 30

22. Question
1 point(s)
Hypotenuse of a right isosceles triangle is $5\sqrt2 $ cm. The value of two side is
- 10 cm
- 6 cm
- 5 cm
- 2.5 cm
Correct

Incorrect

Hint

$BC=5\sqrt2\ cm\\ \\ AB=AC\ (given)$
By Pythagoras theorem
$BC^2=AB^2+AC^2\\ \\ (5\sqrt2)^2=2AB^2\\ \\ \dfrac{50}{2}=AB^2\\ \\ AB=5\ cm$
Longest side $=BC$
Angle opposite to longest side $=\angle BAC$
$\angle BAC=\angle BAD+\angle CAD\\ \\ \angle BAC\rangle\angle BAD\\ \\ \angle BAC\rangle60^o$
Question 23 of 30

23. Question
1 point(s)
According to which congruency criteria, the two triangles are congruent?
- SAS
- ASA
- AAA
- SSS
Correct

Incorrect

Hint

In $\triangle PRS$ and $RQS$
$PR=PS\\ \\ RQ=QS\\ \\ RS=RS\\ \\ \triangle PRS\cong\triangle RQS\ \text{(SSS)}$
Question 24 of 30

24. Question
1 point(s)
If AB || CD || EF || GH || IJ || KL and AC = CE = EG = GI = IK = 1 cm. if OA = 4 cm and AB =3 cm. Find the value of w.
- $\dfrac{8}{2} $ cm
- $6 $ cm
- $\dfrac{4}{3} $ cm
- $9 $ cm
Correct

Incorrect

Hint

$\mathrm{OI=8\ cm\\ \\ \triangle\ OAB\sim\triangle\ OIJ\\ \\ \Rightarrow \dfrac{AB}{IJ}=\dfrac{OA}{OI}\\ \\ \Rightarrow \dfrac{3}{W}=\dfrac{4}{8}\\ \\ \Rightarrow W=\dfrac{24}{4}=6\ cm}$
Question 25 of 30

25. Question
1 point(s)
In $\triangle DEF $, which of the following statement is true?
- $ DE+EF=FD$
- $ DE+EF \langle FD$
- $DE \rangle EF+FD $
- $ DE+EF\rangle FD$
Correct

Incorrect

Hint

$DE+EF\rangle FD$
(:- sum of two sides is greater than its third side)
Question 26 of 30

26. Question
1 point(s)
In a right triangle, the two acute angles are $4:5 $. Find them
- $50^o,40^o $
- $40^o,50^o $
- $45^o,55^o $
- $30^o,60^o $
Correct

Incorrect

Hint

$4x+5x+900=180^o\\ \\ 9x=90\\ \\ x=10^o$
angles are $40^o$ and $50^o$
Question 27 of 30

27. Question
1 point(s)
$\angle Q\angle P $ and $\angle R<\angle S $. Which of the following is true?
- $PS\langle QR $
- $PS\rangle QR $
- $ QR=PS$
- either (b) or (c)
Correct

Incorrect

Hint

$\angle Q<\angle P\\ \\ PO<QO\ ….(1)\\ \\ \angle R<\angle S\\ \\ OS<OR\ ….(2)$
Adding equation (1) and (2)
PO+OS<QO+OR\\ \\ PS
Question 28 of 30

28. Question
1 point(s)
It is not possible to construct a triangle with which of the following sides?
- 8.3 cm, 3.4 cm, 6.1 cm
- 5.4 cm, 2.3 cm, 3.1 cm
- 6 cm, 7 cm, 10 cm
- 3 cm, 5 cm, 5 cm
Correct

Incorrect

Hint

For any triangle sum of any two sides is greater than the third side.
2.3 + 3.1 = 5.4
Question 29 of 30

29. Question
1 point(s)
‘P’ is, the exterior angle of $\triangle \text{PQR}$ and ‘q + r’ is the sum of interior angles opposite to p. which of the following is true?
- p + r = q
- p = q + r
- p + q = r
- r = q – p
Correct

Incorrect

Hint

p = q + r (exterior angle is equal to sum of interior opposite angles)
Question 30 of 30

30. Question
1 point(s)
If DA = DB = DC. Find $\angle $ACB
- $70^o $
- $90^o $
- $100^o $
- $120^o $
Correct

Incorrect

Hint

$DB = DC\\ \\ \Rightarrow \angle 2=\angle 3$
Similarly $\angle1=\angle3\\ \\ So,\ \angle 1=\angle2\\ \\ In\ \triangle ABC\\ \\ \angle1+\angle2+\angle ACB=180^o\\ \\ \angle ACB=90^o\ (\because\angle 1=\angle2)$