Maths Class X Question Bank

Question 1 of 34

1. Question
1 point(s)
In the figure, if $\dfrac{AE}{EB}=\dfrac{AF}{FC} $, then we can conclude that :
- $E $ and $F $ are the mid-points of $AB $ and $AC $ respectively
- $EF||BC $
- $\dfrac{EF}{BC}=\dfrac{AB}{AC} $
- None of the above
Correct

Incorrect

Hint

If $\dfrac{AE}{EB}=\dfrac{AF}{FC}$ then using converse of basic proportionality theorem EF || BC
Question 2 of 34

2. Question
1 point(s)
In the triangle ABC,DE || BC, then the length of DB is :
- 2.5 cm
- 5 cm
- 3.5 cm
- 3 cm
Correct

Incorrect

Hint

$\dfrac{x}{6-x}=\dfrac{10}{14}\\ \\ \Rightarrow 14x=60-10x\Rightarrow 24x=60\\ \\ \Rightarrow x=\dfrac{60}{24}=\dfrac{30}{12}=2.5\ cm$ (approx)
Question 3 of 34

3. Question
1 point(s)
In $\triangle $ABC, if DE || BC, then the value of x is :
- 4
- 6
- 8
- 9
Correct

Incorrect

Hint

$\dfrac{x}{3}=\dfrac{12}{x}\\ \\ x^2=36\\ \\ \Rightarrow x=6$
Question 4 of 34

4. Question
1 point(s)
In the trapezium ABCD, AB||CD, then the value of x is :
- 2
- 3
- -2
- -3
Correct

Incorrect

Hint

$\dfrac{2x}{x-3}=\dfrac{x+3}{x-3}\\ \\ 2x^2-6x=x^2-9\\ \\ \Rightarrow x^2-6x+9=0\\ \\ \Rightarrow x^2-3x-3x+9=0\\ \\ \Rightarrow x(x-3)-3(x-3)=0\\ \\ x=3$
Question 5 of 34

5. Question
1 point(s)
In the given figure, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm, then :
- EF is not parallel to QR
- EF || QR
- Cannot say anything
- None of the above
Correct

Incorrect

Hint

$\dfrac{PQ}{PR}=\dfrac{1.28}{2.56}=\dfrac{1}{2}\\ \\ \dfrac{PE}{PF}=\dfrac{0.18}{0.36}=\dfrac{1}{2}$
EF || QR (by converse of BPT)
Question 6 of 34

6. Question
1 point(s)
In the given figure, if $\triangle ADE\sim\triangle ABC $, then $BC $ is equal to:
- 4.5
- 3
- 3.6
- 2.4
Correct

Incorrect

Hint

$\dfrac{1.5}{3}=\dfrac{1.2}{x}\\ \\ \Rightarrow x=\dfrac{3\times12}{15}=2.4\ cm$
Question 7 of 34

7. Question
1 point(s)
In the given figure, $\triangle ACB\sim\triangle APQ $. If $BC=8\ cm $. $PQ=4\ cm.\ BA=6.5\ cm $ and $AP=2.8\ cm $, then the length of $ AQ$ is
- 3.25 cm
- 4 cm
- 4.25 cm
- 3 cm
Correct

Incorrect

Hint

$\dfrac{8}{4}=\dfrac{6.5}{AQ}\\ \\ AQ=\dfrac{4\times6.5}{8}=3.25\ cm$
Question 8 of 34

8. Question
1 point(s)
If $\triangle ABC\sim\triangle PQR $ and $\angle P=50^o,\ \angle B=60^o $, then $ \angle R$ is:
- $100^o $
- $80^o $
- $70^o $
- Cannot be determined
Correct

Incorrect

Hint

$\angle B=\angle Q=60^o\\ \\ \angle R=180-(60+50)\\ \\ =180-110=70$

Question 9 of 34

9. Question

1 point(s)

$\triangle ABC\sim\triangle DEF $ and be perimeters of $\triangle ABC $ and $\triangle DEF $ are $30\ cm $ and $18\ cm $ respectively. If $BC=9\ cm$, then $EF$ is equal to :

6.3 cm
5.4 cm
7.2 cm
4.5 cm

Correct

Incorrect

Hint

*** QuickLaTeX cannot compile formula:
\mathrm{\dfrac{Perimeter\ of\ \triangle ABC}{Perimeter\ of\ \triangle DEF}=\dfrac{30}{18}=\dfrac{BC}{DE}\\ \\ \Rightarrow \dfrac{30}{18}=\dfrac{9}{DE}\\ \\ \Rightarrow DE=\dfrac{9\times18}{30}=5.4\ cm 

*** Error message:
File ended while scanning use of \math@egroup.
Emergency stop.

Question 10 of 34

10. Question
1 point(s)
$\triangle ABC\sim\triangle DEF $ such that $AB=9.1\ cm $ and $DE=6.5\ cm $. If the perimeter of $\triangle DEF $ is $25\ cm $, then perimeter of $\triangle ABC $ is :
- 35 cm
- 28 cm
- 42 cm
- 40 cm
Correct

Incorrect

Hint

$\dfrac{Perimeter\ of\ \triangle ABC}{Perimeter\ of\ \triangle DEF}=\dfrac{AB}{DE}\\ \\ \dfrac{x}{25}=\dfrac{9.1}{6.5}\\ \\ x=\dfrac{25\times91}{65}=35\ cm}$
Question 11 of 34

11. Question
1 point(s)
If $\triangle ABC\sim\triangle EDF $ and $\triangle ABC $ is not similar to $\triangle DEF $, then which of the following is not true ?
- BC. EF = AC. FD
- AB. EF = AC. DE
- BC. DE = AB. EF
- BC. DE = AB. FD
Correct

Incorrect

Hint

If $\triangle ABC\sim\triangle EDF$
$\dfrac{AB}{ED}=\dfrac{BC}{DF}=\dfrac{AC}{EF}\\ \\ \Rightarrow AB\times DF=BC\times ED$
or $BC\times EF=AC\times OF$
or $AB\times EF=AC\times ED\ BC\times DE\ne AB\times EF$
Question 12 of 34

12. Question
1 point(s)
If in two triangles ABC and PQR. $\dfrac{AB}{QR}=\dfrac{BC}{PR}=\dfrac{CA}{PQ} $, then
- $\triangle PQR\sim\triangle CAB $
- $\triangle PQR\sim\triangle ABC $
- $\triangle CBA\sim\triangle PQR $
- $\triangle BCA\sim\triangle PQR $
Correct

Incorrect

Hint

If $\dfrac{AB}{QR}=\dfrac{BC}{PR}=\dfrac{CA}{PQ}$
then $\triangle PQR\sim\triangle CAB$
Question 13 of 34

13. Question
1 point(s)
In the given figure, two line segments, $AC $ and $BD $ intersect each other at the point $P $ such that $PA=6\ cm,\ PB=3\ cm,\ PC=2.5\ cm,\ PD=5\ cm,\angle APB=50^o$ and $ \angle CDP=30^o$. Then $\angle PBA $ is equal to
- $50^o $
- $30^o $
- $60^o $
- $100^o $
Correct

Incorrect

Hint

$\dfrac{PA}{PD}=\dfrac{PB}{PC}$ (given)
$\angle APB=\angle DPC$ (VOA)
$\triangle APB\sim\triangle DPC$ (SAS)
$\angle A=\angle D=30^o\\ \\ \angle A+\angle B+\angle APB=180^o\\ \\ 30+50+\angle B=180^o\\ \\ \angle B=100^o$
Question 14 of 34

14. Question
1 point(s)
The areas of two similar triangles are $169\ cm^2$ and $121\ cm^2$, if the longest side of the larger triangle is $26\ cm$, then the longest side of the other triangle is :
- 12 cm
- 14 cm
- 19 cm
- 22 cm
Correct

Incorrect

Hint

$\sqrt{\dfrac{169}{121}}=\dfrac{26}{x}\\ \\ \Rightarrow \dfrac{13}{11}=\dfrac{26}{x}\Rightarrow x=22\ cm$
Question 15 of 34

15. Question
1 point(s)
In the following trapezium $ABCD,\ AB||CD $ and $CD=2\ AB $. If area $(\triangle AOB)=84\ cm^2 $, then area $(\triangle COD) $ is:
- $168\ cm^2 $
- $336\ cm^2 $
- $252\ cm^2 $
- None of these
Correct

Incorrect

Hint

$\dfrac{ar\ (\triangle AOB)}{ar\ (\triangle COD)}=\dfrac{AB^2}{CD^2}\\ \\ \dfrac{84}{x}=\dfrac{AB^2}{4AB^2}\Rightarrow x=84\times4=336\ cm^2$
Question 16 of 34

16. Question
1 point(s)
If $\triangle ABC\sim\triangle PQR $ area $(\triangle ABC)=80\ cm^2 $ and area $ (\triangle PQR)=245\ cm^2$, then $\dfrac{AB}{PQ} $ is equal to:
- 0.70069444444444
- 0.17152777777778
- 0.086805555555556
- None of these
Correct

Incorrect

Hint

$\dfrac{80}{125}=\dfrac{AB^2}{PQ^2}\\ \\ \dfrac{16}{25}=\dfrac{AB^2}{PQ^2}\\ \\ \Rightarrow \dfrac{AB}{PQ}=\dfrac{4}{5}$
Question 17 of 34

17. Question
1 point(s)
In the similar triangles, $\triangle ABC $ and $\triangle DEF,\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{3}{4} $. If the median $AL=6\ cm $, then the median $DM $ of $ \triangle DEF$ is:
- $3\sqrt2\ cm $
- $8\sqrt3\ cm $
- $4\sqrt2\ cm $
- $3\sqrt3\ cm $
Correct

Incorrect

Hint

$\dfrac{AL}{DM}=\dfrac{3}{4}\\ \\ \dfrac{6}{DM}=\dfrac{3}{4}\Rightarrow DM=\dfrac{6\times4}{3}=8\ cm$
Question 18 of 34

18. Question
1 point(s)
In the figure, if AC = DE, then the value of EB is :
- $3\sqrt{30}\ cm $
- $2\sqrt{30}\ cm $
- $3\sqrt{15}\ cm $
- $4\sqrt{15}\ cm $
Correct

Incorrect

Hint

$AC^2=12^2+5^2=169\\ \\ DE^2=169\ (\text{As}\ AC=DE)\\ \\ DE=13\ cm\\ \\ BE^2=DE^2-DB^2=169-49=120\\ \\ \Rightarrow BE=2\sqrt{30}\ cm$
Question 19 of 34

19. Question
1 point(s)
In the quadrilateral $ABCD $, if $\angle B=90^o $ and $\angle ACD=90^o $ then $AD^2 $ is
- $AC^2-AB^2+BC^2 $
- $AC^2+DC^2+AB^2 $
- $AB^2+BC^2+CD^2 $
- $AB^2+BC^2+AC^2 $
Correct

Incorrect

Hint

$AB^2+BC^2=AC^2\\ \\ AD^2=AC^2+CD^2\\ \\ =AB^2+BC^2+CD^2$
Question 20 of 34

20. Question
1 point(s)
If diagonals of a rhombus are 12 cm and 16 cm, then the perimeter of the rhombus is :
- 20 cm
- 40 cm
- 28 cm
- 56 cm
Correct

Incorrect

Hint

Side $=\sqrt{36+64}=\sqrt{100}=10\ cm$
Perimeter $=4\times10=40\ cm$
Question 21 of 34

21. Question
1 point(s)
In triangle $ABC $ and $DEF,\ \angle A\ \ne\ \angle C,\angle B=\angle E,\angle F=\angle C $ and $AB=EF $. Then , the two triangle are:
- Neither congruent nor similar
- Congruent as well as similar
- congruent but not similar
- similar but not congruent
Correct

Incorrect

Hint

In triangle $ABC$ and $DEF,\ \angle A\ne C,\ \angle B=\angle E,\ \angle F=\angle C,\ AB=EF$ then two triangle are similar but not congruent.
Question 22 of 34

22. Question
1 point(s)
If $\triangle ABC\sim\triangle QRP,\dfrac{ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac{9}{4} $, perimeter $\triangle ABC=48\ cm,\ AB=18\ cm $ and $BC=18\ cm $, then $PQ $ is equal to:
- $8\ cm $
- $10\ cm $
- $12\ cm $
- $\dfrac{20}{3}\ cm $
Correct

Incorrect

Hint

$AC=48-(18+18)=48-36=12\\ \\ \dfrac{ar\ \triangle ABC}{ar\ \triangle QRP}=\dfrac{AC^2}{QP^2}\\ \\ \dfrac{9}{4}=\dfrac{12^2}{PQ^2}\\ \\ \dfrac{9}{4}=\dfrac{144}{PQ^2}\Rightarrow PQ^2=\dfrac{4\times144}{9}\\ \\ PQ^2=64\Rightarrow PQ=8\ cm$
Question 23 of 34

23. Question
1 point(s)
D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD=2 cm, BD=4 cm, BC=9 cm and DE||BC. Then, length of DE (in cm) is :
- 6
- 5
- 3
- 2.5
Correct

Incorrect

Hint

$\dfrac{AD}{AB}=\dfrac{DE}{BC}\\ \\ AB=AD+DB=2+4=6\ cm\\ \\ \dfrac{2}{6}=\dfrac{DE}{9}\Rightarrow \dfrac{2\times9}{6}=DE\\ \\ \Rightarrow DE=3\ cm$
Question 24 of 34

24. Question
1 point(s)
It is given that $\triangle ABC\sim\triangle PQR $ with $\dfrac{BC}{QR}=\dfrac{1}{4} $, then $\dfrac{ar(PRQ)}{ar(BCA)} $ is equal to:
- $\dfrac{1}{4} $
- $\dfrac{1}{16} $
- $4 $
- $16 $
Correct

Incorrect

Hint

$\dfrac{ar\ (PRQ)}{ar\ (\triangle BCA)}=\dfrac{QR^2}{BC^2}=\dfrac{16}{1}$
Question 25 of 34

25. Question
1 point(s)
The lengths of the diagonals of a rhombus are 24 cm and 32 cm. The perimeter of the rhombus is :
- 9 cm
- 128 cm
- 80 cm
- 56 cm
Correct

Incorrect

Hint

Side $=\sqrt{\left(\dfrac{24}{2}\right)^2+\left(\dfrac{32}{2}\right)^2}=\sqrt{144+256}=20\ cm$
Perimeter $=4\times20=80\ cm$
Question 26 of 34

26. Question
1 point(s)
$\triangle ABC\sim \triangle PQR,\ M $ is the mid-point of $BC $ and $N $ is the mid-point of $QR $. If the area of $\triangle ABC =100\ sq.\ cm $, the area of $\triangle PQR=144\ sq.\ cm $ and $AM=4\ cm $ then $PN $ is:
- 4.8 cm
- 12 cm
- 4 cm
- 5.6 cm
Correct

Incorrect

Hint

$\dfrac{ar\ (\triangle ABC)}{ar\ (\triangle PQR)}=\dfrac{AB^2}{PQ^2}\\ \\ \Rightarrow \dfrac{AB}{PQ}=\dfrac{10}{12}\\ \\ \dfrac{AM}{PN}=\dfrac{AB}{PQ}\Rightarrow PM=4.8\ cm$
Question 27 of 34

27. Question
1 point(s)
$\triangle ABC $ is such that $AB=3\ cm,\ BC=2\ cm $ and $CA=2.5\ cm $. If $\triangle DEF\sim\triangle ABC $ and $EF=4\ cm $, then perimeter of $\triangle DEF $ is:
- 15 cm
- 22.5 cm
- 7.5 cm
- 30 cm
Correct

Incorrect

Hint

$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}\\ \\ \Rightarrow \dfrac{3}{DE}=\dfrac{2}{4}=\dfrac{2.5}{DF}\Rightarrow DE=6\ cm,\ DF=5\ cm$
Perimeter of $\triangle DEF=4+5+6=15\ cm$
Question 28 of 34

28. Question
1 point(s)
A vertical stick 30 m long casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 75 m long on the ground. The height of the tower is :
- 150 m
- 100 m
- 25 m
- 200 m
Correct

Incorrect

Hint

$\dfrac{EB}{BD}=\dfrac{AB}{BC}\\ \\ \Rightarrow \dfrac{30}{15}=\dfrac{x+30}{75}\\ \\ \Rightarrow x=120$
Height $=x+30=150\ m$
Question 29 of 34

29. Question
1 point(s)
$\triangle ABC $ and $\triangle PQR $ are similar triangles such that $\angle A=32^o $ and $\angle R=65^o $, then $\angle B $ is:
- $83^o $
- $32^o $
- $65^o $
- $97^o $
Correct

Incorrect

Hint

$\angle A=\angle P\\ \\ \angle B=\angle Q\\ \\ \angle C=\angle R\\ \\ \angle B=\angle Q=180-(32+65)\\ \\ =180-97=83$
Question 30 of 34

30. Question
1 point(s)
In the figure $\triangle ABC\sim\triangle PQR $, then $y+z $ is:
- $2+\sqrt3 $
- $4+3\sqrt3 $
- $4+\sqrt3 $
- $3+4\sqrt3 $
Correct

Incorrect

Hint

$QR^2=PR^2+PO^2\Rightarrow 6^2=y^2+9\\ \\ \Rightarrow y=\sqrt{27}=3\sqrt3\\ \\ BC^2=AB^2+AC^2\Rightarrow 8^2=z^2+(4\sqrt3)^2\\ \\ \Rightarrow z=4\\ \\ y+z=4+3\sqrt3$
Question 31 of 34

31. Question
1 point(s)
In an isosceles $\triangle ABC $, if $AC=BC $ and $AB^2=2AC^2 $, then $ \angle C$ is equal to:
- $45^o $
- $60^o $
- $30^o $
- $90^o $
Correct

Incorrect

Hint

$AB^2=2AC^2\\ \\ =AC^2+AC^2\\ \\ AB^2=AC^2+BC^2\ (\because AC=BC)\\ \\ \angle C=90$
Question 32 of 34

32. Question
1 point(s)
The areas of two similar triangles $ABC $ and $PQR $ are $25\ cm^2$ and $49\ cm^2$ respectively. If $QR=9.8\ cm$, then $BC $ is :
- 9.8 cm
- 7 cm
- 49 cm
- 25 cm
Correct

Incorrect

Hint

$\dfrac{ar\ (\triangle ABC)}{ar\ (\triangle PQR)}=\dfrac{BC^2}{QR^2}\\ \\ \dfrac{25}{49}=\dfrac{BC^2}{(9.8)^2}\Rightarrow \dfrac{BC}{9.8}=\dfrac{5}{7}\\ \\ \Rightarrow BC=\dfrac{5\times9.8}{7}=7\ cm$
Question 33 of 34

33. Question
1 point(s)
If the ratio of the corresponding sides of two similar triangles is 2 : 3, then the ratio of their corresponding altitude is :
- 3 : 2
- 16 : 81
- 4 : 9
- 2 : 3
Correct

Incorrect

Hint

In two similar triangles, if the corresponding sides are in a particular ratio then altitude will also be in the same ratio.
Question 34 of 34

34. Question
1 point(s)
In the figure PQ||BC and AP : PB = 1 : 2. Find $\dfrac{ar(\triangle APQ)}{ar(\triangle ABC)} $:
- 1 : 4
- 4 : 1
- 1 : 9
- 2 : 9
Correct

Incorrect

Hint

$\dfrac{ar\ (\triangle APQ)}{ar\ (\triangle ABC)}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}$