Mathematics Class IX Questions Bank

Question 1 of 36

1. Question
1 point(s)
If sum of two parallel sides of a trapezium is $15\ cm $ and its area is $30\ cm^2 $, then the height of the trapezium is :
- 2 cm
- 4 cm
- 6 cm
- 8 cm
Correct

Incorrect

Hint

Area $=\dfrac{1}{2}\times$ (Sum of parallel sides) $\times$ heights
$\Rightarrow 30=\dfrac{1}{2}\times15\times h\\ \\ \dfrac{30\times2}{15}=h\\ \\ \Rightarrow h=2\times2=4\ cm$
Question 2 of 36

2. Question
1 point(s)
The altitude of a parallelogram is twice the length of the base and its area if $1250\ cm^2 $. The lengths of the base and the altitude respectively are :
- 20 cm, 40 cm
- 35 cm, 70 cm
- 25 cm, 50 cm
- 15 cm, 30 cm
Correct

Incorrect

Hint

$1250=x\times 2x\\ \\ \dfrac{1250}{2}=x^2\\ \\ \Rightarrow x^2=625\\ \\ \Rightarrow x=25$
Base = 25 cm, height = 50 cm
Question 3 of 36

3. Question
1 point(s)
In the figure PQRS is a parallelogram PM $\bot $ RS and RN $\bot $ PS. If PQ = 12 cm, PM = 6 cm and RN = 8 cm, then the length of PS is equal to :
- 18 cm
- 9 cm
- 4 cm
- 12 cm
Correct

Incorrect

Hint

$PQ\times PM=RN\times PS\\ \\ 12\times6=8\times PS\\ \\ \dfrac{12\times6}{8}=PS\\ \\ PS=9\ cm$
Question 4 of 36

4. Question
1 point(s)
ABCD is a parallelogram one of whose diagonals is AC. Then, which of the following is true ?
- $\mathrm{ar\ (\triangle ADC) > ar\ (\triangle CBA)} $
- $\mathrm{ar\ (\triangle ADC) = ar\ (\triangle CBA)} $
- \mathrm{ar\ (\triangle ABC) < ar\ (\triangle ADC)} [/latex]
- None of these
Correct

Incorrect

Hint

$AD=BC$ (opposite sides of a parallelogram)
Triangles on the same base are equal
Question 5 of 36

5. Question
1 point(s)
Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm, then the distance between the shorter sides is :
- 18 cm
- 16 cm
- 9 cm
- None of these
Correct

Incorrect

Hint

$24\times12=18\times h\\ \\ \dfrac{24\times12}{18}=h\\ \\ h=16\ cm$
Question 6 of 36

6. Question
1 point(s)
Which of the following figures lies on the same base and between the same parallels ?
- only (i)
- both (i) and (ii)
- only (iii)
- only (ii)
Correct

Incorrect

Hint

(ii) lies on the same base and between the parallels
Question 7 of 36

7. Question
1 point(s)
The side of an equilateral triangle is 4 cm. Its area is
- $\dfrac{\sqrt3}{4}\ cm^2 $
- $4\sqrt3\ cm^2 $
- $16\sqrt3\ cm^2 $
- $12\sqrt3\ cm^2 $
Correct

Incorrect

Hint

Area $=\dfrac{\sqrt3}{4}\times a^2$
$=\dfrac{\sqrt3}{4}\times4\times4=4\sqrt3\ cm^2$
Question 8 of 36

8. Question
1 point(s)
The area of the parallelogram ABCD is :
- $10 \ cm^2 $
- $9 \ cm^2 $
- $12 \ cm^2 $
- $15 \ cm^2 $
Correct

Incorrect

Hint

Area of triangle $ABD=\dfrac{1}{2}\times3\times4=6\ cm^2$
Area of parallelogram $=2\times6=12\ cm^2$
Question 9 of 36

9. Question
1 point(s)
In the figure, ABCD is a parallelogram and EFCD is a rectangle. Now which of the following is correct option ?
- ar (|| gm ADCF) = ar (rect. EFCD)
- ar (|| gm ABCD) = ar (rect. EFCD)
- ar (|| gm ADCF) = ar (rect. ABCD)
- None of these
Correct

Incorrect

Hint

$ar\ (||gm\ ABCD)=ar\ (rect\ EFCD)\\ \\ AD=BC\\ \\ ED=FC$
Question 10 of 36

10. Question
1 point(s)
If the sum of the parallel sides of a trapezium is 7 cm and distance between them is 4 cm, then area of the trapezium is :
- $28\ cm^2 $
- $7\ cm^2 $
- $21\ cm^2 $
- $14\ cm^2 $
Correct

Incorrect

Hint

Area $=\dfrac{1}{2}\times7\times4=14\ cm^2$
Question 11 of 36

11. Question
1 point(s)
ABCD is a quadrilateral whose diagonal AC divides it into two parts equal in area then ABCD :
- is a rectangle
- is always a rhombus
- is a parallelogram
- need not be any of (a), (b) or ©
Correct

Incorrect

Hint

If diagonal divides into two parts then it forms a triangle.
It may be kite
Question 12 of 36

12. Question
1 point(s)
The median of a triangle divides it into two :
- Triangle of equal area
- Congruent triangles
- right triangles
- isosceles triangles
Correct

Incorrect

Hint

Median of a triangle is a line segment joining vertex to the mid point of the opposite side. Thus a median of a triangle divides it into two triangles of equal area.
Question 13 of 36

13. Question
1 point(s)
If ABCD is a parallelogram, then which of the following is true ?
- $\mathrm{ar\ (\triangle ABD) = ar(\triangle BCD)} $
- $\mathrm{ar\ (\triangle ABD) = ar(\triangle ABC)} $
- $\mathrm{ar\ (\triangle ABC) = ar(\triangle ACD)} $
- all are true
Correct

Incorrect

Hint

$ar\ \triangle ABC=ar\ (\triangle ACD)$
$BC=AD$ (opposite sides of parallelogram)
Question 14 of 36

14. Question
1 point(s)
In the figure, PQRS and PQLM are parallelogram and X is any point on side side QL. The area of $\triangle $PMX is equal to :
- area of $\triangle $RQL
- Area of || gm PQRS
- area of $\triangle $SPM
- $\dfrac{1}{2} $ area of || gm PQLM
Correct

Incorrect

Hint

Area of $\triangle PMX=\dfrac{1}{2}\ ar$ of $||gm\ PQLM$
Question 15 of 36

15. Question
1 point(s)
In the given figure, ABC is a triangle and AD is one of its medians. The ratio of the areas of triangles ABD and ACD respectively is :
- 2 : 1
- 1 : 2
- 1 : 1
- 3 : 1
Correct

Incorrect

Hint

$BD=CD$ ( $\because$ AD is the median)
So, ratio of areas of triangles $ABD$ and $ACD=1:1$
Question 16 of 36

16. Question
1 point(s)
If the base of an isosceles triangle is 8 cm and one of the equal sides measures 5 cm, then the area of the isosceles triangle is :
- $24\ cm^2 $
- $18\ cm^2 $
- $12\ cm^2 $
- $30\ cm^2 $
Correct

Incorrect

Hint

Base $=8\ cm$
Equal sides $=5\ cm$
Area $=\dfrac{1}{2}\times b\times h$
$h=\sqrt{25-16}\\ \\ \sqrt9=3\ cm$
Area $=\dfrac{1}{2}\times 8\times3=12\ cm^2$
Question 17 of 36

17. Question
1 point(s)
In the figure, point D divide the side BC of $\triangle $ABC in the ratio p : q.The ratio between the ar ($\triangle $ABD) and ar ($\triangle $ADC) is
- $\dfrac{p}{p+q}:\dfrac{q}{p+q} $
- $p:q $
- $q : p $
- None of these
Correct

Incorrect

Hint

$\dfrac{Area\ of \ ABD}{Area\ of\ ADC}=\dfrac{\dfrac{1}{2}\times BD\times AL}{\dfrac{1}{2}\times CD\times AL}\\ \\ =\dfrac{BD}{CD}=\dfrac{p}{q}$
Question 18 of 36

18. Question
1 point(s)
In the figure, ABCD is a trapezium in which AB || CD and its diagonals AC and BD intersect at O. Now ar ($ \triangle$AOD) is equal to :
- $\mathrm{ar\ (\triangle AOB)} $
- $\mathrm{ar\ (\triangle COD)} $
- $\mathrm{ar\ (\triangle BOC)} $
- None of these
Correct

Incorrect

Hint

$ar\ \triangle ABD=ar\ ABC$ (on the)
$ar\ \triangle \ ABD-ar\ \triangle \ AOB=ar\ \triangle ABC-ar\ \triangle\ AOB\\ \\ ar\triangle AOD=ar\ \triangle BOC$
Question 19 of 36

19. Question
1 point(s)
In the figure, DE || BC. Then, which of the following relations is true ?
- $\mathrm{ar\ (\triangle ACD) = ar\ (\triangle BOC)} $
- $\mathrm{ar\ (\triangle ACD) = ar\ (\triangle ABE)} $
- $\mathrm{ar\ (\triangle ACD) = ar\ (\triangle BDE)} $
- $\mathrm{ar\ (\triangle ACD) = ar\ (\triangle CDE)} $
Correct

Incorrect

Hint

area of $\triangle$ BDE = area of $\triangle$ CDE
area of $\triangle$ BDE = $\triangle$ ADE + $\triangle$ CDE
area of $\triangle$ ABE = area of $\triangle$ ADC
Question 20 of 36

20. Question
1 point(s)
P and Q are any two points lying on the sides CD and AD respectively of a parallelogram ABCD. Now which of the two triangles have equal area ?
- $\triangle $APD and $\triangle $BPC
- $\triangle $ABQ and $\triangle $CDQ
- $\triangle $APB and $\triangle $BQC
- None of these
Correct

Incorrect

Hint

area of $\triangle APB=\dfrac{1}{2}\ ar(||gm\ ABCD)$
area of $\triangle BQC=\dfrac{1}{2}\ ar\ (||gm\ ABCD)$
area of $\triangle APB$ = area of $BQC$
Question 21 of 36

21. Question
1 point(s)
The area of the figure formed by joining the mid–points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is :
- $48\ cm^2 $
- $64\ cm^2 $
- $96\ cm^2 $
- $192\ cm^2 $
Correct

Incorrect

Hint

Area of rhombus $=\dfrac{1}{2}\times12\times 16=96\ cm^2$
Area of joining the mid point
$=\dfrac{1}{2}\times96=48\ cm^2$
Question 22 of 36

22. Question
1 point(s)
If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of parallelogram is :
- 1 : 4
- 1 : 3
- 1 : 2
- 1 : 1
Correct

Incorrect

Hint

area of $\triangle AFE$ = area of $\triangle BFD$ = area of $\triangle EDC=ar\ (\triangle DEF)$
area of $DEF=\dfrac{1}{4}$ area of $\triangle ADC$
$AFDE$ = area of $\triangle AFE$ + area of $\triangle DEF$
$=2\times\dfrac{1}{4}$ area of $\triangle ABC$
$=\dfrac{1}{2}$ area of $\triangle ABC$
Question 23 of 36

23. Question
1 point(s)
The mid point of the sides of a triangle ABC along with any one of the vertices as the fourth point makes a parallelogram whose are is equal to :
- $\mathrm{\dfrac{2}{3}ar(ABC)} $
- $\mathrm{\dfrac{1}{4}ar(ABC)} $
- $\mathrm{\dfrac{1}{3}ar(ABC)} $
- $\mathrm{\dfrac{1}{2}ar(ABC)} $
Correct

Incorrect

Hint

Ratio of area of triangle to the area of parallelogram = 1 : 2
Question 24 of 36

24. Question
1 point(s)
ABCD is parallelogram and O is mid point of AB. If area of the parallelogram is 74 sq cm, then area of $ \triangle$DOC is
- 158 sq cm
- 37 sq cm
- 18.5 sq cm
- 222 sq cm
Correct

Incorrect

Hint

Area of $\triangle DOC=\dfrac{1}{2}\times$ area of parallelogram
$=\dfrac{1}{2}\times74=37\$ sq cm
Question 25 of 36

25. Question
1 point(s)
$AD $ is the median of a triangle $ABC $. Area of triangle $ADC=15\ cm^2 $, then $ar\ (\triangle ABC) $ is:
- $15\ cm^2 $
- $22.5\ cm^2 $
- $30\ cm^2 $
- $37.5\ cm^2 $
Correct

Incorrect

Hint

area of $\triangle ADC$ = area of $\triangle ABC$
$=15\ cm^2$
Question 26 of 36

26. Question
1 point(s)
In the figure, AB || DC. Which of the following is true about the figure ?
- $ar\ (AOD) = ar\ (BOC) $
- $ ar\ (AOB) = ar\ (COD)$
- $ ar\ (ADC) = ar\ (ABC)$
- $ ar (AOB) =\dfrac{1}{4}ar\ (ABCD)$
Correct

Incorrect

Hint

$AB||DC\\ \\ AD=BC$
So area of $\triangle AOD$ = area of $BOC$
(Triangles on the same base are equal)
Question 27 of 36

27. Question
1 point(s)
A rectangle and a rhombus are on the same base and between the same parallels. Then the ratio of their areas is :
- 1 : 1
- 1 : 2
- 1 : 3
- 1 : 4
Correct

Incorrect

Hint

Since a rectangle and a rhombus are on the same base and between the same parallels, their areas are equal. So, the ratio is 1:1
Question 28 of 36

28. Question
1 point(s)
ABCD is a trapezium with parallel sides AB = a cm, CD =b cm. E and F are the mid – points of non–parallel sides. The ratio of ar (ABFE) and ar (EFCD) is
- a : b
- (3a + b) : (a + 3b)
- (a + 3b) : (3a + b)
- (2a + b : (3a + b)
Correct

Incorrect

Hint

$EM=\dfrac{1}{2}AB\\ \\ MF=\dfrac{1}{2}CD\\ \\ EM+MF=\dfrac{1}{2}\ (a+b)$
Ratio $=\dfrac{\frac{1}{4}\times(3a+b)h}{\frac{1}{4}(3b+a)h}=\dfrac{3a+b}{3b+a}$
Question 29 of 36

29. Question
1 point(s)
If $E, F, G, H$ are respectively the mid points of the sides of a parallelogram $ABCD$, and $ar\ (EFGH) = 40\ cm^2$, then ar (parallelogram $ABCD$) is :
- $40\ cm^2 $
- $20\ cm^2 $
- $80\ cm^2 $
- $60\ cm^2 $
Correct

Incorrect

Hint

area of $EFGH=\dfrac{1}{2}$ area $ABCD$
$40\times2$ = area of $ABCD$
area of $ABCD=80\ cm^2$
Question 30 of 36

30. Question
1 point(s)
In the figure, D is the midpoint of side BC of $\triangle $ABC . and E is the midpoint of AD. Then the area of $\triangle $ABE is :
- $\dfrac{1}{3} $ area $(\triangle ABC) $
- $\dfrac{1}{2} $ area $(\triangle AEC) $
- $\dfrac{1}{2} $ area $(\triangle BEC) $
- $\dfrac{1}{4} $ area $(\triangle ABC) $
Correct

Incorrect

Hint

$\triangle ABE=\dfrac{1}{2}\times ar\ \triangle\ ABD\\ \\ \dfrac{1}{2}\ ar\ \triangle ABD=\dfrac{1}{2}\times\dfrac{1}{2}\times \ ar\ \triangle ABC\\ \\ =\dfrac{1}{4}\ ar\ \triangle ABC$
Question 31 of 36

31. Question
1 point(s)
In $\triangle $ABC, AD is median of $\triangle $ABC and BE is median of $\triangle $ABD. If ar ($\triangle $ABD) = $15\ \text{cm}^2 $, then ar ($\triangle $ABC) is
- $60\ cm^2 $
- $50\ cm^2 $
- $40\ cm^2 $
- $30\ cm^2 $
Correct

Incorrect

Hint

area of $\triangle ABD=\dfrac{1}{2}\times$ area of $\triangle ABC$
area of $\triangle ADC=2\times15=30\ cm^2$
Question 32 of 36

32. Question
1 point(s)
In the figure, AB || DC, then the triangles that have equal area are :
- $\triangle $ADX, $\triangle $ACX
- $\triangle $ADX, $\triangle $XCB
- $\triangle $ACX, $\triangle $XCB
- All of the above
Correct

Incorrect

Hint

$ar\ (\triangle ACX)=ar\ (\triangle ADX)$
Triangles with same base and between the same parallels are equal in area.
Question 33 of 36

33. Question
1 point(s)
$ D$ and $E $ are the points on the sides $AB $ and $AC $ respectively of triangle $ABC $ such that $DE||BC $. If area of $\triangle DBC=15\ cm^2 $, then area $\triangle EBC $ is:
- $30 \ cm^2 $
- $7.5 \ cm^2 $
- $15 \ cm^2 $
- $20 \ cm^2 $
Correct

Incorrect

Hint

$DE||BC$
$BC$ is base
$ar\ (\triangle DBC)=ar\ (\triangle EBC)\\ \\ ar\ (\triangle BCE)=15\ cm$
Question 34 of 36

34. Question
1 point(s)
If area of parallelogram $ ABCD$ is $25\ cm^2 $ and on the same base $CD $, a triangle $BCD $ is given such that area of $BCD=x\ cm^2 $, then the value of $x $ is:
- $25\ cm^2 $
- $12.5\ cm^2 $
- $15\ cm^2 $
- $20\ cm^2 $
Correct

Incorrect

Hint

area of $\triangle BCD=\dfrac{1}{2}\times25=12.5\ cm^2$
Question 35 of 36

35. Question
1 point(s)
In the figure, if parallelogram ABCD and rectangle ABEF are of equal area, then :
- perimeter of ABCD = perimeter of ABEF
- perimeter of ABCD < perimeter of ABEF
- perimeter of ABCD > perimeter of ABEF
- perimeter of ABCD = $\dfrac{1}{2} $ (perimeter of ABEF)
Correct

Incorrect

Hint

$AB=EF$ …(1)
$AB=CD$ …(2)
$AB+CD=EF+AB\\ \\ BEBE+AF\\ \\ AB+BC+CD+AD>AB+CD+BE+AF\\ \\ AB+BC+CD+AD>AB+BE+EF+AF\\ \\ (\because\ CD=AB=EF)$
Question 36 of 36

36. Question
1 point(s)
The length of the diagonal of the square is 10 cm. The area of the square is :
- $20\ cm^2 $
- $100\ cm^2 $
- $50\ cm^2 $
- $70\ cm^2 $
Correct

Incorrect

Hint

$AO=\dfrac{1}{2}\times10=5\ cm\\ \\ AB=\sqrt{5^2+5^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt2$
Area $=5\sqrt2\times5\sqrt2=25\times2=50\ cm^2$