Mathematics Class IX

Question 1 of 25

1. Question
1 point(s)
If ABCD is a parallelogram, AE $\bot $ DC and CF $\bot $ AD. If AB = 10 cm, AE = 6 cm and CF = 5 cm, then AD is equal to:
- 10 cm
- 6 cm
- 12 cm
- 15 cm
Correct

Incorrect

Hint

AB = CD = 10 cm (Opposite sides of a parallelogram)
CF = 5 cm and AE = 6 cm
Now,
Area of parallelogram = Base × Altitude
$\mathrm{CD \times AE = AD \times CF\\ \\ 10 \times 6 = AD \times 5\\ \\ AD=\dfrac{60}{5}\\ \\ AD=12\ cm}$
Question 2 of 25

2. Question
1 point(s)
If E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively, then ar (EFGH) is equal to:
- $\dfrac{1}{2}\ \text{ar(ABCD)} $
- $\dfrac{1}{4}\ \text{ar(ABCD)} $
- $2\ \text{ar(ABCD)} $
- $\text{ar(ABCD)} $
Correct

Incorrect

Hint

$AD||BC$ and $AD=BC$
$\dfrac{1}{2}\ AD=\dfrac{1}{2}\ BC$
$AH||BF$ and $DH||CF$
$AH=BF$ and $DH=CF$ (H and F are midpoints)
$\therefore, ABFH$ and $HFCD$ are parallelograms.
$\triangle EFH$ and llgmABFH, both lie on a common base, FH.
$\therefore$ area of EFH $=\dfrac{1}{2}$ area of ABFH — 1
area of GHF $=\dfrac{1}{2}$ area of HFCD — 2
Adding eq. 1 and 2 we get;
area of $\triangle EFH$ + area of $\triangle GHF=\dfrac{1}{2}$ (area of ABFH + area of HFCD)
ar (EFGH) $=\dfrac{1}{2}$ ar(ABCD)
Question 3 of 25

3. Question
1 point(s)
If P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD, then:
- ar(APB) > ar(BQC)
- ar(APB) < ar(BQC)
- ar(APB) = ar(BQC)
- None of the above
Correct

Incorrect

Hint

$\triangle \text{APB}$ and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.
ar( $\triangle$ APB) = $\dfrac{1}{2}$ ar(parallelogram ABCD) — 1
ar( $\triangle$ BQC) = $\dfrac{1}{2}$ ar(parallelogram ABCD) — 2
From eq. 1 and 2:
$\mathrm{ar(\triangle APB)=ar(\triangle BQC)}$
Question 4 of 25

4. Question
1 point(s)
If ABCD and EFGH are two parallelograms between same parallel lines and on the same base, then:
- ar (ABCD) > ar (EFGH)
- ar (ABCD) < ar (EFGH)
- ar (ABCD) = ar (EFGH)
- None of the above
Correct

Incorrect
Question 5 of 25

5. Question
1 point(s)
A median of a triangle divides it into two
- Congruent triangles
- Isosceles triangles
- Right triangles
- Equal area triangles
Correct

Incorrect

Hint

Suppose, ABC is a triangle and AD is the median.
AD is the median of $\triangle$ ABC.
$\therefore$ It will divide $\triangle$ ABC into two triangles of equal area.
$\therefore$ ar(ABD) = ar(ACD) — (i)
also,
ED is the median of $\triangle$ ABC.
$\therefore$ ar(EBD) = ar(ECD) — (ii)
Subtracting (ii) from (i),
ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)
$\Rightarrow$ ar(ABE) = ar(ACE)
Question 6 of 25

6. Question
1 point(s)
In a triangle ABC, E is the mid-point of median AD. Then:
- ar(BED) = $\dfrac{1}{4} $ ar(ABC)
- ar(BED) = ar(ABC)
- ar(BED) = $\dfrac{1}{2} $ ar(ABC)
- ar(BED) = 2 ar(ABC)
Correct

Incorrect

Hint

ar(BED) = $\dfrac{1}{2}$ BD.DE
AE = DE (E is the midpoint)
BD = DC (AD is the median on side BC)
DE = $\dfrac{1}{2}$ AD — 1
BD = $\dfrac{1}{2}$ BC — 2
From eq. 1 and 2, we get;
ar(BED) = $\left(\dfrac{1}{2}\right)\times\left(\dfrac{1}{2}\right)$ BC $\times\left(\dfrac{1}{2}\right)$ AD
ar(BED) = $\left(\dfrac{1}{2}\right)\times\left(\dfrac{1}{2}\right)$ ar(ABC)
ar(BED) = $\dfrac{1}{4}$ ar (ABC)
Question 7 of 25

7. Question
1 point(s)
If D and E are points on sides AB and AC respectively of $\triangle $ABC such that ar(DBC) = ar(EBC). Then:
- DE is equal to BC
- DE is parallel to BC
- DE is not equal to BC
- DE is perpendicular to BC
Correct

Incorrect

Hint

$\triangle$ DBC and $\triangle$ EBC are on the same base BC and also having equal areas.
$\therefore$ They will lie between the same parallel lines.
$\therefore$ DE || BC.
Question 8 of 25

8. Question
1 point(s)
If Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Then,
- ar (AOD) = ar (BOC)
- ar (AOD) > ar (BOC)
- ar (AOD) < ar (BOC)
- None of the above
Correct

Incorrect

Hint

$\triangle$ DAC and $\triangle$ DBC lie on the same base DC and between the same parallels AB and CD.
ar( $\triangle$ DAC) = ar( $\triangle$ DBC)
$\Rightarrow$ ar( $\triangle$ DAC) − ar( $\triangle$ DOC) = ar( $\triangle$ DBC) − ar( $\triangle$ DOC)
$\Rightarrow$ ar( $\triangle$ AOD) = ar( $\triangle$ BOC)
Question 9 of 25

9. Question
1 point(s)
If Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar($\triangle $AOD) = ar($\triangle $BOC). Then ABCD is a:
- Parallelogram
- Rectangle
- Square
- Trapezium
Correct

Incorrect

Hint

: ar( $\triangle$ AOD) = ar( $\triangle$ BOC)
ar( $\triangle$ AOD) = ar( $\triangle$ BOC)
⇒ ar( $\triangle$ AOD) + ar( $\triangle$ AOB) = ar( $\triangle$ BOC) + ar( $\triangle$ AOB)
⇒ ar( $\triangle$ ADB) = ar( $\triangle$ ACB)
Areas of $\triangle$ ADB and $\triangle$ ACB are equal.
Therefore, they must lie between the same parallel lines.
Therefore, AB ∥ CD
Hence, ABCD is a trapezium.
Question 10 of 25

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

Incorrect

Hint

$ar\ \triangle ABP=\dfrac{1}{2}ar\ (ABCD)$
For parallelogram ABCD
Area $=AB\times DM$ ….(1)
For $\triangle PAB$
Area $=\dfrac{1}{2}\times AB\times PN$
$=\dfrac{1}{2}\times AB\times DM$ ….(2)
(Distance between the parallel lines are equal)
So, PN = DM
From of (1) and (2)
Area of $\triangle$ PAB $=\dfrac{1}{2}$ ar ABCD
Question 11 of 25

11. Question
1 point(s)
The area of trapezium ABCD in the given figure is
- $62\ cm^2 $
- $93\ cm^2 $
- $124\ cm^2 $
- $155\ cm^2 $
Correct

Incorrect

Hint

ABCD is a trapezium, AB || CD, AE || CD
$EC^2+EB^2=BC^2\\ \\ EC^2=BC^2-EB^2\\ \\ =17^2-15^2=64\\ \\ EC=8\ cm$
Area of trapezium ABCD
$=\dfrac{1}{2}\times(AB+CD)\times EC\\ \\ =\dfrac{1}{2}\times(8+15+8)\times8\\ \\ =124\ cm^2$
Question 12 of 25

12. Question
1 point(s)
In the given figure, ABCD is a parallelogram in which AB = CD = 5 cm and BD $\bot $ DC such that BD = 6.8 cm. The area of parallelogram ABCD is
- $17\ cm^2 $
- $25\ cm^2 $
- $34\ cm^2 $
- $68\ cm^2 $
Correct

Incorrect

Hint

Area of parallelogram $=5\times6.8=34\ cm^2$
Question 13 of 25

13. Question
1 point(s)
In the given figure, ABCD is a parallelogram in which diagonals AC and BD intersect at O. If ar ($\mathrm{||^{gm}} $ ABCD) is $52\ cm^2 $ then the ar($ \triangle$OAB) =
- $26\ cm^2 $
- $18.5\ cm^2 $
- $39\ cm^2 $
- $13\ cm^2 $
Correct

Incorrect

Hint

Area of $\triangle$ OAB = $\dfrac{1}{2}$ ar ( $\triangle$ ABD)
= $\dfrac{1}{2}\times\left(\dfrac{1}{2}\times\text{area of}\ ||^{gm}\ \text{ABCD}\right)\\ \\ =\dfrac{1}{2}\times\dfrac{1}{2}\times52=13\ cm^2$
Question 14 of 25

14. Question
1 point(s)
ABCD is a parallelogram in which DL $\bot $ AB, If AB = 10 cm and DL = 4 cm then the ar ($\mathrm{||^{gm}} $ ABCD is
- $40\ cm^2 $
- $80\ cm^2 $
- $20\ cm^2 $
- $196\ cm^2 $
Correct

Incorrect

Hint

Area of $||^{gm}=AB\times DL$
$=10\times4=40\ cm^2$
Question 15 of 25

15. Question
1 point(s)
The midpoint of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
- $\dfrac{1}{2}\ (ar\ \triangle ABC) $
- $\dfrac{1}{3}\ (ar\ \triangle ABC) $
- $\dfrac{1}{4}\ (ar\ \triangle ABC) $
- $ar\ (\triangle ABC) $
Correct

Incorrect

Hint

Area of parallelogram
$AFDE=ar(\triangle AFE)+ar\ (\triangle DFE)\\ \\ =2(ar\ AFE)\\ \\ =2\times\dfrac{1}{4}\ (ar\ \triangle ABC)\\ \\ =\dfrac{1}{2}\times(ar\ \triangle ABC)$
Question 16 of 25

16. Question
1 point(s)
The length of the diagonals of a rhombus are 12 cm and 16 cm the area of the rhombus is
- $192\ cm^2 $
- $96\ cm^2 $
- $64\ cm^2 $
- $80\ cm^2 $
Correct

Incorrect

Hint

Area of rhombus $=\dfrac{1}{2}\times$ Product of its diagonals
$=\dfrac{1}{2}\times12\times16=96\ cm^2$
Question 17 of 25

17. Question
1 point(s)
In a quadrilateral ABCD it is given that BD = 16 cm. If AL $\bot $ BD and CM $\bot $ BD such that AL = 9 cm and CM = 7 cm then ar(quad ABCD) is
- $256\ cm^2 $
- $128\ cm^2 $
- $64\ cm^2 $
- $96\ cm^2 $
Correct

Incorrect

Hint

Area of quadrilateral ABCD
$ar\ (\triangle ABD)+ar\ (\triangle BCD)\\ \\ =\dfrac{1}{2}\times16\times9+\dfrac{1}{2}\times16\times7\\ \\ =128\ cm^2$
Question 18 of 25

18. Question
1 point(s)
ABCD is a rhombus in which $\angle C=60^o $. Then AC : BD = ?
- $\sqrt3:1 $
- $\sqrt3:\sqrt2 $
- $3:1 $
- $3:2 $
Correct

Incorrect

Hint

ABCD is a rhombus
$BC=CD$
$\Rightarrow\angle BDC=\angle BCD$
In $\triangle BCD$
$\angle BDC+\angle BCD+\angle CBD=180^o$
Question 19 of 25

19. Question
1 point(s)
ABCD and ABFE are parallelogram such that ar(quad EABC) $ =17\ cm^2$ and ar($||^{gm} $ABCD) $ =25\ cm^2$. Then ar ($\triangle $BCF = ?
- $4\ cm^2 $
- $48\ cm^2 $
- $6\ cm^2 $
- $8\ cm^2 $
Correct

Incorrect

Hint

$ar\ (||gm\ ABCD)=ar\ (||gm\ ABFE)=25\ cm^2\\ \\ ar(\triangle BCF)=ar(||gm\ ABFE) - ar(quad EABC)\\ \\ =25-7=8\ cm^2$
Question 20 of 25

20. Question
1 point(s)
$\triangle $ABC and $\triangle $BDE are two equilateral triangles such that D is the midpoint of BC. Then $ar\ (\triangle BDE):ar(\triangle ABC)=? $
- $1:2 $
- $1:4 $
- $\sqrt3:2 $
- $ 3:4$
Correct

Incorrect

Hint

Area of an equilateral $\triangle=\dfrac{\sqrt3}{4}\times\text{(side)}^2$
$\dfrac{ar\ (\triangle BDE)}{ar\ (\triangle ABC)}=\dfrac{\frac{\sqrt3}{4}(BD)^2}{\frac{\sqrt3}{4}(BC)^2}\\ \\ =\dfrac{BC^2}{\dfrac{4}{BC}}=\dfrac{1}{4}$
Question 21 of 25

21. Question
1 point(s)
One of the diagonals of a quadrilateral is 16 cm. The perpendiculars drawn to it from its opposite vertices are 2.6 cm and 1.4 cm. Find its area
- $32\ cm^2 $
- $40\ cm^2 $
- $26\ cm^2 $
- $28\ cm^2 $
Correct

Incorrect

Hint

Area $=\dfrac{1}{2}\times d\times(h_1+h_2)$
$=\dfrac{1}{2}\times16\times(2.6+1.4)\\ \\ =8\times4=32\ cm^2$
Question 22 of 25

22. Question
1 point(s)
The ratio of the bases of two triangles is a : b. If the ratio of their corresponding alttudes is c : d. The ratio of their areas is
- ac : bd
- ad : be
- bd : ac
- be : ad
Correct

Incorrect

Hint

$b_1:b_2=a:b\\ \\ h_1:h_2=c:d$
Ratio of areas $=\dfrac{\frac{1}{2}\times b_1\times h_1}{\frac{1}{2}\times b_2\times h_2}=\dfrac{b_1}{b_2}\times\dfrac{h_1}{h_2}\\ \\ =\dfrac{a}{b}\times\dfrac{c}{d}\\ \\ =\dfrac{ac}{bd}$
Question 23 of 25

23. Question
1 point(s)
The area of the given quadrilateral ABCD is
- $36\ cm^2 $
- $24\ cm^2 $
- $48\ cm^2 $
- $32\ cm^2 $
Correct

Incorrect

Hint

$AC^2+BC^2=AD^2\\ \\ 16+a=AD^2\\ \\ AD=5\ cm$
Area of quadrilateral ABCD
= Area of $\triangle$ ABC + area of $\triangle$ ACD
$=\dfrac{1}{2}\times4\times3+\dfrac{1}{2}\times5\times12\\ \\ =6+30=36\ cm^2$
Question 24 of 25

24. Question
1 point(s)
In the given figure, AB || DC Identify the triangle that have equal areas
- $\triangle ADX,\triangle ACX $
- $\triangle ADX,\triangle XCB $
- $\triangle ACX,\triangle XDB $
- All of the above
Correct

Incorrect

Hint

$\triangle$ AXD and $\triangle$ AXC are on the same base and between the same parallels
Question 25 of 25

25. Question
1 point(s)
Which of the following is not a property of a rhombus?
- All four sides are equal
- Diagonals bisect each other
- Diagonals bisect opposite angles
- One diagonal between the diagonals is 60
Correct

Incorrect