Mathematics Class IX

Question 1 of 30

1. Question
1 point(s)
An isosceles right triangle has area $8\ cm^2 $. The length of its hypotenuse is
- $\sqrt{32}\ cm $
- $\sqrt{16}\ cm $
- $\sqrt{48}\ cm $
- $\sqrt{24}\ cm $
Correct

Incorrect

Hint

$AC=AB$ (given)
$\dfrac{1}{2}\times AC\times AB=8\\ \\ (AC)^2=16\\ \\ AC=AB=4\ cm\\ \\ BC=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}\ cm$
Question 2 of 30

2. Question
1 point(s)
The perimeter of an equilateral triangle is 60 m. The area is
- $10\sqrt3m^2 $
- $15\sqrt3m^2 $
- $20\sqrt3m^2 $
- $100\sqrt3m^2 $
Correct

Incorrect

Hint

Perimeter $=60\ cm$
Side $=\dfrac{60}{3}=20\ m$
Area of an equilateral triangle
$\dfrac{\sqrt3}{4}\times20\times20\\ \\ =100\sqrt3\ m^2$
Question 3 of 30

3. Question
1 point(s)
The sides of a triangle are 56 cm, 60 cm, 52 cm long. The area of the triangle is
- $1322\ cm^2 $
- $1311\ cm^2 $
- $1344\ cm^2 $
- $1392\ cm^2 $
Correct

Incorrect

Hint

$S=\dfrac{56+60+52}{2}=84\ cm\\ \\ \text{Area}=\sqrt{84(84-56)(84-60)(84-52)}\\ \\ =\sqrt{84\times28\times24\times32}=1344\ cm^2$
Question 4 of 30

4. Question
1 point(s)
The area of an equilateral triangle with side $2\sqrt3\ cm^2 $ is
- $ 5.196\ cm^2 $
- $ 0.866\ cm^2 $
- $3.496\ cm^2 $
- $1.732\ cm^2 $
Correct

Incorrect

Hint

$\text{Area}=\dfrac{\sqrt3}{4}\times a^2\\ \\ =\dfrac{\sqrt3}{4}\times2\sqrt3\times2\sqrt3\\ \\ =\dfrac{3}{2}\times2\sqrt3\\ \\ =3\sqrt3\\ \\ =5.196\ cm^2$
Question 5 of 30

5. Question
1 point(s)
The length of each side of an equilateral triangle having an area of $9\sqrt3\ cm^2 $ is
- 8 cm
- 36 cm
- 4 cm
- 6 cm
Correct

Incorrect

Hint

$\text{Area}=\dfrac{\sqrt3}{4}a^2=9\sqrt3\\ \\ a^2=36\\ \\ a=6\ cm$
Question 6 of 30

6. Question
1 point(s)
If the area of an equilateral triangle is $16\sqrt3\ cm^2 $ then perimeter of the triangle is
- 48 cm
- 24 cm
- 12 cm
- 36 cm
Correct

Incorrect

Hint

$\text{Side}=\dfrac{\sqrt3}{4}a^2=16\sqrt3\\ \\ a^2=64\\ \\ a=8\ cm\\ \\ \text{Perimeter}=3\times8=24\ cm$
Question 7 of 30

7. Question
1 point(s)
The sides of a triangle are 35 cm, 54 cm, 61 cm respectively. The length of its largest altitude is
- $16\sqrt5 $ cm
- $10\sqrt5 $ cm
- $24\sqrt5 $ cm
- $28 $ cm
Correct

Incorrect

Hint

$(35-x)^2+h^2=61^2\ ….(1)\\ \\ x^2+h^2=54^2\ …..(2)$
subtracting $\mathrm{eq^n}$ (2) from $\mathrm{eq^n}$ (1)
$(35-x)^2-x^2=61^2-54^2\\ \\ \Rightarrow 1225-68x+x^2-x^2=3721-2916\\ \\ \Rightarrow -70x=-420\\ \\ \Rightarrow x=6$
substituting in $\mathrm{eq^n}$ (2)
$x^2+h^2=54^2\\ \\ h^2=2880\\ \\ h=\sqrt{2880}\\ \\ h=24\sqrt5$
Question 8 of 30

8. Question
1 point(s)
The area of an isosceles triangle having base 2 cm and the length of one of the equal side 4 cm is
- $15\ cm^2 $
- $\sqrt{\dfrac{15}{2}cm^2} $
- $2\sqrt{15}\ cm^2 $
- $4\sqrt{15}\ cm^2 $
Correct

Incorrect

Hint

$S=\dfrac{4+4+2}{2}=5\ cm\\ \\ \text{Area}=\sqrt{5(5-4)(5-4)(5-2)}\\ \\ =\sqrt{5\times1\times1\times3}\\ \\ =\sqrt{15}\ cm^2$
Question 9 of 30

9. Question
1 point(s)
The edges of a triangular board are 6 cm, 8 cm, 10 cm. the cost of painting it at the rate of 9 paise per $\text{cm}^2 $ is
- Rs 2
- Rs 2.16
- Rs 2.48
- Rs 3.00
Correct

Incorrect

Hint

$S=\dfrac{6+8+10}{2}=\dfrac{24}{2}=12\ cm\\ \\ \text{Area}=\sqrt{12\times(12-6)(12-8)(12-10)}\\ \\ =\sqrt{12\times6\times4\times2}=\sqrt{576}=24\ cm^2\\ \\ \text{Cost}=24\times0.09\\ \\ =\text{Rs}\ 2.16$
Question 10 of 30

10. Question
1 point(s)
The area of a trapezium is 108 $\text{cm}^2 $ height is 8 cm. Find the length of its two parallel sides if one side is 3 cm greater than the other.
- 12 cm, 9 cm
- 15.6 cm, 18.6 cm
- 12 cm, 15 cm
- 15 cm, 18 cm
Correct

Incorrect

Hint

$\dfrac{1}{2}\times(x+x+3)\times8=108\\ \\ 2x+3=\dfrac{108\times2}{8}\\ \\ 2x+3=27\\ \\ 2x=24\\ \\ x=12$
Other side $=15\ cm$
Question 11 of 30

11. Question
1 point(s)
The perimeter of a triangular field is 90 and the ratio of the sides is 13:12:5. The area of the field in sq m is
- 270
- 40.249
- 1620
- 21060
Correct

Incorrect

Hint

$13x+12x+5x=90\\ \\ 30x=90\\ \\ x=3$
Sides are $39\ cm, 36\ cm, 15\ cm$
Area $=\sqrt{45\times(45-39)(45-36)(45-15)}\\ \\ =\sqrt{45\times6\times9\times30}\\ \\ =\sqrt{72900}=270\ cm^2$
Question 12 of 30

12. Question
1 point(s)
The sides of a triangular field are 7m, 15 m, 20 m. Find the number of rose beds if each rose bed on an average needs 400 $\text{cm}^2 $ space.
- 1065
- 2100
- 1050
- 1040
Correct

Incorrect

Hint

$S=\dfrac{7+15+20}{2}=21\ m\\ \\ \text{Area}=\sqrt{21\times(21-7)(21-15)(21-20)}\\ \\ =\sqrt{21\times14\times6\times1}\\ \\ =\sqrt{1764}=42\ m^2$
No. of rose bed $=\dfrac{420000}{400}=1050$
Question 13 of 30

13. Question
1 point(s)
BECD is a square, BC=20 cm, CA=7 cm and BD=12 cm. Find the area of the triangle ABC
- $54.6\ cm^2 $
- $ 21\ cm^2$
- $84\ cm^2 $
- $42\ cm^2 $
Correct

Incorrect

Hint

$\text{Area}=\dfrac{1}{2}\times BD\times AC\\ \\ =\dfrac{1}{2}\times12\times7=42\ cm^2$
Question 14 of 30

14. Question
1 point(s)
A parallelogram has a diagonal of 12 cm. The perpendicular distance of this original from an opposite vertex is 6 cm. Find the area of the parallelogram.
- $ 19\ cm^2$
- $ 72\ cm^2$
- $ 36\ cm^2$
- $ 18\ cm^2$
Correct

Incorrect

Hint

$\text{Area of}\ \triangle ADC\\ \\ =\dfrac{1}{2}\times6\times12=36\ cm^2\\ \\ \text{Area of}\ \triangle ABC=\dfrac{1}{2}\times6\times12=36\ cm^2\\ \\ \text{Area of}\ ABCD=36+36=72\ cm^2$
Question 15 of 30

15. Question
1 point(s)
The perimeter of a triangle is 54 cm. one side of a triangle is 7 cm longer then the smallest side and the third side is 5 cm less than 2 times the smallest side. Find the area of the triangle.
- $ 63\ cm^2 $
- $ 151.2\ cm^2$
- $ 27\ cm^2$
- $126\ cm^2 $
Correct

Incorrect

Hint

Let the smallest side be x can
Second side $= (x+7)\ cm$
Third side $= (2x-5)\ cm$
$\text{A/Q}\ x + x + 7 + 2x - 5 = 54\\ \\ 4x+2=54\\ \\ 4x=52\\ \\ x=13\ cm$
Sides are $13\ cm, 20\ cm, 21\ cm$
Area $=\sqrt{27\times(27-13)(27-20)(27-21)}$
$=\sqrt{27\times14\times7\times6}\\ \\ \sqrt{15876}=126\ cm^2$
Question 16 of 30

16. Question
1 point(s)
Find the area and the attitude of the figure below
- $66\ cm^2,\ 16.8\ cm $
- $105.6\ cm^2,\ 16.8\ cm $
- $66\ cm^2,\ 22\ cm $
- $66\ cm^2,\ 12\ cm $
Correct

Incorrect

Hint

$S=\dfrac{13+20+11}{2}=22\ cm\\ \\ \text{Area}=\sqrt{22\times9\times2\times11}=\sqrt{4356}=66\ cm^2\\ \\ \dfrac{1}{2}\times11\times h=66\\ \\ h=\dfrac{66\times2}{11}=12\ cm$
Question 17 of 30

17. Question
1 point(s)
The sides of a quadrilateral are 17 cm, 9 cm, 8 cm, 6 cm. The angle contained by the last two sides is a right angle. Find the area of the quadrilateral.
- $ 36\ cm^2$
- $ 60\ cm^2$
- $ 84\ cm^2$
- $ 120\ cm^2$
Correct

Incorrect

Hint

$\text{Area of}\ \triangle ADC\\ \\ =\dfrac{1}{2}\times6\times8=24\ cm^2\\ \\ AC=\sqrt{8^2+6^2}=\sqrt{64+36}=10\ cm\\ \\ S=\dfrac{10+17+9}{2}=18\ cm\\ \\ \text{Area}=\sqrt{18\times8\times1\times9}\\ \\ =\sqrt{1296}=36\ cm^2$
Areaof ABCD $=36+24=60\ cm^2$
Question 18 of 30

18. Question
1 point(s)
In the parallelogram PQRS, PQ = 20 cm. The altitude corresponding to PQ and PS are 17 cm and 18 cm. Find PS
- 20 cm
- 18.89 cm
- 17 cm
- 21 cm
Correct

Incorrect

Hint

Area of parallelogram having base 20 cm and altitude 17 cm
$\dfrac{1}{2}\times20\times17=170\ cm^2\\ \\ \dfrac{1}{2}\times PS\times18=170\\ \\ PS=\dfrac{170\times2}{18}\\ \\ PS=18.89\ cm$ (approx)
Question 19 of 30

19. Question
1 point(s)
The perimeter of a rhombus is 130 cm and one of its diagonals is 63 cm. what is the length of other diagonal.
- 16 cm
- 504 cm
- 8 cm
- 64 cm
Correct

Incorrect

Hint

Side $=\dfrac{\mathrm{P}}{4}=\dfrac{130}{4}=32.5 \mathrm{~cm}$
Area of rhombus $=\dfrac{1}{2} \times \mathrm{d}_{1} \times \mathrm{d}_{2}$
Area $=\sqrt{\mathrm{S}(\mathrm{S}-32.5)(\mathrm{S}-32.5)(\mathrm{S}-63})$
$\mathrm{S}=\dfrac{2 \times 32.5+63}{2}=\frac{128}{2}=64$
Area $=\sqrt{64 \times(64-32.5)(64-32.5)(64-63)}=504$
$504=\dfrac{1}{2} \times 63 \times \mathrm{d}$
$\mathrm{d}=\dfrac{2 \times 504}{63}=16 \mathrm{~cm}$
Question 20 of 30

20. Question
1 point(s)
The area of an equilateral triangle with a side of pc m is
- $\dfrac{p\sqrt3}{4}\ cm^2 $
- $\dfrac{(p^2\sqrt3)}{4\ cm^2} $
- $\dfrac{p^2}{4\ cm^2} $
- $\dfrac{p^2}{\sqrt3\ cm^2} $
Correct

Incorrect

Hint

Area $=\dfrac{\sqrt3}{4}a^2\\ \\ \dfrac{\sqrt3}{4}p^2$
Question 21 of 30

21. Question
1 point(s)
Find the percentage increase in the area of a triangle if each side increased to 3 times.
- 62
- 15
- 64
- 63
Correct

Incorrect

Hint

Let $\triangle$ ABC be the original triangle and $\triangle$ PQR has sides of 4 times the $\triangle$ ABC
$\mathrm{\dfrac{area\ of\ \triangle ABC}{area\ of\ \triangle PQR}}=\dfrac{a^2}{(4a)^2}=\dfrac{a^2}{16a^2}$
Area of the triangle increases by 1500%
Question 22 of 30

22. Question
1 point(s)
The perimeter of an isosceles triangle is 392 cm and its unequal side is 195 cm. Find the area of the triangle.
- $1379\ cm^2 $
- $1365\ cm^2 $
- $196\ cm^2 $
- None of these
Correct

Incorrect

Hint

$x+x+195=392\\ \\ 2x=277\\ \\ x=138.5\\ \\ S=\dfrac{392}{2}=196\ cm\\ \\ \text{Area}=\sqrt{196\times1\times57.5\times57.5}\\ \\ =\sqrt{196\times\dfrac{330625}{100}}\\ \\ =14\times57.5\\ \\ =805\ cm^2$
Question 23 of 30

23. Question
1 point(s)
A field in the shape of trapezium has its parallel sides as 15 m and 11 m while the non parallel sides are 13 m and 15 m. Find the area of the field
- $156\ m^2 $
- $24\ m^2 $
- $1560\ m^2 $
- $15.6\ m^2 $
Correct

Incorrect

Hint

$S=\dfrac{13+15+4}{2}=16\ cm\\ \\ Area=\sqrt{16\times3\times1\times12}=\sqrt{576}=24\ cm^2\\ \\ \dfrac{1}{2}\times4\times h=24\\ \\ h=12\ cm\\ \\ Area =\dfrac{1}{2}\times(11+15)\times12\\ \\ =\dfrac{1}{2}\times26\times12=156\ cm^2$
Question 24 of 30

24. Question
1 point(s)
Two adjacent sides of a parallelogram are 5 cm and 3.5 cm. one of its diagonals is 6.5 cm. Find the area of the parallelogram.
- $5\sqrt3 $
- $10\sqrt3 $
- $15\sqrt3 $
- $20\sqrt3 $
Correct

Incorrect

Hint

$S=\dfrac{6.5+5+3.5}{2}\\ \\ \text{Area}=\sqrt{7.5\times1\times2.5\times4}\\ \\ =\sqrt{75}=5\sqrt 3$
Area of a parallelogram $=2\times5\sqrt3=10\sqrt3\ cm^2$
Question 25 of 30

25. Question
1 point(s)
If the attitude of an equilateral triangle is $\sqrt6 $ cm. What is its area?
- $2\sqrt3 $
- $2\sqrt2 $
- $3\sqrt3 $
- $6\sqrt2 $
Correct

Incorrect

Hint

Area $=\dfrac{h^2}{\sqrt3}=\dfrac{(\sqrt6)^2}{\sqrt3}=\dfrac{6}{\sqrt3}$
$\dfrac{6\sqrt3}{3}\\ \\ =2\sqrt3\ cm^2$
Question 26 of 30

26. Question
1 point(s)
If the angles of a triangle are in the ratio 5:3:7. What is such a triangle called?
- An acute angled triangle
- An obtuse angled triangle
- A right triangle
- An isosceles triangle
Correct

Incorrect
Question 27 of 30

27. Question
1 point(s)
A triangle has perimeter 32 cm, one side is 11 cm and difference of other two sides is 5 cm. determine its area.
- $4\sqrt{30}\ cm^2 $
- $8\sqrt{30}\ cm^2 $
- $6\sqrt{30}\ cm^2 $
- $5\sqrt{30}\ cm^2 $
Correct

Incorrect

Hint

$a+b+c=32 \ cm \\ \\ a=11\ cm\\ \\ \Rightarrow11+b+c=32\\ \\ \Rightarrow b+c=21\ cm\ ….(1)\\ \\ \text{A/Q},b-c=5\ cm\ …..(2)$
Adding eqn (1) and (2)
$2b=26\Rightarrow b=13\ cm\\ \\ c=8\ cm\\ \\ S=\dfrac{32}{2}=16\ cm\\ \\ Area\sqrt{16\times(16-11)\times(16-13)\times(16-8)}\\ \\ =\sqrt{16\times5\times3\times8}=\sqrt{64\times30}=8\sqrt{30}\ cm^2$
Question 28 of 30

28. Question
1 point(s)
In a quadrilateral ABCD. If AB = 9m BC = 40 m, CD = 28 m, AD = 15 m and $\angle ABC=90^o$. Find its area.
- $136\ m^2 $
- $163\ m^2 $
- $360\ m^2 $
- $306\ m^2 $
Correct

Incorrect

Hint

Area of $\triangle ABC$
$=\dfrac{1}{2}\times9\times40=180\ cm^2\\ \\ AC=\sqrt{(40)^2+9^2}=\sqrt{1600+81}=\sqrt{1681}=41\ m\\ \\ S=\dfrac{15+28+41}{2}=42\ cm\\ \\ Area=\sqrt{42\times1\times14\times27}=\sqrt{15876}=126\ cm^2$
Area of ABCD $=180+126=306\ cm^2$
Question 29 of 30

29. Question
1 point(s)
If the area of an isosceles right triangle is $8\ cm^2 $, what is its perimeter?
- $(8+\sqrt2) $ cm
- $(8+4\sqrt2) $ cm
- $(4+8\sqrt2) $ cm
- $(12\sqrt2) $ cm
Correct

Incorrect

Hint

Sides are a, a, $\sqrt2$ a
Area $=\dfrac{1}{2}\times a^2$
$\Rightarrow 8=\dfrac{1}{2}\times a^2\\ \\ \Rightarrow a=4\ cm$
Perimeter $=a+a+\sqrt2a$
$=4+4+4\sqrt2a\\ \\ =(8+4\sqrt2a)\ cm$
Question 30 of 30

30. Question
1 point(s)
A farmer divided his field which is in the shape of a rhombus of side 100 m between his two sons equally. If the line of division is 160 m. What area of the field did each son get?
- 4840
- 1400
- 4800
- 3600
Correct

Incorrect

Hint

Area of $\triangle ABC=\sqrt{S(S-a)(S-b)(S-c)}$
$S=\dfrac{a+b+c}{2}\\ \\ =\dfrac{100+100+160}{2}\\ \\ =\dfrac{360}{2}=180\ m$
Area $=\sqrt{180\times80\times80\times20}=4800\ m^2$
$\therefore$ Each son gets an area of $4800\ m^2$