Mathematics Class IX Questions Bank

Question 1 of 31

1. Question
1 point(s)
The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the field.
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Hint

$6x+7x+8x=420\\ \\ 21x=420\ \Rightarrow \ x=20$
Sides are $120m,\ 140m,\ 160m$
$S=\dfrac{a+b+c}{2}=\dfrac{420}{2}=210$
Area $=\sqrt{210(210-120)(210-140)(210-160)}$
$=\sqrt{210\times90\times70\times50}\\ \\ =100\sqrt{21\times9\times7\times5}=100\times7\times3\times\sqrt{15}\\ \\ =2100\sqrt{15}\ m^2$
Question 2 of 31

2. Question
1 point(s)
In the figure, find the area of the shaded region.
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Hint

$S=\dfrac{120+122+22}{2}\\ \\ =\dfrac{264}{2}=132$
Area of $\triangle ABC=\sqrt{132(132-120)(132-122)(132-22)}$
$=\sqrt{132\times12\times10\times110}\\ \\ =\sqrt{1742400}=1320\ m^2$
Area of $\triangle CDB=\sqrt{36(36-24)(36-26)(36-28)}$
$=\sqrt{36\times12\times10\times14}=\sqrt{60480}=246\ m^2\ \text{(approx)}$
Area of shaded region $=1320-246=1074\ m^2$
Question 3 of 31

3. Question
1 point(s)
Two sides of a triangle are 50 cm and 78 cm. Its perimeter is 240 cm. Find the altitude on the side of length 50 cm from the opposite vertex.
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Hint

Third side $=240-(78+50)=112\ cm$
$S=\dfrac{240}{2}=120\ cm$
Area $=\sqrt{120(120-78)(120-50)(120-112)}$
$=\sqrt{120\times42\times70\times8}=1680\ cm^2\\ \\ 1680=\dfrac{1}{2}\times50\times h\\ \\ \Rightarrow \dfrac{1680\times2}{50}=h\ \Rightarrow\ h=67.2\ cm$
Question 4 of 31

4. Question
1 point(s)
A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitude.
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Hint

$S=\dfrac{35+54+61}{2}=75\ cm$
Area $=\sqrt{75(75-35)(75-54)(75-61)}$
$=420\sqrt5\ cm^2\\ \\ \dfrac{1}{2}\times35\times h=420\sqrt5\\ \\ h=24\sqrt5\ cm$
Question 5 of 31

5. Question
1 point(s)
The sides of a triangular field are $41\ m,\ 40\ m$ and $9\ m$. Find the number of rose beds that can be prepared in the field, if each rose bed, on an average needs $900\ cm^2$ space.
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Hint

Area $=\sqrt{45(45-41)(45-40)(45-9)}$
$=\sqrt{32400}=180\ m^2\\ \\ =1800000\ cm^2$
Number of flower beds $=\dfrac{1800000}{900}=2000$
Question 6 of 31

6. Question
1 point(s)
A rhombus shaped sheet with perimeter $40\ cm $ and one diagonal $12\ cm $ is painted on both sides at the rate of $Rs\ 5 $ per$cm^2 $. Find the cost of painting.
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Hint

$x+x+x+x=40\\ \\ \Rightarrow 4x=40\Rightarrow x=10\ cm\\ \\ S=\dfrac{10+10+12}{2}=16\ cm$
Area of $\triangle ABC=\sqrt{16(16-10)(16-10)(16-12)}$
$=\sqrt{16\times6\times6\times4}=48\ cm^2$
Area of rhombus $=2\times48=96\ cm^2$
Cost of painting on both sides $=5\times2\times96=₹\ 960$
Question 7 of 31

7. Question
1 point(s)
In the figure, find the area of the trapezium PQRS with height PQ.
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Hint

$TR=\sqrt{169-2S}\\ \\ =\sqrt{144}=12\ m$
Area of $\triangle STR$
$=\dfrac{1}{2}\times12\times5=30\ m^2$
Area of rectangle $PQRT=12\times7=84\ m^2$
Area of trapezium $=30+84=114\ m^2$
Question 8 of 31

8. Question
1 point(s)
The ratio of the equal side to the base of an isosceles triangle is 3 : 2. If the perimeter of the triangle is 32 cm, find its area.
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Hint

$3x+3x+2x=32\\ \\ \Rightarrow8x=32\\ \\ \Rightarrow x=4$
Sides are $12\ cm,12\ cm,8\ cm$
Area $=\sqrt{16(16-12)(16-12)(16-8)}$
$=4\times4\times2\sqrt2=32\sqrt2\ cm^2$
Question 9 of 31

9. Question
1 point(s)
The lengths of the two sides of a right triangle containing the right angle differ by $2\ cm $. If the area of the triangle is $ 24\ cm^2$, find the perimeter of the triangle.
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Hint

$AB=x\ cm\\ \\ BC=(x-2)\ cm$
Area $=\dfrac{1}{2}\times x\times (x-2)$
$24=\dfrac{1}{2}\times x^2-2x\\ \\ \Rightarrow 48=x^2-2x\\ \\ \Rightarrow x^2-2x-48=0\\ \\ \Rightarrow x=8\ cm\\ \\ BC=6\ cm\\ \\ AC=\sqrt{64+36}=10\ cm$
Perimeter $=10+6+8=24\ cm$
Question 10 of 31

10. Question
1 point(s)
Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.
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Hint

$S=\dfrac{18+24+30}{2}=36\ cm$
Area $=\sqrt{36(36-18)(36-24)(36-30)}$
$=\sqrt{46656}=216\ cm^2\\ \\ 216=\dfrac{1}{2}\times18\times\ \text{height}\\ \\ \text{height}=\dfrac{216\times2}{18}=24\ cm$
Question 11 of 31

11. Question
1 point(s)
The sides of a triangular ground are 5m, 7m and 8m respectively. Find the cost of levelling the ground at the rate of Rs. 10 per $m^2 $. (use $\sqrt3=1.73 $)
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Hint

$S=\dfrac{5+7+8}{2}=10\ m$
Area $=\sqrt{10(10-5)(10-7)(10-8)}$
$=\sqrt{10\times5\times3\times2}\\ \\ =10\sqrt3=17.3\ m^2$
cost $=17.3\times10=₹\ 173$
Question 12 of 31

12. Question
1 point(s)
Find the area of a quadrilateral ABCD whose sides AB = 13 cm, BC = 12 cm, CD = 9 cm, DA = 14 cm and diagonal BD = 15 cm.
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Hint

$ar(\triangle ABD)\\ \\ =\sqrt{21(21-13)(21-14)(21-15)}\\ \\ =\sqrt{21\times8\times7\times6}=84\ cm^2\\ \\ ar(\triangle CBD)=\sqrt{18\times(18-12)(18-15)(18-9)}\\ \\ =\sqrt{18\times6\times3\times9}=54\ cm^2$
Area of quadrilateral $ABCD=84+54=138\ cm^2$
Question 13 of 31

13. Question
1 point(s)
A triangular park has sides 120 m, 80 m and 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant ? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gate on one side.
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Hint

$S=\dfrac{50+80+120}{2}=125\ m$
Area $=\sqrt{125(125-50)(125-80)(125-120)}$
$=\sqrt{125\times75\times45\times5}\\ \\ =375\sqrt{15}\ m^2$
Perimeter $=50+80+120=250\ m$
Length of wire needed $=250-3=247\ m$
Cost of fencing $=20\times247=Rs\ 4940$
Question 14 of 31

14. Question
1 point(s)
Sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find its area.
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Hint

$84=13x+14x+15x\\ \\ \Rightarrow 42x=84\Rightarrow x=2$
Sides are $26\ cm,28\ cm,30\ cm$
Area $=\sqrt{42(42-26)(42-28)(42-30)}$
$=\sqrt{42\times16\times14\times12}=336\ cm^2$
Question 15 of 31

15. Question
1 point(s)
The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm and the diagonal AC measures 42 cm. Find the area of the parallelogram.
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Hint

$S=\dfrac{34+20+42}{2}=48\ cm$
Area of $\triangle ACD=\sqrt{48(48-34)(48-20)(48-42)}$
$=\sqrt{48\times14\times28\times6}=336\ cm^2$
Area of parallelogram $=2\times336=672\ cm^2$
Question 16 of 31

16. Question
1 point(s)
The sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
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Hint

$S=\dfrac{540}{2}=270\ cm\\ \\ 12x+17x+25x=540\\ \\ \Rightarrow 54x=540\ \Rightarrow\ x=10$
Sides are $120,170,250$
Area $=\sqrt{270(270-120)(270-170)(270-250)}$
$=\sqrt{270\times150\times100\times20} =9000\ cm^2$
Question 17 of 31

17. Question
1 point(s)
Find the height of the trapezium in which parallel sides are 25 cm and 10 cm and non-parallel sides are 14 cm and 13 cm.
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Hint

Area of $\triangle ADE=\sqrt{21\times7\times8\times6}=84\ cm^2$
$\dfrac{1}{2}\times b\times h=84\\ \\ \Rightarrow height=\dfrac{84\times2}{15}=11.2\ cm$
Question 18 of 31

18. Question
1 point(s)
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use $\pi=3.14 $)
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Hint

$S=\dfrac{8+10+6}{2}=\dfrac{24}{2}=12\ cm$
Area $=\sqrt{12(12-8)(12-10)(12-6)}=24\ m^2$
Area of circle $=\pi\dfrac{(2)^2}{2}=3.14\ m^2$
Area used for growing grasses $=24-3.14=24.86\ m^2$
Question 19 of 31

19. Question
1 point(s)
The perimeter of a triangular ground is 900 m and its sides are in the ratio 3 : 5 : 4. Using Heron’s formula, find the area of the ground.
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Hint

$3x+5x+4x=900\\ \\ \Rightarrow 12x=900\Rightarrow x=75$
Sides are $225m,375m,300m$
Area $=\sqrt{450(450-225)(450-375)(450-300)}$
$=\sqrt{450\times225\times75\times150}=33750\ cn^2$
Question 20 of 31

20. Question
1 point(s)
Find the perimeter of an isosceles right triangle having an area of $200\ cm^2 $.
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Hint

$\dfrac{1}{2}\times b\times b=200\\ \\ \Rightarrow b^2=400\Rightarrow b=20\ cm$
Perimeter $=b+h+\sqrt{b^2+h^2}$
$=20+20+\sqrt{400+400}\\ \\ =40+\sqrt{800}\\ \\ =40+20\sqrt2=40+20\times1.414=68.28\ cm$
Question 21 of 31

21. Question
1 point(s)
Find the area of a rhombus whose perimeter is 200 m and one of the diagonals is 80 m.
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Hint

$9a=200\\ \\ \Rightarrow a=\dfrac{200}{4}=50m\\ \\ a^2=\left(\dfrac{d_1}{2}\right)^2+\left(\dfrac{d_2}{2}\right)^2\\ \\ (50)^2=(40)^2+\left(\dfrac{d_2}{2}\right)^2\\ \\ \dfrac{d_2}{2}=30\ \Rightarrow\ d_2=60m$
Area $=\dfrac{1}{2}\times d_1\times d_2=\dfrac{1}{2}\times60\times80=2400\ m^2$
Question 22 of 31

22. Question
1 point(s)
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 15 cm, 14 cm and 13 cm and the parallelogram stands on the base 15 cm, find the height of parallelogram.
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Hint

$S=\dfrac{15+14+13}{2}=21\ cm$
Area $=\sqrt{21(21-15)(21-14)(21-13)}$
$=\sqrt{21\times6\times7\times8}=84\ cm^2\\ \\ \mathrm{Base\times Height=Area}\\ \\ 15\times h=84\Rightarrow h=\dfrac{84}{15}=5.6\ cm$
Question 23 of 31

23. Question
1 point(s)
A triangular park in a city has dimensions $30\ m \times 26\ m \times 28\ m$. A gardener has to plant grass inside the park at Rs 1.50 per $m^2 $. Find the amount to be paid to the gardener.
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Hint

$S=\dfrac{a+b+c}{2}=\dfrac{30+26+28}{2}=42\ m$
Area $=\sqrt{42(42-30)(42-26)(42-28)}=336\ m^2$
Cost $=336\times1.50=₹504$
Question 24 of 31

24. Question
1 point(s)
A garden is in the shape of a quadrilateral. The sides of the garden are 9 m, 40 m, 28 m and 15 m respectively in consecutive order and the angle between first two sides is a right angle. Find the area of the garden.
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Hint

$AC^2=9^2+40^2=1681\\ \\ AC=41$
Area of $\triangle ABC=\dfrac{1}{2}\times9\times40=180\ m^2$
Area of $\triangle ACD=\sqrt{42(42-28)(42-15)(42-9)}$
$=\sqrt{42\times14\times27\times1}=126\ m^2$
Area of the garden $=180+126=306\ m^2$
Question 25 of 31

25. Question
1 point(s)
If each side of a triangle is doubled, then find the ratio of area of the new triangle thus formed and the given triangle.
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Hint

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Sides of the new triangle $=2a,2b,2c$
Area of the new triangle $=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$
$=4\sqrt{s(s-a)(s-b)(s-c)}\\ \\ =4\times$ area of triangle
Ratio = $4:1$
Question 26 of 31

26. Question
1 point(s)
In the figure, the dimensions of the rectangle ABCD are $51\ cm \times 25\ cm$. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle. If the area of the trapezium PQCD is $\dfrac{5}{6} $ th part of the rectangle, find the lengths QC and PD.
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Hint

Area of rectangle ABCD $=51\times25=1275\ cm^2$ …(1)
Let $QC=9x,PD=8x$
Area of trapezium $PQCD$
$=\dfrac{1}{2}\times(8x+9x)\times25\\ \\ =\dfrac{17x\times25}{2}$ …(2)
Area of $PQCD=\dfrac{5}{6}$ (Area of $ABCD$ ) (given)
$\dfrac{17x\times25}{2}=\dfrac{5}{6}\times1275\\ \\ x=\dfrac{5\times1275\times2}{17\times25\times6}=5\\ \\ QC=9\times5=45,PD=8\times5=40\ cm$
Question 27 of 31

27. Question
1 point(s)
Two parallel sides of a trapezium are 60 cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.
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Hint

$S=\dfrac{25+26+17}{2}=34$
Area $=\sqrt{34(34-25)(34-26)(34-17)}=204\ cm^2\\ \\ \dfrac{1}{2}\times BE\times CF=204\\ \\ \Rightarrow BE=\dfrac{204\times2}{17}=24\ cm$
Area of trapezium $=\dfrac{1}{2}\times(60+77)\times24=1644\ cm^2$
Question 28 of 31

28. Question
1 point(s)
Find the area of the quadrilateral shown in the figure.
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Hint

Area of $\triangle PQS$
$S=\dfrac{34+42+20}{2}=48$
Area $=\sqrt{48(48-34)(48-42)(48-20)}$
$=\sqrt{48\times14\times6\times28}=\sqrt{112896}=336\ cm^2$
Area of $\triangle SQR$
$S=\dfrac{29+20+21}{2}=35$
Area $=\sqrt{35(35-21)(35-20)(35-21)}$
$=\sqrt{35\times6\times15\times14}=\sqrt{44100}=210\ cm^2$
Area of quadrilateral $=210+336=546\ cm^2$
Question 29 of 31

29. Question
1 point(s)
The perimeter of a triangle is 50 cm. One side of the triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.
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Hint

Let the smaller side $=x\ cm$
one side $=(x+4)\ cm$
Third side $=(2x-6)\ cm$
$x+x+4+2x-6=50\\ \\ \Rightarrow 4x=52\Rightarrow x=13\ cm$
Sides are $13cm,17cm,20cm$
$S=\dfrac{50}{2}=25\ cm$
Area $=\sqrt{25(25-13)(25-17)(25-20)}$
$=\sqrt{25\times12\times8\times5}=20\sqrt{30}\ cm^2$
Question 30 of 31

30. Question
1 point(s)
In the figure, ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ABC is constructed. Find the height DF of the parallelogram.
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Hint

$S=\dfrac{7.5+7+6.5}{2}=\dfrac{21}{2}=10.5\ cm$
Area $=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$
$=\sqrt{441}=21\ cm^2$
Area of parallelogram $BCED=BC\times DF=7\times DF$
$\Rightarrow 21=7\times DF\\ \\ \Rightarrow DF=3\ cm$
Question 31 of 31

31. Question
1 point(s)
In the figure, find the area of the quadrilateral PQRS.
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Hint

Area of $\triangle PSQ$
$PQ=\sqrt{(26)^2-(24)^2}\\ \\ =\sqrt{676-576}=10$
Area $=\dfrac{1}{2}\times24\times10=120\ cm^2$
Area of $\triangle SRQ=\dfrac{\sqrt3}{4}\times26\times26$
$=169\sqrt3\\ \\ =292.37\\ \\ =292\ cm^2$ (approx)
Area of $PQRS=292+120=412\ cm^2$