Maths Class X Question Bank

Question 1 of 37

1. Question
1 point(s)
Find the area of the shaded region, if PQ = 24, PR = 7 cm and O is the centre of the circle
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Hint

$QR=\sqrt{576+49}=\sqrt{625}=25$
Radius $=\dfrac{25}{2}$
Area of semicircle $=\dfrac{1}{2}\times\dfrac{22}{7}\times\dfrac{25}{2}\times\dfrac{25}{2}$
$=\dfrac{6875}{28}\ cm^2$
Area of $PQR=\dfrac{1}{2}\times PQ\times PR=84\ cm^2$
Area of shaded region $=\dfrac{6875}{28}-84$
$=\dfrac{4523}{28}\ cm^2$
Question 2 of 37

2. Question
1 point(s)
In the figure, ABCDEF is any regular hexagon. With different vertices A, B, C, D, E and F as the centres of circles with same radius r are drawn. Find area the of the shaded portion.
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Hint

Each interior angle $=\dfrac{n-2}{n}\times180$
$=\dfrac{6-2}{6}\times180\\ \\ =4\times30=120$
Area of 6 shaded region $=6\times\dfrac{120}{360}\times \pi r^2=2\pi r^2$
Question 3 of 37

3. Question
1 point(s)
In the figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm, to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region (Use $ \pi=3.14$)
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Hint

$AE=EC=CD=BD=BF=FA=5\ cm$
Area of sector $CDE=\dfrac{\theta\pi r^2}{360}$
$=\dfrac{60\times3.14}{360}\times 5^2\\ \\ =13.0833\ cn^2$
Area of shaded region $=3\times13.0833=39.25\ cm^2$
Question 4 of 37

4. Question
1 point(s)
In the figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are given. Find the area of shaded region. $\left(use\ \pi=\dfrac{22}{7}\right) $
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Hint

Area of outer sector $=\dfrac{\theta}{360}\times \pi r^2_1$
Area of inner sector $=\dfrac{\theta}{360}\times \pi r^2_2$
Difference $=\dfrac{30}{360}\times \dfrac{22}{7}\times(7^2-3.5^2)=9.625\ cm^2$
Question 5 of 37

5. Question
1 point(s)
The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.
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Hint

$27.2=2r$ + length of arc
$27.2=2\times5.6+5.6\times \theta\\ \\ \Rightarrow \theta=2.857$
Area of sector $=\dfrac{180\times2.857}{\pi\times360}\times \pi\times(5.6)^2$
$=44.79=44.8\ cm^2$
Question 6 of 37

6. Question
1 point(s)
A horse is tied to a peg at one corner of a square shaped grass field of side 25 m by means of a 14 m long rope. Find the area of the part of the field in which the horse can graze $\left[Take\ \pi=\dfrac{22}{7}\right] $
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Hint

Area of the part of the field $=\dfrac{\pi r^2}{4}=\dfrac{\dfrac{22}{7}\times14\times14}{4}=154\ cm^2$
Question 7 of 37

7. Question
1 point(s)
Area of a sector of a circle of radius $36\ cm $ is $54\pi\ cm^2 $. Find the length of corresponding arc of sector.
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Hint

$54\pi =\dfrac{\theta}{360}\times \pi r^2\\ \\ \Rightarrow \dfrac{\theta}{360}=\dfrac{54}{36\times 36}$
length of arc $=2\pi r\left(\dfrac{\theta}{360}\right)$
$=2\pi \times 36\times \dfrac{54}{36\times36}=3\pi\ cm$
Question 8 of 37

8. Question
1 point(s)
The minute hand of a clock is $\sqrt{21} $ cm long. Find the area swept by the minute hand on the face
of the clock from 7.00 am to 7.05 am.
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Hint

$60\ min=360\\ \\ 1\ min=6\\ \\ 5\ min=5\times6=30^o=\theta$
Area $=\dfrac{30}{360}\times\dfrac{22}{7}\times\sqrt{21}\times \sqrt{21}\\ \\ =\dfrac{22}{4}=5.5\ cm^2$
Question 9 of 37

9. Question
1 point(s)
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at the speed of 66 km/hour ?
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Hint

Speed of the car $=66$ km/hr
$=\dfrac{66\times1000}{60}$ m/min
Speed of the car $=1100$ m/min
No. of revolutions $=\dfrac{11000}{0.8\pi}$
$=\dfrac{11000\times7}{0.8\times22}=4375$
Question 10 of 37

10. Question
1 point(s)
Find the area of the quadrant of that circle whose circumference is $22\ cm\ \left(use\ \pi=\dfrac{22}{7}\right) $
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Hint

$2\pi r=22\\ \\ 11=\pi r\Rightarrow r=\dfrac{11}{\pi}=\dfrac{7}{2}\ cm$
Area $=\dfrac{\pi r^2}{4}=\dfrac{1}{4}\times\dfrac{22}{7}\times \left(\dfrac{7}{2}\right)^2$
$=\dfrac{77}{8}\ cm^2$
Question 11 of 37

11. Question
1 point(s)
In the figure, the chord AB of a circle of radius 10 cm subtends an angle of 900 at the centre O. Find the area of the segment ACBA. (Take $\pi=3.14 $)
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Hint

Area of sector $OACB$
$=\dfrac{90}{36}\pi\ (10)^2=25\pi\ cm^2$
Area of $\triangle AOB=\dfrac{1}{2}\times10\times10=50\ cm^2$
Area $=25\pi-50$
$=78.5-50\\ \\ =28.5\ cm^2$
Question 12 of 37

12. Question
1 point(s)
A wheel has diameter 84 cm. Find how many complete revolutions must it make to cover 792 m.
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Hint

Circumference $=2\pi r=264\ cm=2.64\ m$
Total no. of revolutions
$=\dfrac{792}{2.64}=300$
Question 13 of 37

13. Question
1 point(s)
How many times will the wheel of a car rotate in a journey of 2002 m, if the radius of the wheel is 49 cm?
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Hint

Circumference $=2\times\dfrac{22}{7}\times49=308\ cm$
Total rotation $=\dfrac{200200}{308}=650$
Question 14 of 37

14. Question
1 point(s)
A chord of a circle of radius 12 cm subtends an angle of 60° at the centre. Find the area of the corresponding segment of the circle. [use latex]\pi=3.14 [/latex] and $\sqrt3=1.73 $ ]
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Hint

Area of segment $=\dfrac{\theta}{360}\pi r^2$ – Area of $\triangle$
$=\dfrac{60}{360}\pi(12)^2-\dfrac{\sqrt3}{4}(12)^2\\ \\ =75.42-62.28=13.14\ cm^2$
Question 15 of 37

15. Question
1 point(s)
In the figure, a circle of radius 7 cm is inscribed in a square. Find the area of the shaded portion. $\left[use\ \pi=\dfrac{22}{7}\right] $
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Hint

Area of circle $=\pi r^2$
$=\dfrac{22}{7}\times7\times7=154\ cm^2$
Area of square $=14\times14=196\ cm^2$
Area of remaining potion $=196-154=42\ cm^2$
Question 16 of 37

16. Question
1 point(s)
In the given figure, O is the centre of a circle. The area of sector OAPB is $\dfrac{5}{18} $ of the area of the circle. Find x.
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Hint

Area of sector $=\dfrac{5}{18}\times$ area of circle
$\Rightarrow \dfrac{\theta}{360}\times \pi r^2=\dfrac{5}{18}\times \pi r^2\\ \\ \Rightarrow \theta=\dfrac{5}{18}\times 360=100$
Question 17 of 37

17. Question
1 point(s)
Find the perimeter of the given figure, where AED is a semi-circle and ABCD is a rectangle.
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Hint

Perimeter $=AB+BC+CD+AED$
$=20+14+20+\dfrac{2\pi r}{2}\\ \\ =54+\dfrac{22}{7}\times7=76 \ cm$
Question 18 of 37

18. Question
1 point(s)
A horse is placed for grazing inside a rectangular field 70 m by 52 m. It is tethered to one corner by a rope 21 m long. On how much area can it graze? How much area is left ungrazed?
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Hint

Area of grass grazed by hourse
$=\dfrac{\theta}{360}\times \pi r^2\\ \\ =\dfrac{90}{360}\times \dfrac{22}{7}\times21\times\\ \\ =346.5\ cm^2$
Question 19 of 37

19. Question
1 point(s)
The area of a circle inscribed in an equilateral triangle is $154\ cm^2$. Find the perimeter of the triangle. [Take latex]\sqrt3=1.73 [/latex]]
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Hint

$\pi r^2=154\Rightarrow r^2=49\\ \\ r=7\ cm\\ \\ r=\dfrac{h}{3}\Rightarrow h=3r=3\times7=2\ km\\ \\ h=\dfrac{\sqrt3}{2}a\ \Rightarrow a=\dfrac{2h}{\sqrt3}=14\sqrt3\ cm$
Peremeter $=3a=3\times 14\sqrt3$
$=42\sqrt3\\ \\ =42\times1.73\\ \\ =72.66\ cm$
Question 20 of 37

20. Question
1 point(s)
Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshaded area. What area will be left ungrazed? (Take $\pi=3.14 $)
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Hint

Area of 4 quadrant $=\dfrac{\pi r^2}{4}$
$=4\times\left(\dfrac{1}{4}\times\dfrac{22}{7}\times25\times25\right)=1962.5$
Area of square plot $=50\times50=2500$
Area of ungrazed land $=2500-1962.5=537.5\ m^2$
Question 21 of 37

21. Question
1 point(s)
In the figure, $\triangle $ABC is right angled at A, with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Find the value of $r_2 $ the radius of the inscribed circle.
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Hint

$BC=\sqrt{8^2+6^2}=\sqrt{100}=10\ cm$
Area of $\triangle ABC=$ Area of (OAC+OAB+BOC)
$\Rightarrow \dfrac{1}{2}\times6\times8=\dfrac{1}{2}\times8\times r+\dfrac{1}{2}\times6\times r+\dfrac{1}{2}\times10\times r\\ \\ \Rightarrow 12r =24\Rightarrow r=2\ cm$
Question 22 of 37

22. Question
1 point(s)
Find the area of the segment AYB in the figure, if the radius of the circle is 21 cm and $\angle AOB=120^o $
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Hint

Area of circle $=\pi r^2$
$=3.14\times21\times21=1386\ cm^2$
Area of arc $=\dfrac{120}{360}\times1386=462\ cm^2$
Area of triangle $=21\times21\times\dfrac{\sin120}{2}=190.96\ cm^2$
Area of segment $=462-190.96=271.04\ cm^2$
Question 23 of 37

23. Question
1 point(s)
Three equal circles each of radius 6 cm touch one another as shown. Find the area enclosed between them.
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Hint

Area of equilateral $\triangle =\dfrac{\sqrt3}{4}\times a^2$
$=\dfrac{1.73}{4}\times12\times12=62.28\ cm^2$
Area of the arc $=\dfrac{1}{6}\pi r^2$
$=\dfrac{1}{6}\times\dfrac{22}{7}\times6\times6=18.86\ cm^2$
Area of 3 sectors $=3\times18.86=56.67\ cm^2$
Area of shaded portion $=62.28-56.57=5.71\ cm^2$
Question 24 of 37

24. Question
1 point(s)
In an equilateral triangle of side 12 cm, a circle is inscribed touching its sides. Find the area of the portion of the triangle not included in the circle.
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Hint

Area of equilateral $\triangle=\dfrac{\sqrt3}{4}\times12\times12=36\sqrt3\ cm^2$
In $\triangle BDO$
$\tan30=\dfrac{OD}{OB}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{\pi}{6}\\ \\ \Rightarrow r=2\sqrt3\ cm$
Area of circle $=\pi r^2$
$=3.14\times(2\sqrt3)^2=37.68\ cm^2$
Area of shaded region $=36\sqrt3-37.68=24.6\ cm^2$
Question 25 of 37

25. Question
1 point(s)
Prove that the area of a circular path of uniform width k surrounding a circular region of radius r is $\pi\ k(k+2r) $
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Hint

Let radius of circular region $=r$
Radius of circular path of uniform with $k$ surrounding the circular region of radius, $r=r+k$
Area of path $=\pi\ (r+k)^2-\pi r^2$
$=\pi r^2+\pi k^2+2\pi rk-\pi r^2\\ \\ =\pi k(2r+k)$
Question 26 of 37

26. Question
1 point(s)
Find the area of the shaded region in the given figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD. (Use $\pi=3.14 $)
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Hint

Radius of circle $=\dfrac{12}{2}=6\ cm$
Area of shaded region
$=12\times12-\pi\times6^2\\ \\ =144-133.04\\ \\ =30.96\ cm^2$
Question 27 of 37

27. Question
1 point(s)
In the figure, OACBO represents a quadrant of a circle of radius 7 cm with centre at O. If OD = 5 cm, find the area of the shaded region.
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Hint

Area of quadrant $=\dfrac{\pi r^2}{4}$
$=\dfrac{1}{4}\times\dfrac{22}{7}\times7^2=38.5\ cm^2$
Area of $\triangle =\dfrac{1}{2}\times7\times5=17.5\ cm^2$
Area of shaded region $=\dfrac{38.5}{17.5}=21\ cm^2$
Question 28 of 37

28. Question
1 point(s)
In the figure, diameter AB is 12 cm long. AB is trisected at points P and Q. Find the area of the shaded region.
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Hint

$AQ=4+4=8\ cm$
Area of shaded region $=2\times\left[\dfrac{1}{2}\pi(4)^2-\dfrac{1}{2}\pi(2)^2\right]\\ \\ \dfrac{22}{7}(16-4)=37.7\ cm^2$
Question 29 of 37

29. Question
1 point(s)
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle. (Use $\pi=3.14 $)
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Hint

In $\triangle OVS$ ,
$\dfrac{OV}{OS}=\cos60^o\\ \\ \Rightarrow \dfrac{OV}{12}=\dfrac{1}{2}\Rightarrow OV=6\ cm\\ \\ \dfrac{SV}{SO}=\sin60=\dfrac{\sqrt3}{2}\\ \\ \Rightarrow \dfrac{SV}{12}=\dfrac{\sqrt3}{2}\Rightarrow SV=6\sqrt3\ cm\\ \\ ST=2\times6\sqrt3=12\sqrt3\ cm$
Area of $\triangle OST=\dfrac{1}{2}\times12\sqrt3\times6$
$=36\sqrt3=62.28\ cm^2$
Area of sector $OSUT$
$=\dfrac{120}{360}\times\pi(12)^2=150.72\ cm^2$
Area of segment $SUT$
$=150.72-62.28=88.44\ cm^2$
Question 30 of 37

30. Question
1 point(s)
In the figure, ABCD is a square of side 8 cm. CBED and ADFB are quadrants of circle. Find the area of the shaded region. (Use $\pi=3.14 $)
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Hint

Area of quadrant $=\dfrac{1}{4}\times3.14\times8^2=50.24\ cm^2$
Area of $\triangle ABD=\dfrac{1}{2}\times8\times=32\ cm^2$
Shaded area $=2(50.54-32)=36.48\ cm^2$
Question 31 of 37

31. Question
1 point(s)
The length of a rope by which a cow is tethered is increased from 16 m to 23 m. How much additional area can the cow graze now? $\left(use\ \pi=\dfrac{22}{7}\right) $
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Hint

Additional area covered by
cow $=\dfrac{22}{7}(23^2-16^2)$
$=\dfrac{22}{7}\times39\times7=858\ m^2$
Question 32 of 37

32. Question
1 point(s)
The radii of two circles are 4 cm and 3 cm. Find the radius of the circle whose area is equal to the sum of the areas of the two circles. Also find the circumference of this circle.
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Hint

Area of new circle $=\pi r^2_1+\pi r^2_2$
$\Rightarrow \pi r^2=\dfrac{22}{7}(4^2+3^2)\\ \\ \Rightarrow \dfrac{22}{7}\times r^2=\dfrac{22}{7}\times 25\\ \\ \Rightarrow r=5\ cm$
Circumference $=2\times\dfrac{22}{7}\times5$
$=\dfrac{220}{7}=31.43\ cm^2$
Question 33 of 37

33. Question
1 point(s)
Find the area of the shaded region in the figure, if BC = BD = 8 cm, AC = AD = 15 cm and O is the centre of the circle. (Take $\pi=3.14 $)
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Hint

$\mathrm{ar\ \triangle ADB=ar\ \triangle ACB\\ \\ =\dfrac{1}{2}\times15\times8=60\ cm^2\\ \\ AB=\sqrt{15^2+8^2}=17\ cm}$
area of circle $=\dfrac{22}{7}\times\dfrac{17}{2}\times\dfrac{17}{2}=226.87\ cm^2$
area of shaded portion $=226.87-120=106.87\ cm^2$
Question 34 of 37

34. Question
1 point(s)
In a circle of radius 12 cm, an arc subtends an angle of 60° at the centre. Find
(i) Area of sector formed by the arc
(ii) Area of the segment formed by the corresponding chord.
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Hint

(i) Area of sector $=\dfrac{\theta}{360}\times\pi r^2$
$=\dfrac{60}{360}\times\dfrac{22}{7}\times12\times12=75.36$
(ii) Area of segment = Area of sector – area of $\triangle$ OAB
$=75.36-\dfrac{\sqrt3}{4}\times12\times12\\ \\ =75.36-62.28\\ \\ =13.14\ cm^2$
Question 35 of 37

35. Question
1 point(s)
In the figure, ABC is a triangle right angled at A. Find the area of the shaded region if AB = 6cm BC = 10 cm and I is the centre of incircle of $\triangle $ABC
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Hint

In $\triangle BAC$
$BC^2=AB^2+AC^2\\ \\ \Rightarrow 10^2=6^2+AC^2\Rightarrow AC=8\ cm$
Area of $\triangle ABC=\dfrac{1}{2}\times8\times6=24\ cm^2$
Area of $\triangle ABC=\dfrac{1}{2}r\ (AB+BC+CA)\\ \\ \Rightarrow 24=\dfrac{1}{2}r\ (6+8+10)\\ \\ \Rightarrow r=\dfrac{24}{12}=2\ cm$
Area of circle $=\pi r^2$
$=\dfrac{22}{7}\times2^2=12.56\ cm^2$
Area of shaded region $=24-12.56=11.44\ cm^2$
Question 36 of 37

36. Question
1 point(s)
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm and four circles are each of same radius.
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Hint

Area of shaded portion $=a^2-4(\pi r^2)$
$=14\times 14-4\times\dfrac{22}{7}\times\dfrac{7}{2}\times\dfrac{7}{2} \\ \\ =196-154=42\ cm^2$
Question 37 of 37

37. Question
1 point(s)
A square of side 4 cm is inscribed in a circle. Find the area enclosed between the circle and the square. $ \left[\pi=\dfrac{22}{7}\right]$
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Hint

$2r\left(\dfrac{1}{\sqrt2}\right)=4\\ \\ r=2\sqrt2\ cm$
Area of shaded region $=\pi r^2-4^2$
$=\pi (8)-16\\ \\ =25.12-16=9.12\ cm^2$