Areas Related to Circles Class 10 MCQ with Solution Maths

Areas Related to Circles Class 10 MCQ with Solution

1. The area of a quadrant of circle whose circumference is 22 cm, will be :

(A) 9.625 cm²

(B) 8.625 cm²

(D) 12.25 cm²

Ans. (A)

Sol. Circumerence of circle = 22 cm

2r = 22 ⇒ $r\, = \,\frac{7}{2}\,{\rm{cm}}$

Area of a quadrant of the circle = $\frac{1}{4}\,\pi {r^2}$

= $\frac{{77}}{8}\,{\rm{c}}{{\rm{m}}^2}$

= 9.625 cm²

2. A copper wire, when bent in the form of a square, encloses an area of 484 cm². If the same wire is bent in the form of a circle, the radius of the circle will be :

(A) 10 cm

(B) 14 cm

(D) 21 cm

Ans. (B)

Sol. Area of square = 484 cm²

a² = 484

⇒ a = 22 cm

length of wire = circumference of circle = 88 cm

2r = 88 ⇒ r = 14 cm

3. The circumference of two circles are in the ratio of 2 : 3, the ratio of their areas will be :

(A) 4 : 9

(B) 9 : 6

(D) 6 : 9

Ans. (A)

Sol. The ratio of two circumference = 2 : 3

\frac{{2\pi {r_1}}}{{2\pi {r_2}}}

\frac{2}{3}

\frac{{{r_1}}}{{{r_2}}}

\frac{2}{3}

Ratio of their areas = $\frac{{\pi r_1^2}}{{\pi r_2^2}}\, = \,{\left( {\frac{2}{3}} \right)^2}\, = \,\frac{4}{9}$

4. The length of the arc that subtends an angle 30° at the centre of a circle of radius 4 cm :

(A) $\frac{{2\pi }}{3}$ cm

(B) $\frac{{4\pi }}{8}$ cm

(D) $\frac{\pi }{3}$ cm

Ans. (A)

Sol. Radius of the circle (r) = 4 cm

Angle subtended by the arc (θ) = 30°

Length of the arc l = $\frac{\theta }{{360}}\, \times 2\,\pi r\, = \,\frac{{2\pi }}{3}$ cm

5. An arc of length 20 cm subtends an angle of 144° at the centre of a circle. The radius of the circle will be :

(A) 15 cm

(B) 20 cm

(D) 30 cm

Ans. (C)

Sol. Length of arc = l = $\frac{\theta }{{360}}\, \times 2\,\pi r\,$

⇒ $20\pi = \frac{{144^\circ }}{{360^\circ }} \times 2\pi r$ ⇒ r = 25 cm.

6. If a square is inscribed in a circle. The ratio of areas of circle to the area of the square will be :

(A) : 2

(B) 2 :

(D) 4 :

Ans. (A)

Sol. $\frac{{{\rm{area}}\,{\rm{of}}\,{\rm{circle}}}}{{{\rm{area}}\,{\rm{of}}\,{\rm{square}}}}$ = $\frac{{\pi {r^2}}}{{{a^2}}}\, = \,\frac{{\pi {r^2}}}{{{{(\sqrt 2 \,r)}^2}}}\, = \,\frac{{\pi {r^2}}}{{2{r^2}}}\, = \,\frac{\pi }{2}\, = \,\pi \,:\,2$

7. A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. The radius of its wheels is :

(A) 45.35 cm

(B) 36.35 cm

(D) None of these

Sol. (C)

Circumference of wheel = 1 complete revolution distance

2r = $\frac{{20}}{9}\,m$

r = $\frac{{1000 \times \,7}}{{198}}$

r = 35.35 cm

8. An arc of a circle is of length 5 cm and the sector it bounds has an area of 20 cm², the radius of the circle will be :

(A) 6 cm

(B) 8 cm

(D) None of these

Ans. (B)

Sol. area = $\frac{{\pi {r^2}}}{{2\pi r}}\, \times \,l$

area = $\frac{{rl}}{2}$ $\frac{{20\,\pi \, \times \,2}}{{5\pi }}$ = r = 8 cm

9. Area of a sector of angle P° of a circle with radius R is :

(A) $\frac{{\rm{P}}}{{270^\circ }} \times 2\pi {\rm{R}}$

(B) $\frac{{\rm{P}}}{{180^\circ }} \times 2\pi {\rm{R}}$

(D) $\frac{{\rm{P}}}{{720^\circ }} \times 2\pi {{\rm{R}}^{\rm{2}}}$

Ans. (D)

Sol. Area of sector of angle P° of a circle with radius R is = $\frac{P}{{720}} \times \,2\pi {R^2}$

10. In the given, if ABCD is a square of side 14 cm, then area of shaded region will be :

(A) 42 cm²

(B) 56 cm²

(D) 70 cm²

Ans. (A)

Sol. Area of shaded region = Area of square – area of circle

= 14 × 14 – r²

= $196\, - \,\frac{{22}}{7} \times \,7 \times \,7$

= 196 – 154 = 42 cm²

11. Points A, B, C & D are the centres of four circles such that each have a radius of length one unit. If a point is selected at random from the interior of square ABCD, what is the probability that the point will be chosen from the shaded diagram.

(A) $\left( {\frac{\pi }{4}\,\, - \,\,1} \right)$

(B) $\left( {\frac{\pi }{2}\,\, - \,\,1} \right)$

(D) $\left( {1\,\, - \,\frac{\pi }{4}} \right)$

Ans. (D)

Sol. Radius = 1

Area of square = 2 × 2 = 4 sq. unit

Area of shaded region = (4 – ) unit

Probability that the point chosen from shaded region

= $\left( {\frac{{4-\pi }}{4}} \right)$ = $\left( {1\,\, - \,\frac{\pi }{4}} \right)$

12. Two concentric circles with centre O have A, B, C, D as the points of intersection with line l. If AD = 12 cm and BC = 8 cm, then BD =

(A) 8 cm

(B) 9 cm

(D) None of these

Ans. (C)

Sol. AD = 12 cm

BC = 8 cm

BD = ?

Drop the perpendicular from centre to given line segment AD.

It will bisect the chard AD or BC.

CE = 4

DE = 6

CD = DE – CE = 6 – 4 = 2

BD = BC + CD = 8 + 2 = 10 cm.

13. In the given figure, a regular hexagon is circumscribed by a circle of radius 5 units. Then the side of hexagon is :

(A) $\frac{{5\sqrt 3 }}{2}$ units

(B) 6 units

(D) 5 units

Ans. (D)

Sol. Angle made by the side of regular hexagon at centre of circle is 60°

AB = 5.

14. In the figure below, RSTV is a square inscribed in a circle with centre O and radius r. The total area of shaded region is

(A) ${r^2}(\pi -2)$

(B) $2r(2-\pi )$

(D) $\pi {r^2}-8r$

Ans. (A)

Sol. VT = ST = $2r/\sqrt 2$

Area of square = $\frac{{2r}}{{\sqrt 2 }} \times \frac{{2r}}{{\sqrt 2 }} = 2{r^2}$

Area of circle = r²

Area of shaded region = r² – 2r²

= r² ( – 2)

15. Find the area of circle whose radius is 5 cm :

(A) 25 cm²

(B) 26 cm²

(D) None of these

Ans. (A)

Sol. Area = × 5 × 5

= 25 cm²

16. Two non intersecting circles,one lying inside the other are of radii x and y ( x > y). If the minimum distance between circumferences is z, the distance between their centres is :

(A) x – y + z

(B) x – y – z

(D) x – y

Ans. (B)

Sol. CD = Z

x = AB + BC + CD

AB = x – BC – CD = x – y – z

17. The distance between the centres of the two circles of radii r₁and r₂ is d (r₁ > r₂). They will touch each other internally if:

(A) d = r₁ or r₂

(B) d = r₁ + r₂

(D) d = $\sqrt {{r_1}{r_2}}$

Ans. (C)

Sol. AB + BC = AC

AB = AC – BC = r₁ – r₂

⇒ d = r₁ – r₂

18. The length of the circumference of a circle equals the perimeter of a triangle of equal sides. This is also equal to the perimeter of a square. The areas covered by the circle, triangle and square are c, t and s respectively. Then:

(A) s > t > c

(B) c > t > s

(D) s > c > t

Ans. (C)

Sol. 2r = 3a = 4b = K

(c) Area of Circle = $\pi {r^2} = \pi \times \frac{{{K^2}}}{{4{\pi ^2}}} = \frac{{{K^2}}}{{4\pi }}$ $= \frac{{{K^2}}}{{4 \times 3.14}} = \frac{{{K^2}}}{{12.56}}$

(t) Area of triangle = $\frac{{\sqrt 3 }}{4} \cdot {a^2} = \frac{{\sqrt 3 }}{4} \cdot \frac{{{K^2}}}{9} = \frac{{{K^2}}}{{12\sqrt 3 }}$ $= \frac{{{K^2}}}{{12 \times 1.73}} = \frac{{{K^2}}}{{20.76}}$

(s) Area of square = b² = $\frac{{{K^2}}}{{16}}$

Area of c >s > t

19. In figure, a circle touches the side BC of ΔABC at P and also touches AB and AC produced in Q and R respectively. If AQ = 5 cm, find the perimeter of ΔABC.

(A) 14 cm

(B) 15 cm

(D) 10 cm

Ans. (D)

Sol. Let BQ = BP = x and CR = PC = y

BC = x + y, AC = 5 – y, AB = 5 – x

Perimeter of ΔABC = 5 – y + 5 – x + x + y = 10 cm.

20. Two chords AB, CD of lengths 6cm, 12cm respectively of a circle are parallel. If the perpendicular distance between AB and CD is 3cm, find the radius of the circle.

(A) $5\sqrt 3 \,cm$

(B) $3\sqrt 5 \,cm$

(D) $2\sqrt 3 \,cm$

Ans. (B)

Sol. LM = 3 cm

LB = 3 cm

MD = 6 cm

LM = LO – MO

= $\sqrt {O{B^2}-L{B^2}} -\sqrt {O{D^2}-M{D^2}}$

3 = $\sqrt {{r^2}-9} -\sqrt {{r^2}-36}$

On solving get r = $3\sqrt 5$ cm

21. In figure, O is the centre of circle of radius 28 cm. Find the area of minor segment ASB :

(A) 30.81 cm²

(B) 15.25 cm²

(D) 45.75 cm²

Ans. (A)

Sol. Area of segment ASB = Area of sector AOB – Area of ΔAOB

= $\frac{{45}}{{360}} \times \pi \times 28 \times 28-\frac{1}{2} \times 28 \times 28 \times \sin 45^\circ$

= 30.81 cm²

22. If the area of a sector is one-twelfth that of the complete circle. Then the angle of the sector will be :

(A) 30°

(B) 40°

(D) 60°

Ans. (A)

Sol. angle of sector = $\frac{{360}}{{12}}$

= 30°.

23. In the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle, then area of the ΔPQR will be :

(A) 72 cm²

(B) 104 cm²

(D) 96 cm²

Ans. (C)

Sol. ΔPQR will be right angled triangle

So, Area of ΔPQR = $\frac{1}{2} \times 24 \times 7$

= 84 cm²

24. In the given figure, ABCD is a rectangle, having AB = 20 cm and BC = 14 cm. Two sectors of 180° have been cut off. The area of shaded region will be :

(A) 96 cm²

(B) 144 cm²

(D) 112 cm²

Ans. (C)

Sol. Area of shaded region = $20 \times 14 - \left( {\frac{1}{2}\pi {r^2}} \right) \times 2$

here r = $\frac{{14}}{2}$ = 7

Area of shaded region = $20 \times 14-\frac{1}{2} \times \frac{{22}}{7} \times 7 \times 7 \times 2$

= 126 cm²

25. The area of the right angled triangle, if the radius of its circumcircle is 5 cm and altitude drawn to the hypotenuse is 4 cm :

(A) 25 cm²

(B) 21 cm²

(D) 20 cm²

Ans. (D)

Sol. Circumcentre of right angled triangle will be mid point of hypotenuse.

So, AC = 2 × 5 = 10

Area = $\frac{1}{2} \times 10 \times 4 = 20$ cm²

26. A Sector of a circle of radius 4 cm contains an angle of 30°. Then area of sector is :

(A) 3.6 cm²

(B) 4.19 cm²

(D) None of these

Ans. (B)

Sol. Area of sector = $\frac{{30^\circ }}{{360^\circ }} \times \pi \times 4 \times 4$

= 4.19 cm²

27. The area of a sector of a circle of radius 5 cm is 5 $\pi$ cm². Then the angle contained by the sector is :

(A) 75°

(B) 36°

(D) 57°

Ans. (C)

Sol. $\frac{\theta }{{360}} \times \pi {r^2}$ = 5 $\pi$

$\frac{\theta }{{360}} \times \pi \times 5 \times 5$ = 5 $\pi$

θ = 72°

28. The sum of the radii of two circles is 140 cm and the difference of their circumference is 88 cm. The diameters of circles are :

(A) 154, 126

(B) 145, 162

(D) 190, 154

Ans. (A)

Sol. r₁ + r₂ = 140 cm ……….(1)

2 $\pi$ r₁ – 2 $\pi$ r₂ = 88 ……….(2)

from (1) & (2)

r₁ = 77

r₂ = 63

D₁ = 154 cm

D₂ = 126 cm

29. A chord PQ of length 12 cm subtends on of 60° at the centre of a circle. Then the arc length of minor segment cut off by chord PQ is :

(A) 15

(B) 16.8

(D) None of these

Ans. (D)

Sol. Triangle POB will be equilateral triangle

So r = 12

Arc of minor segment = $\frac{{60}}{{360}} \times 2\pi \times 12$

= 4 $\pi$

= $\frac{{4 \times 22}}{7}$

= 12.6

30. The area of shaded region is :

(A) 437.5 cm²

(B) 385.6 cm²

(D) None of these

Ans. (C)

Sol. Area of shaded Region = ${50^2}-4 \times \frac{1}{4}\pi \times {(25)^2}$

= 2500 – 1964.3 = 535.7 cm²

31. A sector of 56° cut out from a circle contains area 4.4 cm². Then the radius of the circle is :

(A) 2 cm

(B) 3.5 cm

(D) 4 cm

Ans. (C)

Sol. $\frac{{56^\circ }}{{360^\circ }} \times \frac{{22}}{7} \times {r^2}$ = 4.4

r = 3 cm.

32. From a circular sheet of paper with a radius 20 cm, four circles of radius 5 cm each are cut out. What is the ratio of the cut to the reamining circular portion :

(A) 1 : 3

(B) 4 : 1

(D) 4 : 3

Ans. (A)

Sol. Total Area cut out = 4 ×(5)² = 100

Initial Area = (20)² × $\pi$ = 400 $\pi$

Remaining Area = (400 – 100) $\pi$ = 300 $\pi$

Ratio = $\frac{{100\pi }}{{300\pi }}$ = 1/3

33. The diagram represents the area swept by the wiper of a car. With the dimensions given in the figure, calculate the shaded area swept by the wiper :

(A) 38.5 cm²

(B) 205.34 cm²

(D) 208.16 cm²

Ans. (A)

Sol. Area of shaded Region =

= 38.5 cm²

34. If the perimeter of a semi-circular protractor is 66 cm. Then diameter of protractor is :

(A) 42 cm

(B) 21 cm

(D) $\frac{{77}}{3}$ cm

Ans. (D)

Sol. $\pi$ r + 2r = 66

r [ $\pi$ + 2] = 66

r = $\frac{{66}}{{\frac{{22}}{7} + 2}}$

r = $\frac{{66}}{{36}} \times 7$

= $\frac{{77}}{6}$

so, diameter = 2r = $\frac{{77}}{3}$ cm

35. Find the value of q (in degree) in given sector of the circle, if area of sector is $\frac{{22}}{7}\,\,c{m^2}$ and radius is 3 cm :

(A) 55^o

(B) 60^o

(D) 70^o

Ans. (C)

Sol. $\frac{\theta }{{360}} \times \pi \times {3^2}$ = $\frac{{22}}{7}$

θ = 40°

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