Chapter 11 – Area Related To Circles

Class 10 Mathematics · 69 questions · 61 with answers

Solved examples

Ex. 1Multiple choice

If the area of a circle is 154 cm2, then its perimeter is

  • (A)11 cm
  • (B)22 cm
  • (C)44 cm
  • (D)55 cm
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Solution : Answer (C)

Ex. 2Multiple choice

If θ is the angle (in degrees) of a sector of a circle of radius r, then area of the sector is r 2 θ r 2 θ 2 r θ 2 r θ

  • (A)
  • (B)
  • (C)
  • (D)360 180 360 180
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Solution : Answer (A)

Ex. 1True / FalseExercise 11.1

Is the following statement true? Give reasons for your answer. Area of a segment of a circle = area of the corresponding sector – area of the corre- sponding triangle.

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Solution : Statement is not true. It is true only for a minor segment. In the case of a major segment, area of the triangle will have to be added to the corresponding area of the sector.

Ex. 2Short answerExercise 11.1

In Fig. 11.2, a circle is inscribed in a square of side 5 cm and another circle is circumscribing the square. Is it true to say that area of the outer circle is two times the area of the inner circle? Give reasons for your answer.

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Solution : It is true, because diameter of the inner circle = 5 cm and that of outer circle = diagonal of the square = 5 2 cm. 5 2 5 A1 So, A1 = π and A2 = π , giving A2 = 2 2 2

Ex. 1Short answerExercise 11.2

Find the diameter of the circle whose area is equal to the sum of the areas of the two circles of diameters 20 cm and 48 cm.

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Solution : Here, radius r1 of first circle = cm = 10 cm and radius r2 of the second circle = cm = 24 cm Therefore, sum of their areas = r 12 + r 22 = 2 + 2 = × (1) Let the radius of the new circle be r cm. Its area = π r2 (2) Therefore, from (1) and (2), π r2 = π × 676 or r2 = 676 i.e., r = 26 Thus, radius of the new circle = 26 cm Hence, diameter of the new circle = 2×26 cm = 52 cm

Ex. 2Short answerExercise 11.2

Find the area of a sector of circle of radius 21 cm and central angle 120°. θ

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Solution : Area of the sector = × r 2 120 22 = × × (21) 2 cm 2 360 7 = 22 × 21 cm 2 = 462 cm 2

Ex. 3Short answerExercise 11.2

In Fig 11.4, a circle of radius 7.5 cm is inscribed in a square. Find the area of the shaded region (Use π = 3.14) AREA RELATED TO CIRCLES 125

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Solution : Area of the circle = π r2 = 3.14 × (7.5)2 cm2 = 176. 625 cm2 Clearly, side of the square = diameter of the circle = 15 cm So, area of the square = 152cm2 = 225 cm2 Therefore, area of the shaded region = 225 cm2 – 176.625 cm2 = 48.375 cm2

Ex. 4Short answerExercise 11.2

Area of a sector of a circle of radius 36 cm is 54 π cm2. Find the length of the corresponding arc of the sector.

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Solution : Let the central angle (in degrees) be θ. π × (36) 2 So, = 54 π

Ex. 1Short answerExercise 11.3

A chord of a circle of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding major segment of the circle. (Use π = 3.14). AREA RELATED TO CIRCLES 129

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Solution : Let A B be the chord of a circle of radius 10 cm, with O as the centre of the circle (see Fig. 11.14). Here, ∠A O B = 90° and we have to find the area of the major segment (which is shaded). As ∠AOB= 90°, therefore angle of the major sector = 360° – 90° = 270° 270 2 So, area of the major sector = × π × (10) cm2 3 2 = × 3.14 × 100 cm 2 2 = 75 × 3.14 cm = 235.5 cm Now, to find the area of ∆ OAB, draw OM ⊥ AB.

Ex. 2Short answerExercise 11.3

With the vertices A, B and C of a triangle ABC as centres, arcs are drawn with radii 5 cm each as shown in Fig. 11.15. If AB = 14 cm, BC = 48 cm and CA = 50 cm, then find the area of the shaded region. (Use π = 3.14).

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Solution : Area of the sector with angle A ∠A ∠A = ×π r 2 = ×π× (5)2 cm2 360° 360° Area of the sector with angle B ∠B ∠B = ×π r 2 = ×π× (5)2 cm2 360° 360° AREA RELATED TO CIRCLES 131 ∠C and the area of the sector with angle C = × π × (5)2 cm2 360° Therefore, sum of the areas (in cm ) of the three sectors ∠A 2 ∠B ∠C = × π × (5) + ×π× (5) 2 + ×π× (5) 2 360° 360° 360° ∠A + ∠B + ∠C = × 25 π 360° 180° = × 25 π cm 2 (Because ∠A +∠B + ∠C = 180°) 360° π 2 2 2 = 25 × cm = 25 × 1.57 cm = 39.25 cm Now, to find area of ∆ ABC, we find a+b+c 48 + 50 + 14 s= = cm = 56 cm

Ex. 3Short answerExercise 11.3

A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase in area of the grassy lawn in which the calf can graze.

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Solution : Let the calf be tied at the corner A of the square lawn (see Fig. 11.16) Then, the increase in area = Difference of the two sectors of central angle 90° each and radii 11.5 m (6 m + 5.5 m) and 6 m, which is the shaded region in the figure. So, required increase in area 90 90 = × × 2 − × 2 2 360 360 = × (11.5 + 6) (11.5 − 6) m 2 = × 17.5 × 5.5m2 7×4 = 75.625 m .

Questions

Q1Multiple choiceExercise 11.1

If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then

  • (A)R1 + R2 = R
  • (B)R12 + R 22 = R2
  • (C)R1 + R2 < R
  • (D)R12 + R 22 < R 2
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(B) R12 + R 22 = R2

Q2Multiple choiceExercise 11.1

If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then

  • (A)R1 + R2 = R
  • (B)R1 + R2 > R
  • (C)R1 + R2 < R
  • (D)Nothing definite can be said about the relation among R1, R2 and R. AREA RELATED TO CIRCLES 121
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(A) R1 + R2 = R

Q3Multiple choiceExercise 11.1

If the circumference of a circle and the perimeter of a square are equal, then

  • (A)Area of the circle = Area of the square
  • (B)Area of the circle > Area of the square
  • (C)Area of the circle < Area of the square
  • (D)Nothing definite can be said about the relation between the areas of the circle and square.
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(B) Area of the circle > Area of the square

Q4Multiple choiceExercise 11.1

Area of the largest triangle that can be inscribed in a semi-circle of radius r units is 1 2

  • (A)r2 sq. units
  • (B)r sq. units
  • (C)2 r2 sq. units
  • (D)2 2 r sq. units
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(A) r2 sq. units

Q5Multiple choiceExercise 11.1

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

  • (A)22 : 7
  • (B)14 : 11
  • (C)7 : 22
  • (D)11: 14
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(B) 14 : 11

Q6Multiple choiceExercise 11.1

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

  • (A)10 m
  • (B)15 m
  • (C)20 m
  • (D)24 m
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(A) 10 m

Q7Multiple choiceExercise 11.1

The area of the circle that can be inscribed in a square of side 6 cm is

  • (A)36 π cm2
  • (B)18 π cm2
  • (C)12 π cm2
  • (D)9 π cm2
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(D) 9 π cm2

Q8Multiple choiceExercise 11.1

The area of the square that can be inscribed in a circle of radius 8 cm is

  • (A)256 cm2
  • (B)128 cm2
  • (C)64 2 cm2
  • (D)64 cm2
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(B) 128 cm2

Q9Multiple choiceExercise 11.1

The radius of a circle whose circumference is equal to the sum of the circum- ferences of the two circles of diameters 36cm and 20 cm is

  • (A)56 cm
  • (B)42 cm
  • (C)28 cm
  • (D)16 cm
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(C) 28 cm

Q10Multiple choiceExercise 11.1

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is

  • (A)31 cm
  • (B)25 cm
  • (C)62 cm
  • (D)50 cm
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(D) 50 cm

Q1Short answerExercise 11.2

Is the area of the circle inscribed in a square of side a cm, πa2 cm2? Give reasons for your answer.

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No, radius of the circle is

Q2Short answerExercise 11.2

Will it be true to say that the perimeter of a square circumscribing a circle of radius a cm is 8a cm? Give reasons for your answer.

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Yes, side of the square is 2a cm

Q3Short answerExercise 11.2

In Fig 11.3, a square is inscribed in a circle of diameter d and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer. AREA RELATED TO CIRCLES 123

This question refers to a figure in the original PDF.

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No, side of the outer square = diagonal of the inner square

Q4Short answerExercise 11.2

Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?

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No, it is only true for minor segment.

Q5Short answerExercise 11.2

Is it true that the distance travelled by a circular wheel of diameter d cm in one revolution is 2 π d cm? Why?

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No, it is πd.

Q6Short answerExercise 11.2

In covering a distance s metres, a circular wheel of radius r metres makes 2 r revolutions. Is this statement true? Why?

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Yes, distance covered in one revolution = 2π r

Q7Short answerExercise 11.2

The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?

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No, it will depend on the value of radius. tt ©

Q8Short answerExercise 11.2

If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2 r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?

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Yes, it will be true for the arcs of the same circle.

Q9Short answerExercise 11.2

The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?

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No, it will be true for the arcs of the same circle.

Q10Short answerExercise 11.2

The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?

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No, it will be true for arcs of the same circle.

Q11Short answerExercise 11.2

Is the area of the largest circle that can be drawn inside a rectangle of length a cm and breadth b cm (a > b) is π b2 cm2? Why?

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Yes, radius of the circle breadth of the rectangle.

Q12Short answerExercise 11.2

Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?

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Yes, their radii are equal

Q13Short answerExercise 11.2

Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?

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Yes, their radii are equal

Q14Short answerExercise 11.2

Is it true to say that area of a square inscribed in a circle of diameter p cm is p2 cm2? Why?

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No, diagonal of the square is p cm.

Q54Short answerExercise 11.2

× 360 or θ= = 15 36 × 36 θ Now, length of the arc = × 2 πr = × 2 π × 36 cm = 3 π cm

Q1Short answerExercise 11.3

Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm.

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33 cm

Q2Short answerExercise 11.3

In Fig. 11.5, a square of diagonal 8 cm is inscribed in a circle. Find the area of the shaded region.

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(16π – 32 ) cm2

Q3Short answerExercise 11.3

Find the area of a sector of a circle of radius 28 cm and central angle 45°.

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308 cm2

Q4Short answerExercise 11.3

The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?

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500.

Q5Short answerExercise 11.3

A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m. Find the area of the field in which the cow can graze.

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154 m2

Q6Short answerExercise 11.3

Find the area of the flower bed (with semi-circular ends) shown in Fig. 11.6.

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(380 + 25π )cm 2

Q7Short answerExercise 11.3

In Fig. 11.7, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use π = 3.14).

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54.5 cm2

Q8Short answerExercise 11.3

Find the area of the shaded field shown in Fig. 11.8. AREA RELATED TO CIRCLES 127

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(32 + 2π )m2

Q9Short answerExercise 11.3

Find the area of the shaded region in Fig. 11.9.

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(248 – 4π )m2

Q10Short answerExercise 11.3

Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.

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– 49 3 cm2

Q11Short answerExercise 11.3

Find the area of the shaded region in Fig. 11.10, 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 π = 3.14).

This question refers to a figure in the original PDF.

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30.96 cm2

Q12Short answerExercise 11.3

In Fig. 11.11, 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 π = 3.14).

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39.25 cm2

Q13Short answerExercise 11.3

In Fig. 11.12, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.

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308 cm2

Q14Short answerExercise 11.3

A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

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15246 m2

Q15Short answerExercise 11.3

In Fig. 11.13, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

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1386 cm2

Q16Short answerExercise 11.3

A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

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cm π

Q1Long answerExercise 11.3

1 So, AM = A B and ∠AOM = × 90° = 45°.

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33 cm

Q2Long answerExercise 11.3

2 AM 1 Now, = sin 45° = OA 2 So, AM = 10 × cm. Therefore, A B = 10 2 cm and OM = OA cos 45° = 10× cm = 5 2 cm So, area of ∆ OAB = base × height

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(16π – 32 ) cm2

Q1Long answerExercise 11.3

2 2 = 10 2 × 5 2 cm = 50 cm Therefore, the area of the required major segment

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33 cm

Q2Long answerExercise 11.3

2 2 = 235.5 cm + 50 cm = 285.5 cm Another method for the area of ∆ OAB As, ∠AOB = 90°, Therefore, area of ∆ OAB= OA × OB

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(16π – 32 ) cm2

Q1Long answerExercise 11.3

2 2 = 10 × 10 cm = 50 cm

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33 cm

Q2Long answerExercise 11.3

2 By Heron’s Formula, ar (ABC) = s(s–a ) (s–b ) (s–c) = 56 × 8 × 6 × 42 cm = 336 cm So, area of the shaded region = area of the ∆ ABC – area of the three sectors 2 2 = (336 – 39.25) cm = 296.75 cm Alternate Method for ar (ABC) 2 2 2 2 2 2 Here, AB + BC = (14) + (48) = 2500 = (50) = (CA) So, ∠B = 90° (By converse of Pythagoras Theorem)

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(16π – 32 ) cm2

Q1Long answerExercise 11.3

1 2 2 Therefore, ar (ABC) = AB × BC = × 14 × 48 cm = 336 cm

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33 cm

Q2Long answerExercise 11.3

2

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(16π – 32 ) cm2

Q1Short answerExercise 11.4

The area of a circular playground is 22176 m . Find the cost of fencing this ground at the rate of Rs 50 per metre.

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Rs 26400

Q2Short answerExercise 11.4

The diameters of front and rear wheels of a tractor are 80 cm and 2 m respec- tively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.

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560

Q3Short answerExercise 11.4

Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals. AREA RELATED TO CIRCLES 133

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24 21 – 77 m 2 75.36 – 36 3 cm2

Q4Short answerExercise 11.4

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60° (Use π = 3.14).

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pu T

Q5Short answerExercise 11.4

A circular pond is 17.5 m is of diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m2

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Rs 3061.50

Q6Short answerExercise 11.4

In Fig. 11.17, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

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196 cm2

Q7Short answerExercise 11.4

Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

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1.967 cm2 (approx)

Q8Short answerExercise 11.4

Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.

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8.7 cm2

Q9Short answerExercise 11.4

Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

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42 cm2 168 cm2 11. 4.3 m2 800 cm2 1:3:5

Q10Short answerExercise 11.4

On a square cardboard sheet of area 784 cm , four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.

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12.

Q11Short answerExercise 11.4

Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Fig. 11.18. Find the area of floor that remains uncovered with tiles. (Use π = 3.14)

This question refers to a figure in the original PDF.

Q12Short answerExercise 11.4

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm . (Use π = 3.14).

Q13Short answerExercise 11.4

An archery target has three regions formed by three concentric circles as shown in Fig. 11.19. If the diameters of the concentric circles are in the ratio 1: 2:3, then find the ratio of the areas of three regions.

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5 1 154 2 44

Q14Short answerExercise 11.4

The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 a m and 6 : 40 a m.

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45 cm2

Q15Short answerExercise 11.4

Area of a sector of central angle 200° of a circle is 770 cm . Find the length of the corresponding arc of this sector. AREA RELATED TO CIRCLES 135

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73 cm , Areas: cm , 154 cm2 ; Arc lengths: cm ; 6 3 3 3 Arc lengths of two sectors of two different circles may be equal, but their area need not be equal. 25π 25 17. 180 – 8π cm2 18. 40 19. + cm 2 20. 462 cm 2 4 2 tt ©

Q16Short answerExercise 11.4

The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

Q17Short answerExercise 11.4

Find the area of the shaded region given in Fig. 11.20.

This question refers to a figure in the original PDF.

Q18Short answerExercise 11.4

Find the number of revolutions made by a circular wheel of area 1.54 m in rolling a distance of 176 m.

Q19Short answerExercise 11.4

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Q20Short answerExercise 11.4

Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.