Important Questions Class 10 Maths Chapter 12

Important Questions Class 10 Mathematics Chapter 12 – Areas Related to Circles

Mathematics is a subject which requires both theoretical as well as practical knowledge of the concepts . The formulas need to be learned in the theoretical format thoroughly, whereas the numerals should be solved using a practical approach accurately. When they combine both these skills, they easily score well in the examinations.

You come across a lot of things in your daily life that are circular in shape, like a wheel, a cake, a washer, etc. Ever wondered how you can calculate the Area of these objects. Class 10 Mathematics Chapter 12 will help you learn the calculation of areas of circular objects. The essential topics covered in this chapter include perimeter and area of the circle, areas of sector and segment of the circle, areas of combinations of plane figures, and all the details about the circle.

You can find the overview of the complete chapter along with chapter-related NCERT textbook questions as well as additional questions in the Chapter 12 Class 10 Mathematics important questions. It also covers questions from popular books by best-selling authors. Thus, students can easily comprehend the level of questions that would be asked in school as well as in competitive examinations.

Extramarks always focuses on the practice-oriented approach while studying. As a result, it trains students to include practice sessions in their preparation by providing them with a good set of questions. You can access it anytime from the Extramarks’ official website.

Important Questions Class 10 Mathematics Chapter 12 – With Solutions

The following important questions and their solutions are included in Class 10 Mathematics Chapter 12 important questions:

Question 1: If the addition of the areas of two circles having radii R1 and R2 is equal as the Area of a circle of radius R, then

(A) R = R1 + R2

(B) R12 + R22 = R2 

(C) R1 + R2 <R 

(D)R12 + R22<R22

Answer 1: (B) R12 + R22 = R2

Explanation: Radius of first circle= R1

Area of the first circle =πR1

The radius of the second circle =R2

Area of the second circle =πR2

The Radius of the third circle = R

Area of the third circle=πR2

According to question

πR12 + πR22=πR2

π(R12 + R22)=πR2

R12 + R22=R2

Hence, option B is correct.

 

Question 2: When the sum of the circumferences of two circles having radii R1 and R2 is equal as the Circumference of the circle of radius R, then

(A) R1 + R2=R

(B) R1 + R2>R

(C) R1 + R2<R

(D) No relation among R1, R2, and R.

Answer 2: (A) R1 +R2=R

Explanation: Radius of first circle= R1

circumference of the first circle =2πR1

The radius of the second circle =R2

circumference of the second circle =2πR2

The Radius of the third circle = R

circumference of the third circle=2πR

According to question

2πR1 + 2πR2=2πR

2π(R1 + R2)=2πR

R1 + R2=R

Hence option A is correct.

 

Question 3: When the Circumference of the circle and the perimeter of a square are equal, we get

(A) Area of a circle = the Area of the square

(B) area of a circle > the Area of the square

(C) Area of a circle < the Area of the square

(D) No relation between the areas of the circle and square.

Answer 3: (B) Area of the circle > Area of the square

Explanation: Let r be the radius of the circle and a be the side of the square.

Given that,

Circumference of a circle = Perimeter of a square

⇒ 2πr = 4a

⇒ (22/7)r=2a

⇒ 11r = 7a

⇒ r=(7/11)a                            …..(1)

then, the Area of the circle, A1 = πr2, and the Area of the square, A2 = a2

From (1),

A1 = π × ((7/11) x a)2

      = 22/7 × (49/121) a2

      = 14/11 x a2

∴ A1 = (14/11) A2                   

(∵ A2 = a2)

⇒ A1 > A2

∴ Area of a circle > Area of the square.

Hence, the correct answer is option B.

 

Question 4: The Area of the largest triangle that can be inside in a semicircle of radius r units is

(A) r2 sq. units

(B) 1/2r2 sq. units

(C) 2r2 sq. units

(D) √2r2 sq. units

Answer 4: (A) r2 sq. units

Explanation: The largest triangle, that can be covered in a semicircle of radius r units is the triangle having its base as the diameter of the semicircle, and the two other sides have been taken by considering a point C on the circumference of the semicircle and joining it by the endpoints of the diameter A and B.

therefore, ∠C = 90° (by the properties of the circle)

So, △ABC is a right-angled triangle with the base as diameter AB of the circle and height be CD.

Now, the height of the triangle = r

∴ Area of largest △ABC = (1/2)× Base × Height

                                         = (1/2) × AB × CD

                                         = (1/2) × 2r × r

                                        = r2 sq. units

therefore, the correct answer is an option (A).

 

Question 5: When 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

Answer 5: (B) 14: 11

Explanation: Let r be the radius of the circle, and a be the side of the square.

Given that,

The perimeter of a circle = Perimeter of a square

⇒ 2πr = 4a

⇒ a = πr/2

then, Area of the circle = r2 and Area of the square = a2

Therefore,

Ratio of their areas = Area of Circle/Area of Square

                               = πr2/a2

                               = πr2/(πr/2)2

                               = πr2/(π2r2/4)

                               = 14/11

Hence, the correct answer is option B

 

Question 6: It is proposed to build the single circular park equal in the Area to the sum of areas for the two circular parks of the diameters of 16 m and 12 m in a locality. The radius of the new park will be

(A) 10 m 

(B) 15 m 

(C) 20 m 

(D) 24 m

Answer 6: (A) 10 m

Explanation: Let assume, D1 be the diameter of the first circular park = 16 m

therefore, Radius of first circular park = 8 m

Let assume D2 be the diameter of the second circular park = 12 m

therefore, Radius of second circular park = 6 m

Area of first circular park = πr2 = π(8)2 = 64π m2

Area of second circular park = πr2 = π(6)2 = 36π m2

here Given that,

Area of single circular park = Area of first circular park + Area of second circular park

therefore,πR2 = 64π + 36π

                           = 100π          (here, the radius of the single circular park is R)

⇒ πR2 = 100π

⇒ R2 = 100

⇒ R = 10

therefore,Radius of the single circular park will be 10 m.

Hence, the correct answer is option A.

 

Question 7: The Area of the circle which can be inscribed in a square of side 6 cm is

(A) 36π cm2

(B) 18π cm2

(C) 12π cm2

(D) 9π cm2

Answer 7:(D) 9 cm2

Explanation :Side of the square = 6 cm

therefore, Diameter of a circle = Side of square = 6 cm

⇒ Radius of the circle = 3 cm

therefore,  Area of the circle = πr2 

                                 = π(3)2

                                 = 9π cm2

Hence, the correct answer is option D.

 

Question 8: The Area of the square which can be inscribed in a circle of Radius 8 cm is

(A) 256 cm2

(B) 128 cm2

(C) 64√2 cm2

(D) 64 cm2

Answer 8: (B) 128 cm2

Explanation : Area of square =a2

Diagonal of square = Diameter of circle

Diagonal of square =8 2 =16cm

Let the side of the square = a cm

Using Pythagoras theorem in ABC

(16)2=a2+a2

2a2=256

a2=128

Area of square ABCD= a2

=128 cm2

 

Question 9: The Radius of the circle whose Circumference is the same as the sum of the circumferences of the two circles of the diameters of 36 cm and 20 cm is

(A) 56 cm 

(B) 42 cm 

(C) 28 cm 

(D) 16 cm

Answer 9: (C) 28 cm

Explanation : Diameter of first circle, d1 = 36 cm

Diameter of second circle, d2 = 20 cm

therefore, Circumference of first circle = πd1 = 36π cm

Circumference of second circle = πd2 = 20π cm

then, let D be the diameter of the new circle formed.

here Given that,

Circumference of the circle = Circumference of first circle + Circumference of second circle

πD = πd1 + πd2

⇒ πD = 36π + 20π

⇒ πD = 56π

⇒ D = 56

⇒ Radius = 56/2 = 28 cm

Hence, the correct answer is option (C).

 

Question 10: The diameter of a circle which has an 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

Answer 10: (D) 50 cm

Explanation : Area of first circle = πr12 = π(24)2

                                        = 576π cm2

Area of second circle = πr22 = π(7)2

                                               = 49π cm2

Given that,

Area of the circle = Area of first circle + Area of second circle

∴ πR2 = 576π + 49π                (where R is the radius of the new circle)

⇒ πR2 = 625π

⇒ R2 = 625

⇒ R = 25                 (∵ radius cannot be negative)

∴ Radius of the circle = 25 cm

Thus, diameter of the circle = 2R = 50 cm.

Hence, the correct answer is an option (D).

 

Question 11: Tick the correct option in the following and justify your choice: When the perimeter and the Area of the circle are numerically equal, then the radius of the circle is

(A) Two units

(B) π units

(C) Four units

(D) Seven units

Answer 11: (A) 2 units

Explanation: As we know the perimeter of the circle = the Area of the circle,

Therefore, 2πr = πr2

Or, r = 2

Hence, option (A) is correct that is the radius of the circle is two units

 

Question 12: Tick the correct solution in the following:

The Area of the sector of the angle p (in degrees) of a circle with radius R is

(A) p/180 × 2πR

(B) p/180 × π R2

(C) p/360 × 2πR

(D) p/720 × 2πR2

Answer 12: (D) p/720 × 2πR2.

Explanation: The area of a sector = (θ/360°)×πr2

Given that θ = p

So, Area of sector = p/360×πR2

Multiplying and dividing by 2 simultaneously,

= (p/360)×2/2×πR2

= (2p/720)×2πR2

So, option (D) is correct.

 

Question 13: Find the Area for the sector of a circle of Radius 5 cm if the corresponding arc length is 3.5 cm.

Answer 13 :

Radius of a circle = r = 5 cm

The arc length of the sector = l = 3.5 cm

Let the central angle (in radians) be θ.

As Arc length = Radius × Central angle (in radians)

∴ Central angle (θ) = Arc length / Radius = l / r = 3.5/5 = 0.7 radians

Then, Area of the sector = (½) × r2θ = (½) × 25 × 0.7 = 8.75 cm2

Thus, the required area of the sector of a circle is 8.75 cm2.

 

Question 14: Is the Area of the circle inscribed in a square of side a cm, a2 cm2? Give reasons for your answer.

Answer 14: False

Explanation:

Assume a be the side of the square.

We are given that the circle is inscribed in the square.

Diameter of circle = Side of square = a

Radius of the circle = a/2

Area of the circle = πr2 = π(a/2)2 = (πa2)/4 cm2.

Therefore, the Area of the circle is (πa2)/4 cm2.

Hence, the Area of the circle inscribed in a square of side a cm is not a2 cm2

 

Question 15: 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.

Answer 15: True

Explanation:

We can assume r be the radius of the circle = a cm.

Thus, the diameter of the circle = d = 2 × Radius = 2a cm.

As the circle is inside the square, hence,

the Side of a square = Diameter of the circle = 2a cm. Therefore, the Perimeter of a square = 4 × (side) = 4 × 2a = 8a cm.

So, it will be true to say that the perimeter of a square circumscribing a circle of radius a cm is 8a cm.

 

Question 16: Is it true to say that the Area of a segment of a circle is less than the Area of its corresponding sector? Why?

Answer 16: False

Explanation:

This is not true in the case of a major segment, and the Area is always greater than the Area of the corresponding sector. It is true only for the case of the minor segment. 

Hence, we can conclude that it is false to say that Area of a segment of a circle is less than the Area of its corresponding sector.

 

Question 17: Is it true that the Distance travelled by a circular wheel of diameter d cm in one revolution is 2 d cm? Why?

Answer 17: The statement is False

Explanation:

Distance travelled by the circular wheel of radius r in one revolution equals the Circumference of a circle.

We get that,

Circumference of the circle = 2πd; here, d is the diameter of the circle. 

Hence, it is not true that the Distance travelled by a circular wheel of diameter d cm in one revolution is 2 d cm.

 

Question 18: In covering a distance of s metres, a circular wheel of radius r metres makes s/2πr revolutions. Is this statement true? Why?

Answer 18: True

Explanation:

The Distance travelled by the circular wheel of radius r m in one revolution is the same as the Circumference of a circle = 2πr

No. of revolutions taken in 2πr m distance = 1

No. of revolutions taken in 1 m distance = (1/2πr)

No. of revolutions taken in s m distance = (1/2πr) × s = s/2πr

Hence, the statement “in covering a distance s metres, a circular wheel of radius r metres makes s/2πr revolutions” is true.

 

Question 19: 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.

Answer 19:

Radius of the first circle = r1 = 15 cm

Radius of the second circle = r2 = 18 cm

∴ Circumference of the first circle = 2πr1 = 30π cm

Circumference of the second circle = 2πr2 = 36π cm

assume, the radius of the circle = R

As per the question,

Circumference of circle = Circumference of the first circle + Circumference of the second circle

2πR= 2πr1+ 2πr2

⇒ 2πR = 30π + 36π

⇒ 66π ⇒ R = 33

⇒ Radius = 33 cm

Hence, the required radius of a circle is 33 cm.

 

Question 20: As given in Fig, a square of diagonal 8 cm is inscribed in a circle. Find the Area of the shaded region.

Answer 20:

Assume a be the side of the square.

∴ the Diameter of the circle = the Diagonal of the square = 8 cm

In the right-angled triangle ABC,

Using the Pythagoras theorem,

(AC)2 = (AB)2 + (BC)2

Therefore, (8)2= a2 +a2

⇒ 64= 2a2

⇒ a2= 32

Therefore,

The area of a square = a2= 32 cm2

The radius of the circle = Diameter/2 = 4 cm

The area of the circle = πr2 = π(4)2 = 16 cm2

hence, the area of the shaded region = Area of circle – Area of square

The area of the shaded region = 16π – 32

= 16 × (22/7) – 32

= 128/7

= 18.286 cm2

 

Question 21: What is the Area of a circle whose Circumference is 44 cm?

Answer 21:

Circumference of the circle = 2πr

From the question,

2πr = 44

Or, r = 22/π

then, area of circle = πr2 = π × (22/π)2

thus, area of circle = (22×22)/π = 154 cm2

 

Question 22: Find the Area of a sector of a circle of radius 28 cm and a central angle of 45°.

Answer 22:

Area of a sector of the circle = (1/2)r2θ,

(Where r = radius and θ = angle in radians subtended by the arc at the center of the circle)

Here, Radius of circle = 28 cm

Angle subtended at the center = 45°

Angle subtended at the centre (in radians) = θ 45π/180 = π/4

∴ Area of a sector of the circle = ½ r2θ

= ½ × (28)2 × (π/4)

= 28 × 28 × (22/8×7)

= 308 cm2

Therefore, the required Area of a sector of a circle is 308 cm2.

 

Question 23: The wheel of a motorcycle is of the radius of 35 cm. How many revolutions per minute should the wheel make so as to keep a speed of 66 km/h?

Answer 23:

The Radius of the wheel = r = 35 cm

1 revolution of a wheel = Circumference of a wheel

= 2πr

= 2 × (22/7) × 35

= 220 cm

For, we get, 

Speed of a wheel = 66 km/hr

= (66×1000×100)/60 cm/min

= 110000 cm/min

∴ No. of the revolutions in 1 min = 110000/220 = 500

Therefore, the required number of revolutions per minute is 500.

 

Question 24: A cow is tied with a rope having a length of 14 m at the corner for a rectangular field of the dimensions 20m × 16m. Find out the Area of the field for which the cow can graze.

Answer 24:

Let ABCD be a rectangular field.

Length of field = 20 m

The breadth of the field = 16 m

According to the question,

A cow has tied at a point A.

Let the length of rope be AE = 14 m = l.

The angle passing through the center of the sector = 90°

Angle subtended at the centre (in radians) θ = 90π/180 = π/2

∴ Area of a sector of the circle = ½ r2θ

= ½ × (14)2 × (π/2)

= 154 m2

Therefore, the required Area of a sector of a circle is 154 m2.

 

Question 25: The radii of the two circles are 19 cm and 9 cm, respectively. Find out the radius of the circle, which has a circumference same to the sum of the circumferences of the two circles.

Answer 25:

The radius of the first circle = 19 cm (given)

∴ Circumference of the first circle = 2π×19 = 38π cm.

The radius of the second circle = 9 cm (given)

∴ Circumference of the second circle = 2π×9 = 18π cm. Thus,

The sum of the Circumference of the two circles = 38π+18π = 56π cm.

Then, let the Radius of the third circle = R

∴ The Circumference of the third circle = 2πR.

It is given that the sum for the Circumference of the two circles = Circumference of the third circle

Hence, 56π = 2πR

Or R = 28 cm.

 

Question 26: The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having an area the same as the addition of the areas of the two circles.

Answer 26:

Radius of 1st circle = 8 cm (given)

∴ Area of 1st circle = π(8)2 = 64π

Radius of 2nd circle = 6 cm (given)

∴ Area of 2nd circle = π(6)2 = 36π.

So,

The sum of the 1st and 2nd circles will be = 64π+36π = 100π.

Then, assume that the radius of the 3rd circle = R;

therefore, the Area of the circle 3rd circle = πR2.

We know that the Area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle

and, πR2 = 100πcm2

R2 = 100cm2

So, R = 10cm

 

Question 27: Find the Area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°.

Answer 27:

We know that the angle for the sector is 60°.

As We know that the area of sector = (θ/360°)×πr2

∴ Area of the sector with an angle of 60° = (60°/360°)×πr2 cm2

= (36/6)π cm2

= 6×22/7 cm2 = 132/7 cm2

 

Question 28: Calculate the Area of a sector of angle 60°. Given that the circle has a radius of 6 cm.

Answer 28:

We know that

The angle of the sector = 60°.

We know the formula,

The Area of sector = (θ/360°)×π r2

= (60°/360°) × π r2 cm2

and, the area of the sector = 6 × 22/7 cm2 = 132/7 cm2

 

Question 29: Find out the Area of a quadrant of a circle whose Circumference is 22 cm.

Answer 29:

Circumference of the circle, C = 22 cm (as given).

We get that a quadrant of a circle is a sector which is making an angle of 90°.

assume the radius of the circle = r

As C = 2πr = 22,

R = 22/2π cm = 7/2 cm

∴ Area of the quadrant = (θ/360°) × πr2

Here, θ = 90°

So, A = (90°/360°) × π r2 cm2

= (49/16) π cm2

= 77/8 cm2 = 9.6 cm2

 

Question 30: The length of the minute hand of a clock is 14 cm. Find out the Area swept by the minute hand in 5 minutes.

Answer 30:

The length of the minute hand = radius of the clock (circle).

Therefore, the Radius (r) of the circle = 14 cm (given).

Angle made by the minute hand in 60 minutes = 360°.

Thus, the angle covered by the minute hand in 5 minutes = 360° × 5/60 = 30°.

We get

the area of a sector = (θ/360°) × πr2.

Then, the Area of the sector, making an angle of 30° = (30°/360°) × πr2 cm2

= (1/12) × π142

= (49/3)×(22/7) cm2

= 154/3 cm2

 

Question 31: If the Radius of a circle is 4.2 cm, compute its Area and Circumference.

Answer 31:

Area of a circle = πr2

thus, area = π(4.2)2 = 55.44 cm2

Circumference of a circle = 2πr

thus, circumference = 2π(4.2) = 26.4 cm

 

Question 32: As given in Fig, a square is inside a circle of diameter d and another the square is circumscribing a circle. Is the Area for the outer square four times more than the Area for the inner square? Give reasons for your answer.

Answer 32: False

Explanation:

Diameter of the circle = d

Therefore,

the Diagonal of inner square EFGH = Side of the outer square ABCD = Diameter of circle = d

Let the side of the inner square EFGH be a

Then, in the right-angled triangle EFG,

(EG)2 = (EF)2 + (FG)2

By the Pythagoras theorem,

⇒ d2 = a2 +a2

⇒ d2 = 2a2

⇒ a2 = d2/2

∴ Area of the inner circle = a2 = d2/2

Also, the Area of the outer square = d2

∴ the Area of the outer circle is just two times the Area of the inner circle.

Hence, the Area of the outer square is not equal to four times the Area of the inner square.

 

Question 33: On a square cardboard sheet of Area 784 cm2, 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 out the Area of the square sheet not covered by the circular plates.

Answer 33:

As per the question,

The area of the square = 784 cm2,

and The side of the square = √Area = √784 = 28 cm

The four circular plates are congruent. Then the diameter of each circular plate = 28/2 = 14 cm.

Therefore, the radius of each circular plate is 7 cm.

We get that

The Area of the sheet which is not covered by plates = Area of the square – Area of the four circular plates.

As all the four circular plates are congruent,

We find the Area of all four plates equal.

∴ Area of one circular plate = πr2 = (22/7) × 72 = 154 cm2

Then,

Area of four plates = 4×154 = 616 cm2

Area of the sheet not covered by plates = 784 – 616 = 168 cm2

Question 34: A chord of a circle of Radius 10 cm subtends a right angle at the center. Find the Area of the corresponding:

(i) minor segment

(ii) major sector. (Use π = 3.14)

 Answer 34:

AB is the chord that subtends an angle of 90° at the center O.

It is seen that the radius (r) of the circle = 10 cm

(i) Area of the minor sector = (90/360°)×πr2

= (¼)×(22/7)×102

Or, Area of the minor sector = 78.5 cm2

Also, the Area of the ΔAOB = ½×OB×OA

Where, OB and OA are the radii of the circle i.e., = 10 cm

Thus, the Area of ΔAOB = ½×10×10

= 50 cm2

Then, Area of minor segment = area of the minor sector – Area of ΔAOB

= 78.5 – 50

= 28.5 cm2

(ii) Area of  the major sector = Area of circle – Area of minor sector

= (3.14×102)-78.5

= 235.5 cm2

 

Question 35. In the circle of Radius 21 cm, an arc makes an angle of 60° at the center. Find out:

(i) the length of the arc

(ii) the Area of the sector formed by the arc

(iii) the Area of the segment formed by the corresponding chord

Answer 35:

We know that,

Radius = 21 cm

θ = 60°

(i) The length of an arc = θ/360°×Circumference(2πr)

The length of an arc AB = (60°/360°)×2×(22/7)×21

= (1/6)×2×(22/7)×21

Or Arc AB Length = 22cm

(ii) It is given that the angle subtends by the arc = 60°

Thus, the Area of a sector making an angle of 60° = (60°/360°)×π r2 cm2

= 441/6×22/7 cm2

therefore, the Area of the sector formed by the arc APB is 231 cm2

(iii) The Area of the segment APB = Area of sector OAPB – Area of ΔOAB

If the two arms of the triangle are the radii of the circle and thus are the same, and one angle is 60°, ΔOAB is an equilateral triangle. Hence, its Area will be √3/4×a2 sq. Units.

The area of the segment APB = 231-(√3/4)×(OA)2

= 231-(√3/4)×212,

and, Area of segment APB = [231-(441×√3)/4] cm2

 

Question 36. A chord of the circle of Radius 15 cm makes the angle of 60° at the center. Find out the areas for the corresponding minor and the major segments of the circle. (Use π = 3.14 and √3 = 1.73)

Answer 36:

We know that

Radius = 15 cm

θ = 60°.

Hence,

The Area of the sector OAPB = (60°/360°)×πr2 cm2

= 225/6 πcm2.

Then, ΔAOB is equilateral as two sides are the circle’s radii and, 

Therefore, equal, and one angle is 60°.

So, the Area of ΔAOB = (√3/4) ×a2

And (√3/4) ×152

∴ Area of ΔAOB = 97.31 cm2.

Then, the Area of minor segment APB = Area of OAPB – Area of ΔAOB

And, area of minor segment APB = ((225/6)π – 97.31) cm2 = 20.43 cm2

Or

The Area of major segment = Area of the circle – Area of segment APB

And area of major segment = (π×152) – 20.4 = 686.06 cm2

 

Question 37: In the figure, AB and CD are the two diameters of the circle (with center O) perpendicular to the each other, and OD is the diameter of the smaller circle. If OA = 7 cm, find out the Area of the shaded region (pink and yellow regions together).

Answer 37:

Radius for the larger circle, R = 7 cm

Radius for the smaller circle, r = 7/2 cm

Height for the ΔBCA = OC = 7 cm

Base for the ΔBCA = AB = 14 cm

Area for the ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm2

Area for the larger circle = πR2 = 22/7 × 72 = 154 cm2

Area for the larger semicircle = 154/2 cm2 = 77 cm2

Area for the smaller circle = πr2 = 22/7 × 7/2 × 7/2 = 77/2 cm2

Area for the shaded region = Area for the larger circle – Area for the triangle – Area for the larger semicircle + Area for the smaller circle

Area for the shaded region = (154 – 49 – 77 + 77/2) cm2

= 66.5 cm2

Question 38. A chord of the circle of Radius 12 cm subtends the angle of 120° at the centre. Find out the Area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Answer 38:

Radius, r = 12 cm.

Then, draw a perpendicular OD on chord AB, and it will bisect chord AB.

So, AD = DB

Now, the area for the minor sector = (θ/360°)×πr2

= (120/360)×(22/7)×122

= 150.72 cm2

Considering the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

Now, cos 30° = AD/OA

√3/2 = AD/12

Or, AD = 6√3 cm

We know that OD bisects AB. So,

AB = 2×AD = 12√3 cm

Then,, sin 30° = OD/OA

Or, ½ = OD/12

 OD = 6 cm

So, the area for the ΔAOB = ½ × base × height

Here, base = AB = 12√3 and

Height = OD = 6

Thus, area of ΔAOB = ½×12√3×6 = 36√3 cm = 62.28 cm2

The area of the corresponding Minor segment = the area for the Minor sector – Area for ΔAOB

So, we get,

= 150.72 cm2– 62.28 cm2 

= 88.44 cm2

 

Question 39. A horse is tied with the peg at one corner of the square-shaped grass field of side 15 m by means of 5 m long rope (see the given Fig.). Find out

(i) an area of a part of the field where the horse can graze.

(ii) the increase in the grazing Area for the rope was 10 m long instead of 5 m. (Use π = 3.14)

Answer 39:

As we know, the horse is tied at one end of the square field. It will graze only a quarter (i.e., sector with θ = 90°) of the field with a radius of 5 m, where the length of the rope will form the radius of the circle, r = 5 m.

We also know that the side of the square field = 15 m

(i) Area of the circle = πr2 = 22/7 × 52 = 78.5 m2

Then, an area of a part of the field in which the horse could graze = ¼ (the Area of the circle) = 78.5/4 = 19.625 m2

(ii) When the rope is increased to 10 m,

The area of the circle will be = πr2 =22/7×102 = 314 m2

Then, the Area of the part of the field in which the horse could graze = ¼ (the Area of the circle)

= 314/4 = 78.5 m2

Therefore, Increment in the grazing area = 78.5 m2 – 19.625 m2 = 58.875 m2

 

Question 40:  In the figure, O is the center for the circle such that the diameter AB = 13 cm as well as AC = 12 cm. BC is joined. Find out the Area for the shaded region. (take π = 3.14)

Answer 40:

We are aware that the angle in the semicircle is the right angle.

So, ∠ACB = 9° 

By the Pythagoras theorem,

BC2 + AC2 = AB2

BC2 = AB2 – AC2

= (13)2 – (12)2

= 169 – 144

= 25

⇒ BC = 5 cm

From the given condition,

Diameter for the circle = AB = 13 cm

Radius for the semicircle = AB/2 = 13/2 cm

Area for the shaded region = Area for the semicircle – Area for the right triangle ABC,

= (1/2)πr2 – (1/2) × BC × AC

= (1/2) × 3.14 × (13/2) × (13/2) – (1/2) × 5 × 12

= 66.33 – 30

= 36.33 cm2

Question 41. A brooch is made up of silver wire in the form of the circle with a diameter of 35 mm. The wire is also used for making the 5 diameters which divide the circle into 10 equal sectors as shown in given Fig. Find out:

(i) the total length for the silver wire required.

(ii) the Area of each sector for the brooch.

Answer 41:

Diameter (D) = 35 mm

Total number of the diameters to be considered= 5

Then, the total length of 5 diameters that would be needed = 35×5 = 175

Circumference of the circle = 2πr

And, C = πD = 22/7×35 = 110

Area of the circle = πr2

And, A = (22/7)×(35/2)2 = 1925/2 mm2

(i) Total length of the silver wire needed = The Circumference of the circle + The length of 5 diameter

= 110+175 = 285 mm

(ii) Total Number of the sectors in the brooch = 10

Thus, the Area of each sector = the total Area of the circle/number of sectors

Therefore, Area of the each sector = (1925/2)×1/10 = 385/4 mm2

 

Question 42. Find out the perimeter of the equilateral triangle if it inscribes a circle whose Area is 154 cm2.

Answer 42:

where, as the equilateral triangle is inscribed in a circle, the circle is an incircle.

then, the radius of the incircle is given by

r = Area of triangle/semi-perimeter

As per question, it is given that Area of the incircle = 154 cm2

thus, π × r2 = 154

and, r = 7 cm

then, assume the length of each arm of the equilateral triangle to be “x” cm

thus, the semi-perimeter of the equilateral triangle = (3x/2) cm

or, the Area of the equilateral triangle = (√3/4) × x2

We know that, r = Area of triangle / semi-perimeter

thus, r = [x2(√3/4) / (3x/2)]

=> 7 = √3x/6

Or, x = 42/√3

Multiply both numerator and denominator by √3

thus, x = 42√3 / 3 = 14√3 cm

then, the perimeter of an equilateral triangle will be = 3x = 3 × 14√3 = 72.7 cm.

 

Question 43. An umbrella has 8 ribs which are symmetrically spaced (as per given Fig.). Let the umbrella be a flat circle of radius 45 cm, and find out the Area between the two consecutive ribs for the umbrella.

Answer 43:

The radius (r) of the umbrella if flat = 45 cm

thus, the area of the circle (A) = πr2 = (22/7)×(45)2 =6364.29 cm2

Total number of the ribs (n) = 8

thus, The Area between two consecutive ribs of the umbrella = A/n

6364.29/8 cm2

and, The Area between the two consecutive ribs of the umbrella = 795.5 cm2

 

Question 44. A car having two wipers which do not overlap. Each wiper has a blade of the length of 25 cm sweeping through an angle of 115°. Find out the total Area cleaned at each sweep of the blades.

Answer 44:

We know that,

Radius (r) = 25 cm

Sector angle (θ) = 115°

Hence, there are 2 blades,

Thus, The total area of the sector made by the wiper = 2×(θ/360°)×π r2

= 2×(115/360)×(22/7)×252

= 2×158125/252 cm2

= 158125/126 = 1254.96 cm2

 

Question 45. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of an angle of 80° to the Distance of 16.5 km. Find out the Area of the sea over which the ships are warned.

(Use π = 3.14)

Answer  45:

Assume O bet the position of the Lighthouse.

Where the radius will be the Distance over which light spreads.

Given, radius (r) = 16.5 km

Sector angle (θ) = 80°

Then, the total Area of the sea on which the ships are warned = Area made by the sector

And, Area of sector = (θ/360°)×πr2

= (80°/360°)×πr2 km2

= 189.97 km2

 

Question 46: The cost of fencing the circular field at a rate of Rs. 24 per meter is Rs. 5280. The field is to be ploughed at a rate of Rs. 0.50 per m2. Find out the cost of ploughing the field (Take π = 22/7).

Answer 46:

The length of the fence (in metres) = Total cost/Rate = 5280/24 = 220

Thus, the circumference of the field = 220 m

When r metres is the radius of the field, then 2πr = 220

2 × (22/7) × r = 220

r = (220 × 7)/ (2 × 22)

r = 35

Therefore, the radius of the field = 35 m

Area of the field = πr2

= (22/7) × 35 × 35

= 22 × 5 × 35 m2

= 3850 sq. m.

The cost of ploughing 1 m2 of the field = Rs. 0.50

Hence, the total cost for ploughing the field = 3850 × Rs. 0.50 = Rs. 1925

 

Question 47. A round table cover consists of six same designs as shown in given Fig. If the radius for the cover is 28 cm, find out the cost of making designs at a rate of ₹ 0.35 per cm2. (Use √3 = 1.7)

Answer 47:

Total number of the same designs = 6

AOB= 360°/6 = 60°

Radius of the cover = 28 cm

Cost of making the design = ₹ 0.35 per cm2

If the two arms of the triangle are the radii of the circle and thus are same, and one angle is 60°, ΔAOB is an equilateral triangle. Therefore, its Area will be (√3/4)×a2 sq. units

Where, a = OA

∴ Area of the equilateral ΔAOB = (√3/4)×282 = 333.2 cm2

Area of the sector ACB = (60°/360°)×πr2 cm2

= 410.66 cm2

Thus, area of a single design = area of the sector ACB – area of ΔAOB

= 410.66 cm2 – 333.2 cm2 

Thus, = 77.46 cm2

therefore, Area of 6 designs = 6×77.46 cm2 = 464.76 cm2

Thus, total cost of making design = 464.76 cm2 ×Rs.0.35 per cm2

= Rs. 162.66

 

Question 48. A square is inscribed in a circle. Find out the ratio of the Area of the circle and the square.

Answer 48:

As the square is inscribed in the circle, a diagonal for the square will be = the diameter for the circle.

Let “r” be the radius for the circle, and “d” be the length for each diagonal for the square.

We know,

Length for the diagonal of a square = side (s) × √2

So,

d = 2r

And, s × √2 = 2r

Or, s = √2r.

We know the Area for the square = s2.

Thus, the area for the square = (√2r)2 = 2r2.

Now, the Area for the circle = π × r2

∴ Area of the circle: an area for the square = π × r2: 2r2 = π: 2.

So, the ratio of the Area between the circle and the square is π: 2.

 

Question 49. Find out the Area for the shaded region is given in Fig if PQ = 24 cm, PR = 7 cm and O is the circle’s centre.

Answer 49:

where P is in the semicircle and so,

P = 90°;

thus, it can be concluded that QR is the hypotenuse of the circle and is equal to the diameter of the circle.

∴ QR = D

Using the Pythagorean theorem,

QR2 = PR2+PQ2

Or, QR2 = 72+242

QR= 25 cm = Diameter

therefore, the radius of the circle = 25/2 cm

then, the Area of the semicircle is = (πR2)/2

Thus, = (22/7)×(25/2)×(25/2)/2 cm2

= 13750/56 cm2 = 245.54 cm2

hence, the Area of the ΔPQR = ½×PR×PQ

= (½)×7×24 cm2

= 84 cm2

therefore, the area of the shaded region = 245.54 cm2-84 cm2

= 161.54 cm2

 

Question 50: The wheels of a car are of diameter 80 cm each. Find out the complete revolutions of each wheel made in 10 minutes if the car is travelling at the speed of 66 km per hour?

Answer 50:

The radius of the car’s wheel = 80/2 = 40 cm (as D = 80 cm)

thus, the Circumference of wheels = 2πr = 80 π cm

then, in one revolution, the Distance will be covered = Circumference of the wheel = 80 π cm

We observe that the Distance covered by the car in 1 hr = 66km

Converting km into cm, we get,

car covered by the Distance in 1hr = (66 × 105) cm

In 10 minutes, the distance will be covered = (66 × 105 × 10)/60 = 1100000 cm/s

∴ Distance covered by car = 11 × 105 cm

then, the number of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) = 11 × 105 /80 π = 4375.

 

Question 51. Find out the Area for the shaded region in given Fig. when radii of the two concentric circles having centre O are 7 cm and 14 cm respectively, and the angle AOC = 40°.

Answer 51:

We know that,

Angle made by the sector = 40°,

and, the radius of the outer circle = R = 14 cm

And, the Radius of the inner circle = r = 7 cm

As We know that,

Area of the sector = (θ/360°)×πr2

thus, Area for OAC = (40°/360°)×πr2 cm2

= 68.44 cm2

Area for the sector OBD = (40°/360°)×πr2 cm2

= (1/9)×(22/7)×72 

Thus, = 17.11 cm2

hence, the Area of the shaded region ABDC 

    = The Area of the OAC – the Area of OBD

And

Thus, = 68.44 cm2 – 17.11 cm2 = 51.33 cm2

 

Question 52: Find out the Area for the sector of a circle with the radius of 4 cm and the angle of 30°. Also, find out the Area of the corresponding major sector (Use π = 3.14)

Answer 52:

Let OAPB be the sector.

Area of the major sector = [(360 – θ)/ 360] × πr2

=[(360 – 30)/360] × 3.14 × 4 × 4

= (330/360) × 3.14 × 16

= 46.05 cm2

Thus, = 46.1 cm2 (approx)

 

Question 53. Find out the Area for the shaded region in below Fig. when ABCD is a square for the side of 14 cm, and APD and BPC are semicircles.

Answer 53:

Side of the square ABCD (as given as) = 14 cm

thus, Area of ABCD = a2

= 14×14 cm2 = 196 cm2

As We know that the side of the square = diameter of the circle = 14 cm

thus, side of the square = diameter of the semicircle = 14 cm

hence, Radius of the semicircle = 7 cm

then, area for the semicircle = (πR2)/2

= (22/7×7×7)/2 cm2

= 77 cm2

Hence, 

Area for two semicircles = 2×77 cm2 = 154 cm2

here, area of the shaded region = Area of the Square – Area of two semicircles

= 196 cm2 -154 cm2

= 42 cm2

Question 54. A chord subtends an angle of 90°at the center of the circle whose radius having 20 cm. Find out the Area of the corresponding major segment of a circle.

Answer 54:

as we know:

the Area of the sector = θ/360 × π × r2

and, Base and height of the triangle formed will be = radius of the circle

and the Area of the minor segment = the Area of the sector – the Area of the triangle formed

the Area of the major segment = the Area of the circle – the Area of the minor segment

then,

Radius of circle = r = 20 cm and

subtended Angle = θ = 90°

the Area of the sector = θ/360 × π × r2 = 90/360 × 22/7 × 202

and, area of the sector = 314.2 cm2

the Area of the triangle = ½ × base × height = ½ × 20 × 20 = 200 cm2

the Area of the minor segment = 314.2 – 200 = 114.2 cm2

the Area of the circle = π × r2 = (22/7) × 202 = 1257.14

the Area of the major segment = 1257.14 – 114.2 = 1142 .94 cm2

thus, the area of the corresponding major segment of the circle = 1142 .94 cm2

 

Question 55. In given Fig. AB is the diameter of the circle, AC = 6 cm and BC = 8 cm. Find out the Area of the shaded region (Use π= 3.14).

Answer 55:

As per the question,

AC = 6cm and BC = 8 cm

A triangle in the semi-circle with hypotenuse as diameter is the right angled triangle.

Using the Pythagoras theorem in right angled triangle ACB,

(AB)2 = (AC)2 + (CB)2

(AB)2 = (6)2 + (8)2

⇒(AB)2 = 36 + 64

⇒(AB)2 = 100 ⇒(AB)= 10

The diameter of the circle = 10 cm

Thus, Radius of the circle = 5 cm

Area of circle = πr2

= π(5)2

= 25π cm2

= 25 × 3.14 cm2

= 78.5 cm2

We get that,

The area of the right angled triangle = ( ½ ) × Base × Height

= (½) × AC × CB

= (½) × 6 × 8

= 24 cm2

Then, Area of the shaded region = Area of the circle – Area of the triangle

= (78.5-24)cm2

= 54.5cm2

Question 56: the round table cover has six equal designs as presents in the figure. When the radius of the cover is 28 cm, find out the cost of making the designs at the rate of Rs. 0.35 per cm2. (Use 3 = 1.7)

Answer 56:

Total number of equal designs = 6

∠AOB = 360°/6 = 60°

The radius of the cover = 28 cm

Cost of making design = Rs. 0.35 per cm2.

hence the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its Area will be √3/4 × a2

Here, a = OA

therefore, Area of equilateral ΔAOB = √3/4 × 282 = 333.2 cm2

Area of sector ACB = (60°/360°) × π r2 cm2

= 410.66 cm2

Area of a single design = area of sector ACB – Area of ΔAOB

= 410.66 cm2 – 333.2 cm2 = 77.46 cm2

∴ Area of 6 designs = 6 × 77.46 cm2 = 464.76 cm2

So, the total cost of making the design = 464.76 cm2 × Rs. 0.35 per cm2

= Rs. 162.66

Question 57. In a circular table cover of the radius 32 cm, a design is formed, leaving the equilateral triangle ABC in the middle as given in Fig. Find out the Area of the design.

Answer 57:

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

⇒ BD = AB/2

hence, AD is the median of the triangle

∴ AO = Radius for the circle = (2/3) AD

⇒ (2/3)AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

By the Pythagoras theorem,

AB2 = AD2 +BD2

⇒ AB2 = 482+(AB/2)2

⇒ AB2 = 2304+AB2/4

⇒ 3/4 (AB2)= 2304

⇒ AB2 = 3072

⇒ AB= 32√3 cm

Area for ΔADB = √3/4 ×(32√3)2 cm2 = 768√3 cm2

Area for circle = πR2 = (22/7)×32×32 = 22528/7 cm2

Area for the design = Area for circle – Area for ΔADB

= (22528/7 – 768√3) cm2

 

Question 58: Find out the Area of the segment AYB shown in the figure if the radius for the circle has 21 cm and ∠ AOB = 120°. (Use π = 22/7).

Answer 58:

The Area for the segment AYB = Area for the sector OAYB – Area for the ∆ OAB …..(1)

The area of the sector OAYB = (120/360) × (22/7) × 21 × 21 = 462 cm2 ……(2)

As we know, Draw OM ⊥ AB.

And, OA = OB (radius)

hence, by RHS congruence, ∆ AMO ≅ ∆ BMO.

the mid-point of AB is M and ∠ AOM = ∠ BOM = (1/2) × 120° = 60°

assume, OM = x cm

In triangle OMA,

OM/OA = cos 60°

x/21 = ½

x = 21/2

OM = 21/2 cm

Same as,

AM/OA = sin 60°

AM/21 = √3/2

AM = 21√3/2 cm

Thus, AB = 2 AM = 2 (21√3/2) = 21√3 cm

The area of the triangle OAB = (½) × AB × OM

= (½) × 21√3 × (21/2)

= (441/4)√3 cm2 ………..(3)

From (1), (2) and (3),

The area for the segment AYB = [462 – (441/4)√3] cm2  = 271.04 cm2 

Question 59. An area of the equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as the centre, the circle is drawn with the radius same as half the length of the side for the triangle (see in given Fig.). Find out the Area for the shaded region. (Use π = 3.14 and √3 = 1.73205)

Answer 59:

ABC is the equilateral triangle.

 ∠ A = ∠ B = ∠ C = 60°

There are three sectors each making 60°.

the Area of ΔABC = 17320.5 cm2

Here,  √3/4 ×(side)2 = 17320.5

= (side)2 =17320.5×4/1.73205

= (side)2 = 4×104

= side = 200 cm

Thus, the radius for the circles = 200/2 cm = 100 cm

the Area of the sector = (60°/360°)×π r2 cm2

= 1/6×3.14×(100)2 cm2

= 15700/3cm2

the Area of 3 sectors = 3×15700/3 = 15700 cm2

so, area for the shaded region = Area for equilateral triangle ABC – Area for 3 sectors

= 17320.5-15700 cm2 = 1620.5 cm2

Question 60: Find out the Area of the shaded design in the given figure. Here ABCD is the square of 10 cm, and semicircles are drawn with the each side of the square as a diameter. (Use π = 3.14).

Answer 60:

Let us assume I, II, III, and IV for the unshaded regions.

We know that, the side for the square ABCD = 10 cm

sides of the square are also the diameters of semicircles.

The radius of the semicircle = 10/2 = 5 cm

Then, area for the region I +III = Area for square ABCD – Area for two semicircles of radius 5 cm

= (10)2 – 2 × (½) π ×(5)2

= 100 – 3.14 × 25

= 100 – 78.5

= 21.5 cm2

Same as,

Area for the region II + Iv = 21.5 cm2

Area for the shaded region = Area for square ABCD – Area for the region (I + II + III + IV)

= 100 – 2× 21.5

= 100 – 43

= 57 cm2

 

Question 61. Below Fig. depicts an archery target marked with its five scoring regions from the centre to the outwards as Gold, Red, Blue, Black, and White. The diameter for the region representing Gold score is 21 cm, as well as each of the other bands is 10.5 cm wide. Find out the Area of each of the five scoring regions.

Answer 61:

The radius for the 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm)

So, area for the gold region = π r12 = π(10.5)2 = 346.5 cm2

We know that each of the other bands is 10.5 cm wide,

Hence, the radius for the 2nd circle, r2 = 10.5cm+10.5cm = 21 cm

Thus,

So,  Area for the red region = Area for the 2nd circle − Area for the gold region = (πr22−346.5) cm2

= (π(21)2 − 346.5) cm2

= 1386 − 346.5

= 1039.5 cm2

In the same way,

The radius for the 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm

The radius for the 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm

The Radius for the 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm

For the area for the nth region,

A = Area for the circle n – Area for the circle (n-1)

So, Area for the blue region (n=3) = Area for the third circle – Area for the second circle

= π(31.5)2 – 1386 cm2

= 3118.5 – 1386 cm2

= 1732.5 cm2

Area for the black region (n=4) = Area for the fourth circle – Area for the third circle

= π(42)2 – 1386 cm2

= 5544 – 3118.5 cm2

= 2425.5 cm2

Thus, Area for the white region (n=5) = Area for the fifth circle – Area for the fourth circle

= π(52.5)2 – 5544 cm2

= 8662.5 – 5544 cm2

= 3118.5 cm2

Question 62. Find out the Area for the shaded region in given Fig. here, a circular arc for Radius of 6 cm has been drawn with vertex O for the equilateral triangle OAB for the side 12 cm as the centre.

Answer 62:

We know that OAB is an equilateral triangle having each angle as 60°

Area for the sector is common in both.

Radius for the circle = 6 cm.

Side for the triangle = 12 cm.

Area for the equilateral triangle = (√3/4) (OA)2= (√3/4)×122 = 36√3 cm2

Area for the circle = πR2 = (22/7)×62 = 792/7 cm2

Area for the sector making angle 60° = (60°/360°) ×πr2 cm2

= (1/6)×(22/7)× 62 cm2 = 132/7 cm2

Area for the shaded region = Area for the equilateral triangle + Area for the circle – Area for the sector

= 36√3 cm2 + 792/7 cm2 – 132/7 cm2

= (36√3+660/7) cm2

 

Question 63. On the square handkerchief, nine circular designs, each for a radius of 7 cm are made (see Fig. 12.29). Find out the area for the remaining portion of the handkerchief.

Answer 63:

Number for the circular designs = 9

Radius for the circular design = 7 cm

There are three circles in one side for the square handkerchief.

 Side for the square = 3×diameter for the circle 

= 3×14 

= 42 cm

Area for the square = 42×42 cm2 = 1764 cm2

Area for the circle = π r2 = (22/7)×7×7 = 154 cm2

Total area for the design = 9×154 = 1386 cm2

Area for the remaining portion of the handkerchief 

= Area for the square – Total area for the design 

= 1764 – 1386 = 378 cm2

Question 64. In Given Fig, OACB is a quadrant for the circle with the centre O as well as the radius 3.5 cm. If OD = 2 cm, find out the Area of the

(i) the quadrant OACB,

(ii) the shaded region.

Answer 64:

Radius for the quadrant = 3.5 cm = 7/2 cm

(i) Area for the quadrant OACB = (πR2)/4 cm2

= (22/7)×(7/2)×(7/2)/4 cm2

= 77/8 cm2

(ii) Area for the triangle BOD = (½)×(7/2)×2 cm2

= 7/2 cm2

Area for the shaded region = Area for the quadrant – Area for the triangle BOD

= (77/8)-(7/2) cm2 = 49/8 cm2

= 6.125 cm2

 

Question 65. In given Fig. the square, OABC is inside the quadrant OPBQ. If OA = 20 cm, find out the Area for the shaded region. (Use π = 3.14)

Answer 65:

Side for the square = OA = AB = 20 cm

Radius for the quadrant = OB

OAB is the right angled triangle

By the Pythagoras theorem in ΔOAB,

OB2 = AB2+OA2

OB2 = 202 +202

 OB2 = 400+400

 OB2 = 800

OB= 20√2 cm

Area for the quadrant = (πR2)/4 cm2 = (3.14/4)×(20√2)2 cm2 = 628cm2

Area for the square = 20×20 = 400 cm2

Area for the shaded region = Area for the quadrant – Area for the square

= 628-400 cm2 = 228cm2

Question 66. AB and CD are respectively arcs for the two concentric circles of radii 21 cm and 7 cm as well as the centre O (see given Fig.). If ∠AOB = 30°, find out the Area for the shaded region.

Answer 66:

Radius for the larger circle, R = 21 cm

Radius for the smaller circle, r = 7 cm

Angle made by sectors for the both concentric circles = 30°

Area for the the larger sector 

= (30°/360°)×πR2 cm2

= (1/12)×(22/7)×212 cm2

= 231/2cm2

Area for the smaller circle 

= (30°/360°)×πr2 cm2

= 1/12×22/7×72 cm2

=77/6 cm2

Area for the shaded region 

= (231/2) – (77/6) cm2

= 616/6 cm2 = 308/3cm2

 

Question 67. In given Fig, ABC is the quadrant for the circle of Radius 14 cm as well as a semicircle is drawn with BC as the diameter. Find out the Area for the shaded region.

Answer 67:

Radius for the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is the diameter for the semicircle.

ABC is the right-angled triangle.

By the Pythagoras theorem in ΔABC,

BC2 = AB2 +AC2

 BC2 = 142 +142

 BC = 14√2 cm

Radius for the semicircle = 14√2/2 cm = 7√2 cm

Area for the ΔABC =( ½)×14×14 = 98 cm2

Area for the quadrant = (¼)×(22/7)×(14×14) = 154 cm2

Area for the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm2

Area for the shaded region =Area for the semicircle + Area for the ΔABC – Area for the quadrant

= 154 +98-154 cm2 = 98cm2

Question 68. Calculate the Area for the designed region in given Fig. common between the two quadrants for the circles of Radius 8 cm each.

Answer 68:

AB = BC = CD = AD = 8 cm

Area for the ΔABC = Area for the ΔADC = (½)×8×8 = 32 cm2

Area for the quadrant AECB = Area for the quadrant AFCD = (¼)×22/7×82

= 352/7 cm2

Area for the shaded region = (Area for the quadrant AECB – Area for the ΔABC) = (Area for the quadrant AFCD – Area for the ΔADC)

= (352/7 -32)+(352/7- 32) cm2

= 2×(352/7-32) cm2

= 256/7 cm2

Benefits of Solving Important Questions Class 10 Mathematics Chapter 12

Today, it is very easy to get resources online for various competitive examinations but it is quite hard to get quality resources out of these available study materials. As a result, it is quite necessary that students judge the quality of the material and the resources before availing it. This can be done if they have the right guidance from qualified mentors. Hence, Extramarks believes in providing overall guidance as well as top-notch study material.

Our important questions of Class 10 Mathematics Chapter 12 has all the study material and questions for students to prepare well as well as test themselves during their preparation, ensuring a better score in the examinations. You can get it from the official website of Extramarks.

 Few benefits of availing NCERT curriculum-based questions for Class 10 Mathematics are given below:

• It is planned, structured, formulated, re-checked, tested according to the CBSE curriculum and then designed as per the needs of the students. Thus, before handing it over to the students, it is verified by our knowledgeable teachers and mentors.
• It is trusted, used and appreciated by lakhs of students, teachers and various educational authorities. Hence, ensuring students that they have access to the right quality study material.
• The questions are covered from the NCERT textbooks, NCERT Exemplar and other reference books and sources for students to have a collection of various questions ranging from easy level to complex level. This gives students a well-rounded exposure to different formats of questions which would help them become more confident during the final board exams.
• The solutions are given from the NCERT textbook after analysing the CBSE latest curriculum and then are written by the experts and scholars in the field of Mathematics aiding students to solve with right and accurate steps in their board examinations.
• The key points provided in the important questions Class 10 Mathematics Chapter 12 help students to quickly recall whatever they have studied in the chapter from the NCERT textbook. After proper revision, they can solve the questions and test their level of preparation. Thereby giving students a complete guidance  to excel in their examinations.