NCERT Solutions for Class 10 Maths Chapter 11 Areas Related To Circles

Class 10 Maths Chapter 11, Areas Related to Circles, is a crucial chapter that connects geometry with real-world applications. This chapter introduces students to the concepts of circles, their parts (arcs, sectors, segments), and combinations with other plane figures. Let’s see the key concepts, and areas related to circle class 10 solutions and solved examples to master this chapter effectively.

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Class 10 Maths Chapter 11:

The chapter on “Areas Related to Circles” in the NCERT Class 10 Maths textbook covers the following key topics:

• Basic properties of circles, including radius, diameter, and circumference.

• Formulas to find the area of a circle.

• Concepts of sectors and segments, including how to calculate their areas.

• The ring's area is the region between two concentric circles.

• Applications of these concepts in solving real-life problems.

1. What Is a Circle?

Areas related to circles class 10 talked about a circle is a set of points that are all the same distance from a fixed point. This fixed point is called the center (usually labeled O), and the constant distance is called the radius (r). The circle is written as.

Points to remember:

Center (O): The middle point of the circle.

Radius (r): The distance from the center to any point on the circle.

Diameter (d): Twice the radius, so.

Circumference and Area of a Circle

Check Out: CBSE Class 10th Sample Papers

Class 10 Maths Chapter 11 Exercise 11.1 Areas Related To Circles

Below is the NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 

Areas Related To Circles - 

1. Find the area of a sector of a circle with a radius of 6 cm. 

if the angle of the sector is 60°. Solution: It is given that the angle of the sector is 60° 

We know that the area of sector = (θ/360°)×πr 2 

∴ Area of the sector with angle 60° = (60°/360°)×πr 2 cm2 = (36/6)π cm2

= 6×22/7 cm2 = 132/7 cm2 

2. Find the area of a quadrant of a circle whose circumference is 22 cm. 

Solution: Circumference of the circle, C = 22 cm (given) It should be noted that a quadrant of a circle is a sector which is making an angle of 90°. 

Let the radius of the circle = r As C = 2πr = 22, R = 22/2π cm = 7/2 cm 

∴ Area of the quadrant = (θ/360°) × πr 2 Here, θ = 90° 

So, A = (90°/360°) × π r 2 cm2 = (49/16) 

π cm2 = 77/8 cm2 = 9.6 cm2

3.  The length of the minute hand of a clock is 14 cm. 

Find the area swept by the minute hand in 5 minutes. Solution: Length of minute hand = radius of the clock (circle)

∴ Radius (r) of the circle = 14 cm (given) Angle swept by minute hand in 60 minutes = 360° 

So, the angle swept by the minute hand in 5 minutes = 360° × 5/60 = 30° 

We know, Area of a sector = (θ/360°) × πr 2 

Now, area of the sector making an angle of 30° = (30°/360°) × πr 2 cm2 = (1/12) × π14 2 

= (49/3)×(22/7) cm2

= 154/3 cm2

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

(i) minor segment

(ii) major sector. (Use π = 3.14)

Solution:

Here AB be the chord which is subtending an angle 90° at the center O. It is given that the radius (r) of the circle = 10 cm

(i) Area of minor sector = (90/360°)×πr 2

= (¼)×(22/7)×10 2 Or, Area of minor sector = 78.5 cm 2 Also, area of ΔAOB = ½×OB×OA Here, OB and OA are the radii of the circle i.e. = 10 cm So, area of ΔAOB = ½×10×10 = 50 cm 2 Now, area of minor segment = area of minor sector – area of ΔAOB = 78.5 – 50 = 28.5 cm 2

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

= (3.14×10 2 )-78.5 = 235.5 cm 2

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5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

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

Solution:

Given, Radius = 21 cm θ = 60°

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

∴ 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 subtend by the arc = 60°

So, area of the sector making an angle of 60° = (60°/360°)×π r 2 cm 2 = 441/6×22/7 cm 2 Or, the area of the sector formed by the arc APB is 231 cm 2

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

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔOAB is an equilateral triangle. So, its area will be √3/4×a 2 sq. Units. Area of segment APB = 231-(√3/4)×(OA) 2 = 231-(√3/4)×21 2 Or, Area of segment APB = [231-(441×√3)/4] cm 2

Class 10 Maths Exercise 11.2 NCERT Solutions 

1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer:

Given: A circle whose center is O and radius is 6 cm and a point P is 10 cm away from its center.

To construct: To construct the pair of tangents to the circle and measure their lengths.

Steps of Construction:

(a) Join PO and bisect it. Let M be the mid-point of PO.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.

(c) Join PQ and PR.

Then PQ and PR are the required two tangents.

By measurement, PQ = PR = 8 cm

Justification:

Join OQ and OR.

∵ ∠OQP and ∠ORP are the angles in semi circles.

∠OQP = 90° = ∠ORP

Also, since OQ, OR are radii of the circle, PQ and PR will be the tangents to the circle at Q and R respectively.

2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Answer: To construct: To construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its lengths. Also to verify the measurements by actual calculation.

Steps of Construction:

1. With ‘O’ as centre two circles are constructed with radii 4 cm and 6 cm.

2. Draw OM radius for small circle and drawn perpendicular at M which meets big circle at P and Q.

3. Now PQ is the tangnet drawn for small circle.

Tangent PMQ = 9 cm

Verification : In ⊥∆OMP, ∠M = 90°.

Similarly, MQ = 4.5 cm.

∴ PQ = PM + MQ = 4.5 + 4.5

∴ PQ = 9 cm.

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3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its center. Draw tangents to the circle from these two points P and Q.

Answer: To construct: A circle of radius 3 cm and take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre and then draw tangents to the circle from these two points P and Q.

Steps of Construction:

1. Draw a line segment PQ of 14 cm.

2. Now, mark the midpoint O of PQ.

3. Draw the perpendicular bisectors of PO and OQ which intersects at points R and S on PQ.

4. With centre R and radius RP draw a circle.

5. With centre S and radius, SQ draw a circle.

6. And now, with centre O and radius 3 cm draw another circle which intersects the previous circles at the points A, B, C, and D.

7. Finally, join PA, PB, QC and QD. Thus, PA, PB, QC, and QD are the required tangents.

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 600.

Answer:

To construct: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 600.

Steps of Construction:

Step I: Take a point O on the plane of the paper and draw a circle of radius OA = 5 cm.

Step II: Produce OA to B such that OA = AB = 5 cm.

Step III: Taking A as the centre draw a circle of radius AO = AB = 5 cm.

Suppose it cuts the circle drawn in step I at P and Q.

Step IV: Join BP and BQ to get the desired tangents.

Justification: In OAP, we have

OA = OP = 5 cm (= Radius) Also,

AP = 5 cm (= Radius of circle with centre A)

∴ ∆OAP is equilateral.

⇒ ∠PAO = 60º ⇒ ∠BAP = 120º

In ∆BAP, we have

BA = AP and ∠BAP = 120º

∴ ∠ABP = ∠APB = 30º ⇒ ∠PBQ = 60º

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer:

To construct: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle.

Steps of Construction:

The tangents can be constructed on the given circles as follows

Step 1

Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.

Step 2

Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents.

Justification:

The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.

∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠ASB = 90°

⇒ BS ⊥ AS

Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

Check Out: CBSE Class 10th Previous Year Papers

Tips for Students

1. Visualize Diagrams: Sketch figures to understand sectors, segments, and combinations.

2. Memorize Formulas: Focus on the circumference, area, and sector-related formulas.

3. Practice Combinations: Break complex shapes into circles, triangles, and rectangles.

Summary of Important Formulas

Chapter 11 maths class 10 helps build a strong foundation in understanding circles and their properties. By learning to calculate the circumference, area, and parts of a circle, you will be better prepared to tackle both school problems and real-life applications. Practice these problems regularly to gain confidence in solving them.

Class 10 Maths Chapter 11 FAQs

1. How do I calculate the circumference of a circle?

The circumference is the distance around the circle. You can calculate it using the formula: Circumference=2πr, where r is the radius of the circle.

2. Can these circle formulas be used in real-life problems?

Yes, these formulas help solve everyday problems such as calculating the distance a wheel travels in one rotation, finding the area of circular fields, and determining the cost of fencing or plowing a circular field.

3. How is the area of a circle determined?
The area of a circle is given by:

Area=πr^2

4. What is a sector of a circle?
A sector is a “slice” of a circle, enclosed by two radii and the arc between them.