Master Class 10 Maths Chapter 12: Areas Related to Circles – Formulas, NCERT Solutions & CBSE Questions Explained

✅ Class 10 Maths Chapter 12: Areas Related to Circles

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📚 Chapter Overview:

This chapter helps students apply the properties of circles to find the area and perimeter (circumference) of sectors, segments, and composite figures involving circles. It blends geometry and mensuration skills for real-life applications.

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🧠 Key Concepts, Formulas, and Theorems:

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🔹 1. Circle:

A closed two-dimensional figure where all points are equidistant from a fixed point (center).

• Radius (r): Distance from center to any point on the circle.

• Diameter (d): Twice the radius, i.e., $d = 2r$

• Circumference: Total boundary length of the circle

  $$\text{Circumference} = 2\pi r$$

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🔹 2. Area of a Circle:

$$\text{Area} = \pi r^2$$

Where $\pi = \frac{22}{7} \text{ or } 3.14$

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🔹 3. Sector of a Circle:

A sector is a portion of a circle enclosed by two radii and the arc between them.

• Area of Sector:

  $$\frac{\theta}{360^\circ} \times \pi r^2$$

• Length of Arc:

  $$\frac{\theta}{360^\circ} \times 2\pi r$$

  where $\theta$ is the central angle.

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🔹 4. Segment of a Circle:

A segment is a region bounded by an arc and a chord.

• Area of Segment = Area of Sector – Area of Triangle

  • For minor segment (smaller one)

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🔹 5. Area of Combined Figures:

Involves use of:

• Semi-circles

• Quadrants (¼ circle)

• Rings (area between two concentric circles)

• Circle inside square, square inside circle, etc.

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📘 NCERT Exercise Solutions

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✅ Exercise 12.1 – Basic Formulas

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Q1. Find the circumference of the circle of radius 10 cm.

$$\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 10 = \frac{440}{7} \approx 62.86 \text{ cm}$$

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Q2. The radius of a circular garden is 28 m. Find its area.

$$\text{Area} = \pi r^2 = \frac{22}{7} \times 28^2 = \frac{22}{7} \times 784 = 2464 \text{ m}^2$$

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Q3. Find the length of arc of a sector with angle 60° and radius 21 cm.

$$\text{Arc length} = \frac{60}{360} \times 2\pi r = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = 22 \text{ cm}$$

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Q4. Area of sector with radius 14 cm and angle 45°.

$$\frac{45}{360} \times \pi r^2 = \frac{1}{8} \times \frac{22}{7} \times 14^2 = \frac{1}{8} \times \frac{22}{7} \times 196 = 77 \text{ cm}^2$$

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✅ Exercise 12.2 – Segments and Shaded Regions

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Q1. Find the area of the segment of a circle of radius 10.5 cm with central angle 60°.

Step 1: Area of sector:

$$= \frac{60}{360} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 10.5^2 = 57.75 \text{ cm}^2$$

Step 2: Area of equilateral triangle using:

$$= \frac{\sqrt{3}}{4} a^2 = \frac{1.732}{4} \times 10.5^2 = 47.63 \text{ cm}^2$$

Segment = Sector – Triangle

$$= 57.75 – 47.63 = \boxed{10.12 \text{ cm}^2}$$

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Q2. Find area of minor segment in circle with radius 14 cm and angle 90°.

Sector area:

$$= \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14^2 = 154 \text{ cm}^2$$

Triangle area (right-angled):

$$= \frac{1}{2} r^2 = \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2$$

Segment area:

$$= 154 – 98 = \boxed{56 \text{ cm}^2}$$

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✅ Exercise 12.3 – Application-Based Problems

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Q1. A path of 2.5 m width is constructed around a circular garden of radius 15.5 m. Find area of path.

Outer radius = $r_2 = 15.5 + 2.5 = 18$ m
Inner radius = $r_1 = 15.5$ m

$$\text{Area} = \pi (r_2^2 - r_1^2) = \frac{22}{7}(18^2 - 15.5^2) = \frac{22}{7}(324 - 240.25) = \frac{22}{7} \times 83.75 = \boxed{263.5 \text{ m}^2}$$

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Q2. Four sectors each of radius 7 cm are cut from a circle and placed in the corners of a square (side 14 cm). Find area not covered.

Area of square = $14^2 = 196 \text{ cm}^2$
Area of 4 sectors = Area of complete circle = $\pi r^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2$

Shaded area = $196 - 154 = \boxed{42 \text{ cm}^2}$

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📝 CBSE Previous Year Questions (PYQs)

Year	Question	Marks
2023	Find area and arc length of sector with radius 21 cm, angle 120°	3
2021	Path 2 m wide around a circular garden of radius 28 m	3
2020	A circle with radius 7 m has 90° unplanted flower bed – find remaining area	3
2019	Find segment area of circle with radius 14 cm, angle 60°	3
2018	A lawn in shape of quadrant of radius 14 m. Find its area.	3
2016	Find shaded area formed by circle and square overlapping	4
2015	Area of sector with radius 28 cm and angle 45°	3

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📌 Summary of Important Formulas

Concept	Formula
Area of Circle	$\pi r^2$
Circumference	$2\pi r$
Area of Sector	$\frac{\theta}{360} \times \pi r^2$
Length of Arc	$\frac{\theta}{360} \times 2\pi r$
Area of Segment	Area of sector – Area of triangle
Area of Ring	$\pi (R^2 - r^2)$

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