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Measuring Space: Perimeter and Area - Area of a Circle

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A circle is defined as the set of all points in a plane that are at a fixed distance (radius) from a fixed point (center).

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The radius (rr) is the distance from the center to any point on the boundary, while the diameter (dd) is twice the radius (d=2rd = 2r).

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The Area of a circle is the total region enclosed within its boundary, measured in square units (e.g., cm2\text{cm}^2, m2\text{m}^2).

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The constant π\pi (pi) is the ratio of the circumference to the diameter, approximately taken as 227\frac{22}{7} or 3.143.14.

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A semicircle is half of a circle, so its area is 12πr2\frac{1}{2} \pi r^2.

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A quadrant is one-fourth of a circle, so its area is 14πr2\frac{1}{4} \pi r^2.

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The area of a circular ring (or annulus) formed by two concentric circles with radii RR and rr (where R>rR > r) is the difference between the areas of the two circles.

📐Formulae

Area of a Circle (A)=πr2\text{Area of a Circle (A)} = \pi r^2

Diameter (d)=2r\text{Diameter (d)} = 2r

Circumference (C)=2πr\text{Circumference (C)} = 2\pi r

Area of Semicircle=12πr2\text{Area of Semicircle} = \frac{1}{2} \pi r^2

Area of Quadrant=14πr2\text{Area of Quadrant} = \frac{1}{4} \pi r^2

Area of Circular Ring=π(R2−r2)\text{Area of Circular Ring} = \pi(R^2 - r^2)

💡Examples

Problem 1:

Find the area of a circular sheet of paper whose radius is 14 cm14 \text{ cm}. (Use π=227\pi = \frac{22}{7})

Solution:

Given: Radius (rr) = 14 cm14 \text{ cm}. Using the formula: A=πr2A = \pi r^2 A=227×14×14A = \frac{22}{7} \times 14 \times 14 A=22×2×14A = 22 \times 2 \times 14 A=44×14A = 44 \times 14 A=616 cm2A = 616 \text{ cm}^2

Explanation:

To find the area, substitute the given radius into the formula πr2\pi r^2 and simplify the calculation by cancelling 1414 with the denominator 77.

Problem 2:

The area of a circle is 154 cm2154 \text{ cm}^2. Find its radius and circumference.

Solution:

Given: Area (AA) = 154 cm2154 \text{ cm}^2.

  1. Find radius (rr): πr2=154\pi r^2 = 154 227×r2=154\frac{22}{7} \times r^2 = 154 r2=154×722r^2 = \frac{154 \times 7}{22} r2=7×7=49r^2 = 7 \times 7 = 49 r=49=7 cmr = \sqrt{49} = 7 \text{ cm}

  2. Find circumference (CC): C=2πrC = 2\pi r C=2×227×7C = 2 \times \frac{22}{7} \times 7 C=44 cmC = 44 \text{ cm}

Explanation:

First, use the area formula to solve for r2r^2, then take the square root to find rr. Finally, use the radius to calculate the circumference.

Problem 3:

A circular track has an outer radius of 21 m21 \text{ m} and an inner radius of 14 m14 \text{ m}. Find the area of the track.

Solution:

Given: Outer radius (RR) = 21 m21 \text{ m}, Inner radius (rr) = 14 m14 \text{ m}. Area of track = Area of outer circle - Area of inner circle A=πR2−πr2=π(R2−r2)A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) A=227(212−142)A = \frac{22}{7}(21^2 - 14^2) Using the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b): A=227(21−14)(21+14)A = \frac{22}{7}(21 - 14)(21 + 14) A=227×7×35A = \frac{22}{7} \times 7 \times 35 A=22×35A = 22 \times 35 A=770 m2A = 770 \text{ m}^2

Explanation:

The area of a track (ring) is calculated by subtracting the area of the smaller inner circle from the larger outer circle. Using the algebraic identity (a2−b2)(a^2-b^2) makes the calculation easier.