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Mensuration - Area and Circumference of a Circle

Grade 9ICSE

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

🔑Concepts

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The basic parts of a circle include the radius (rr), the distance from the center to the edge; the diameter (dd), which is 2r2r and passes through the center; and the circumference (CC), which is the perimeter of the circle.

A circle showing radius r and diameter d passing through center O.
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A semi-circle is exactly half of a circle. Its area is half the area of a circle, but its perimeter includes the curved arc length (πr\pi r) plus the straight diameter (2r2r).

A semi-circle highlighting the straight diameter and the curved arc.
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A circular ring (or annulus) is the region between two concentric circles. The area of the ring is calculated by subtracting the area of the inner circle from the area of the outer circle: Area=πR2−πr2Area = \pi R^2 - \pi r^2.

Concentric circles showing the outer radius R and inner radius r.
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The distance covered by a rotating wheel in one complete revolution is equal to its circumference (2πr2\pi r). To find the total distance, multiply the circumference by the number of rotations.

📐Formulae

d=2rd = 2r

Circumference(C)=2πr=πdCircumference (C) = 2 \pi r = \pi d

Area(A)=πr2Area (A) = \pi r^2

Area of a Semi−circle=12πr2Area \ of \ a \ Semi-circle = \frac{1}{2} \pi r^2

Perimeter of a Semi−circle=πr+2rPerimeter \ of \ a \ Semi-circle = \pi r + 2r

Area of a Ring(Annulus)=πR2−πr2=π(R2−r2)Area \ of \ a \ Ring (Annulus) = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)

Radius(r)=AπRadius (r) = \sqrt{\frac{A}{\pi}}

💡Examples

Problem 1:

Find the circumference and the area of a circle whose radius is 77 cm. (Take π=227\pi = \frac{22}{7})

Solution:

Step 1: Identify the given values. Given, radius r=7r = 7 cm.

Step 2: Calculate the circumference using the formula C=2πrC = 2 \pi r. C=2×227×7C = 2 \times \frac{22}{7} \times 7 C=2×22=44C = 2 \times 22 = 44 cm.

Step 3: Calculate the area using the formula A=πr2A = \pi r^2. A=227×7×7A = \frac{22}{7} \times 7 \times 7 A=22×7=154A = 22 \times 7 = 154 cm2^2.

Explanation:

This problem demonstrates the direct application of basic circle formulas. We substitute the known radius into the standard equations for circumference and area to find the results.

Problem 2:

The circumference of a circle is 8888 cm. Find its area.

Solution:

Step 1: Use the circumference to find the radius rr. 2πr=882 \pi r = 88 2×227×r=882 \times \frac{22}{7} \times r = 88 447×r=88\frac{44}{7} \times r = 88 r=88×744=2×7=14r = 88 \times \frac{7}{44} = 2 \times 7 = 14 cm.

Step 2: Use the radius to find the area. Area=πr2Area = \pi r^2 Area=227×14×14Area = \frac{22}{7} \times 14 \times 14 Area=22×2×14Area = 22 \times 2 \times 14 Area=44×14=616Area = 44 \times 14 = 616 cm2^2.

Explanation:

In this problem, the radius is not given directly. We must first use the given circumference to solve for the unknown radius. Once the radius is found, we can then proceed to calculate the area.

Problem 3:

A circular path of width 77 m is built around a circular park of radius 2121 m. Find the area of the path.

Diagram showing a circular park with radius 21m and a surrounding path of 7m.

Solution:

  1. Let the radius of the inner park be r=21r = 21 m.
  2. The width of the path is 77 m, so the outer radius R=r+width=21+7=28R = r + width = 21 + 7 = 28 m.
  3. The area of the path is the area of the ring: Area=πR2−πr2Area = \pi R^2 - \pi r^2 Area=π(R2−r2)=227×(282−212)Area = \pi(R^2 - r^2) = \frac{22}{7} \times (28^2 - 21^2)
  4. Using identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b): Area=227×(28−21)(28+21)Area = \frac{22}{7} \times (28-21)(28+21) Area=227×7×49Area = \frac{22}{7} \times 7 \times 49 Area=22×49=1078 m2Area = 22 \times 49 = 1078 \text{ m}^2

Explanation:

To find the area of a path around a circle, we calculate the difference between the areas of the larger outer circle and the smaller inner circle.

Problem 4:

Find the perimeter of a semi-circular plate whose diameter is 1414 cm. (Take π=227\pi = \frac{22}{7})

A semi-circular plate with a diameter of 14 cm.

Solution:

  1. Given diameter d=14d = 14 cm, so radius r=142=7r = \frac{14}{2} = 7 cm.
  2. Perimeter of a semi-circle is given by the sum of the arc length and the diameter: Perimeter=πr+dPerimeter = \pi r + d
  3. Substitute the values: Perimeter=(227×7)+14Perimeter = (\frac{22}{7} \times 7) + 14 Perimeter=22+14=36 cmPerimeter = 22 + 14 = 36 \text{ cm}

Explanation:

The perimeter of a semi-circle consists of the curved boundary (half the circumference) plus the straight base (the diameter).