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Shape and Space - Parts and Properties of Circles

Grade 6IB

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

🔑Concepts

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The center is the fixed point in the middle of the circle, and every point on the boundary is at an equal distance from this center. The radius is the line segment connecting the center to any point on the boundary.

A circle showing the center and the radius.
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The diameter is a line segment passing through the center with both endpoints on the circle's boundary. It is the longest chord and its length is exactly twice the radius (d=2rd = 2r).

A circle showing the diameter passing through the center.
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The circumference is the total distance around the edge of the circle (the perimeter). The ratio of the circumference to the diameter is always a constant called π\pi (approximately 3.143.14 or 227\frac{22}{7}).

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A chord is any line segment joining two points on a circle. A chord that passes through the center is a diameter. An arc is any part of the circumference.

A circle illustrating a chord and an arc.

📐Formulae

d=2×rd = 2 \times r

r=d2r = \frac{d}{2}

C=π×dC = \pi \times d

C=2×π×rC = 2 \times \pi \times r

A=π×r2A = \pi \times r^2

💡Examples

Problem 1:

If a circle has a radius of 7 cm7\text{ cm}, what is its diameter?

Solution:

d=2×r=2×7=14 cmd = 2 \times r = 2 \times 7 = 14\text{ cm}

Explanation:

Since the diameter is twice the length of the radius, we multiply the given radius (7 cm7\text{ cm}) by 22.

Problem 2:

Calculate the circumference of a circle with a diameter of 10 cm10\text{ cm} using π=3.14\pi = 3.14.

Solution:

C=π×d=3.14×10=31.4 cmC = \pi \times d = 3.14 \times 10 = 31.4\text{ cm}

Explanation:

Using the formula C=πdC = \pi d, we substitute 3.143.14 for π\pi and 1010 for dd to get the perimeter of the circle.

Problem 3:

A circular garden has a diameter of 14 m14\text{ m}. Find its radius and the distance around the garden (circumference) using π=227\pi = \frac{22}{7}.

Solution:

r=142=7 mr = \frac{14}{2} = 7\text{ m} C=π×d=227×14=22×2=44 mC = \pi \times d = \frac{22}{7} \times 14 = 22 \times 2 = 44\text{ m}

Explanation:

First, we find the radius by dividing the diameter by 22. Then, we calculate the circumference using the fractional value of π\pi to simplify the multiplication with the diameter.

Problem 4:

Find the area of a circle with a radius of 3 cm3\text{ cm} (Take π=3.14\pi = 3.14).

Solution:

A=π×r2=3.14×32=3.14×9=28.26 cm2A = \pi \times r^2 = 3.14 \times 3^2 = 3.14 \times 9 = 28.26\text{ cm}^2

Explanation:

The area is calculated by squaring the radius (3×3=93 \times 3 = 9) and then multiplying the result by π\pi (3.143.14).

Problem 5:

Calculate the area of a circular tabletop that has a diameter of 20 cm20\text{ cm}. Use π=3.14\pi = 3.14.

A circle with a diameter labeled 20 cm.

Solution:

  1. Find the radius: r=d2=202=10 cmr = \frac{d}{2} = \frac{20}{2} = 10\text{ cm}.
  2. Use the area formula: A=π×r2A = \pi \times r^2.
  3. Substitute the values: A=3.14×(10)2A = 3.14 \times (10)^2.
  4. A=3.14×100=314 cm2A = 3.14 \times 100 = 314\text{ cm}^2.

Explanation:

To find the area, we must first convert the diameter into a radius. Since the radius is half the diameter, we use 10 cm10\text{ cm} in the area formula.

Problem 6:

A bicycle wheel has a radius of 35 cm35\text{ cm}. How far does the wheel travel in one full rotation? (Use π=227\pi = \frac{22}{7})

A circle representing a wheel with a radius of 35 cm on a flat surface.

Solution:

  1. The distance traveled in one rotation is equal to the circumference (CC).
  2. Use the formula: C=2×π×rC = 2 \times \pi \times r.
  3. Substitute the values: C=2×227×35C = 2 \times \frac{22}{7} \times 35.
  4. Simplify: C=2×22×5C = 2 \times 22 \times 5.
  5. C=44×5=220 cmC = 44 \times 5 = 220\text{ cm}.

Explanation:

One full rotation of a wheel is equivalent to its circumference. We use the radius provided and the fraction value of π\pi to simplify the calculation.

Parts and Properties of Circles Grade 6 Notes & Examples