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Basic Geometrical Ideas - Circles

Grade 6CBSE

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

🔑Concepts

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A circle is a simple closed curve where every point on the boundary is at an equal distance from a fixed point inside it called the Center (OO).

A circle showing the center point O.
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The Radius (rr) is the line segment connecting the center to any point on the circle. The Diameter (dd) is a line segment passing through the center with both endpoints on the circle; d=2rd = 2r.

A circle illustrating the radius and diameter.
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A Chord is a line segment joining any two points on the circle. The diameter is the longest chord. An Arc is a portion of the circumference.

A circle showing a chord.
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A Sector is the region in the interior of a circle enclosed by an arc and a pair of radii. A Segment is the region enclosed by a chord and an arc.

A circle showing a shaded sector.

📐Formulae

Diameter(d)=2×Radius(r)\text{Diameter} (d) = 2 \times \text{Radius} (r)

Radius(r)=d2\text{Radius} (r) = \frac{d}{2}

Circumference(C)=2×π×r\text{Circumference} (C) = 2 \times \pi \times r

💡Examples

Problem 1:

If the radius of a circle is 5 cm5\text{ cm}, calculate the length of its longest chord.

Solution:

  1. Identify that the longest chord of a circle is its diameter.
  2. Use the formula: d=2×rd = 2 \times r
  3. Substitute the given radius: d=2×5 cmd = 2 \times 5\text{ cm}
  4. d=10 cmd = 10\text{ cm}

Explanation:

Since the diameter is the longest possible chord in any circle, we simply multiply the given radius by 22 to find the answer.

Problem 2:

A circle has a diameter of 14 cm14\text{ cm}. Find the length of the radius.

Solution:

  1. Use the relationship: r=d2r = \frac{d}{2}
  2. Substitute the diameter: r=14 cm2r = \frac{14\text{ cm}}{2}
  3. r=7 cmr = 7\text{ cm}

Explanation:

The radius is always half the length of the diameter. By dividing the diameter by 22, we find the distance from the center to the edge.

Problem 3:

In the given figure, identify the points that lie in the interior, exterior, and on the circle.

Circle with points O, P, Q, A, B at different positions.

Solution:

Interior: P,O\text{Interior: } P, O Exterior: Q\text{Exterior: } Q On the Circle: A,B\text{On the Circle: } A, B

Explanation:

Points PP and OO are inside the boundary (Interior). Point QQ is outside (Exterior). Points AA and BB are exactly on the boundary of the circle.

Problem 4:

If the diameter of a circular park is 20 m20\text{ m}, find the distance from the center to any point on the fence.

Circle with a diameter marked as 20m and a radius marked as r.

Solution:

Distance from center to boundary=Radius(r)\text{Distance from center to boundary} = \text{Radius} (r) r=d2r = \frac{d}{2} r=202=10 mr = \frac{20}{2} = 10\text{ m}

Explanation:

The distance from the center to any point on the boundary of a circle is called the radius. Since the diameter is 20 m20\text{ m}, the radius is half of it, which is 10 m10\text{ m}.