Example 1 - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

Transcript

Example 1 Two circles of equal radius are located such that each circle passes through the centre of the other circle (Fig. 6.12). Given that the radius of each circle is r units, find the perimeter of the shape formed by the two circles in terms of r units. Since both circles have same radius, we can write Perimeter of shape = 2 × Circumference of circle – Length of arc CAD – Length of arc CBD By Symmetry, both arcs would be the same So, we can write Perimeter of shape = 2 × Circumference of circle – 2 × Length of arc CBD Finding Length of arc CBD Now, AB = AC = BC = Radius = r So, ∆ ABC is an equilateral triangle Thus, all angles are 60° ∴ ∠ CAB = 60° By symmetry, ∠ CAD = 60° For arc CBD in left circle Angle = 𝜃 = 60° + 60° = 120° Radius = r So, we can write Length of arc CBD = 2𝜋𝑟 × 𝜃/(360° ) = 𝟐𝝅𝒓 × (𝟏𝟐𝟎° )/(𝟑𝟔𝟎° ) = 2𝜋𝑟 × 1/3 = 𝟐𝝅𝒓/𝟑 Now, Perimeter of shape = 2 × Circumference of circle – 2 × Length of arc CBD = 2 × 𝟐𝝅𝒓− 2 × 𝟐𝝅𝒓/𝟑 = 𝟒𝝅𝒓− 𝟒𝝅𝒓/𝟑 So, we can write Length of arc CBD = 2𝜋𝑟 × 𝜃/(360° ) = 𝟐𝝅𝒓 × (𝟏𝟐𝟎° )/(𝟑𝟔𝟎° ) = 2𝜋𝑟 × 1/3 = 𝟐𝝅𝒓/𝟑 Now, Perimeter of shape = 2 × Circumference of circle – 2 × Length of arc CBD = 2 × 𝟐𝝅𝒓− 2 × 𝟐𝝅𝒓/𝟑 = 4𝜋𝑟− 4𝜋𝑟/3 = 4𝜋𝑟(1−1/3) = 4𝜋𝑟 ×2/3 = 𝟖𝝅𝒓/𝟑