Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

Last updated at June 3, 2026 by

What fraction of the rectangle is covered by the circles? [Ganita Man] - End-of-Chapter Exercises

part 2 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9
part 3 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9 part 4 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9 part 5 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9 part 6 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9 part 7 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9 part 8 - Question 17 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9

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Question 17 – Part 1 What fraction of the rectangle is covered by the circles? Let Length of Rectangle = l Breadth of Rectangle = b Labelling our figure Now, Diameter of circle = b ∴ Radius of circle = 𝒃/𝟐 And, 3 Γ— Diameter = Length of rectangle 3 Γ— b = l l = 3b We need to find fraction of the rectangle is covered by the circles So, we need to find (𝑨𝒓𝒆𝒂 𝒐𝒇 πŸ‘ π’„π’Šπ’“π’„π’π’†π’”)/(𝑨𝒓𝒆𝒂 𝒐𝒇 π‘Ήπ’†π’„π’•π’‚π’π’ˆπ’π’†) Now, (𝑨𝒓𝒆𝒂 𝒐𝒇 π’„π’Šπ’“π’„π’π’†)/(𝑨𝒓𝒆𝒂 𝒐𝒇 π‘Ήπ’†π’„π’•π’‚π’π’ˆπ’π’†) Now, (𝑨𝒓𝒆𝒂 𝒐𝒇 πŸ‘ π’„π’Šπ’“π’„π’π’†π’”)/(𝑨𝒓𝒆𝒂 𝒐𝒇 π‘Ήπ’†π’„π’•π’‚π’π’ˆπ’π’†)=(3 Γ— πœ‹ Γ— π‘…π‘Žπ‘‘π‘–π‘’π‘ ^2)/(πΏπ‘’π‘›π‘”π‘‘β„Ž Γ— π΅π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž) =(3 Γ— πœ‹ Γ— (𝑏/2)^2)/(𝑙 Γ— 𝑏) =(πŸ‘ Γ— 𝝅 Γ— (𝒃/𝟐)^𝟐)/(πŸ‘π’ƒ Γ— 𝒃) =(πœ‹ Γ— (𝑏/2)^2)/𝑏^2 =(πœ‹ ×𝑏^2/4)/𝑏^2 =(πœ‹ Γ— 𝑏^2)/(𝑏^2 Γ— 4) =𝝅/πŸ’ Thus, Fraction covered is 𝝅/πŸ’ Question 17 – Part 2 What fraction of the rectangle is covered by the circles? Let Length of Rectangle = l Breadth of Rectangle = b Labelling our figure Now, Diameter of circle = b ∴ Radius of circle = 𝒃/𝟐 And, 3 Γ— Diameter = Length of rectangle 4 Γ— b = l l = 4b We need to find fraction of the rectangle is covered by the circles So, we need to find (𝑨𝒓𝒆𝒂 𝒐𝒇 πŸ’ π’„π’Šπ’“π’„π’π’†π’”)/(𝑨𝒓𝒆𝒂 𝒐𝒇 π‘Ήπ’†π’„π’•π’‚π’π’ˆπ’π’†) Now, (𝑨𝒓𝒆𝒂 𝒐𝒇 πŸ’ π’„π’Šπ’“π’„π’π’†π’”)/(𝑨𝒓𝒆𝒂 𝒐𝒇 π‘Ήπ’†π’„π’•π’‚π’π’ˆπ’π’†)=(4 Γ— πœ‹ Γ— π‘…π‘Žπ‘‘π‘–π‘’π‘ ^2)/(πΏπ‘’π‘›π‘”π‘‘β„Ž Γ— π΅π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž) =(4 Γ— πœ‹ Γ— (𝑏/2)^2)/(𝑙 Γ— 𝑏) =(πŸ’ Γ— 𝝅 Γ— (𝒃/𝟐)^𝟐)/(πŸ’π’ƒ Γ— 𝒃) =(πœ‹ Γ— (𝑏/2)^2)/𝑏^2 =(πœ‹ ×𝑏^2/4)/𝑏^2 =(πœ‹ Γ— 𝑏^2)/(𝑏^2 Γ— 4) =𝝅/πŸ’

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