Question 19 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

Transcript

Question 19 The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm^2. Find the perimeter of each small rectangle. Let the Longer side of smaller rectangle be a And Shorter side of smaller rectangle be b Labelling our figure Thus, 4a = 5b a = 𝟓/𝟒b Writing everything in terms of b And, Length of big rectangle = 5b Breadth of big rectangle = b + a = b + 5/4b = b(1+5/4) = b((4 + 5)/4) = 𝟗𝒃/𝟒 Given that Area of large rectangle is 72 〖𝒄𝒎〗^𝟐 Now, Area of large rectangle = 72 Length × Breadth = 72 𝟓𝐛 × 𝟗𝒃/𝟒=𝟕𝟐 (45𝑏^2)/4=72 b^2=72 ×4/45 b^2=24 ×4/15 b^2=8 ×4/5 b^2=4 × 2 × 4 × 1/5 b^2=4^2 ×2/5 b=√(4^2 ) × √(2/5) 𝐛=𝟒 × √(𝟐/𝟓) And, b=√(4^2 ) × √(2/5) 𝐛=𝟒 × √(𝟐/𝟓) We need to find perimeter of small rectangle Perimeter of small rectangle = 2 × (Length + Breadth) = 2(a + b) Putting a = 5/4b = 2 × (5/4b + b) = 2 × 9/4b = 𝟗/𝟐b = 9/2 × 𝟒 × √(𝟐/𝟓) = 18√(𝟐/𝟓) = 𝟏𝟖√(𝟎.𝟒) cm