Question 4 - Page 170 - Areas in Real Life - Chapter 7 Class 8 - Area (Ganita Prakash II)
Last updated at March 16, 2026 by Teachoo
Areas in Real Life
Question 1 - Page 170
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Class 8 (Ganita Prakash - 1, 2 & Old NCERT)
Chapter 7 Class 8 - Area (Ganita Prakash II)
- Area of Rectangle and Square
- Perimeter vs Area
- Figure it out - Page 150-152
- Area of Triangle
- Applications of Area of Triangle Formula
- Triangles between Parallel Lines with a Common Base
- Figure it out - Page 157-159
- Area of any Polygon
- Figure it out - Page 160
- Area of Parallelogram
- Area of Rhombus
- Area of Trapezium
- Figure it out - Page 169-170
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Areas in Real Life
Transcript
Question 4 - Page 170 Express the following lengths in inches: (i) 5.08 cm (ii) 11.43 cmWe know that 1 in = 2.54 cm 2.54 cm = 1 in 1 cm = 1/2.54 in 1 cm = 1/(254/100) in 1 cm = ๐๐๐/๐๐๐ in Now, solving each part (i) 5.08 cm Now, 5.08 cm = 5.08 ร 100/254 in = ๐๐๐/๐๐๐ ร ๐๐๐/๐๐๐ = 508/254 ร 100/100 = 2 ร 1 = 2 in (i) 11.43 cm Now, 11.43 cm = 11.43 ร 100/254 in = ๐๐๐๐/๐๐๐ ร ๐๐๐/๐๐๐ = 1143/254 ร 100/100 = 1143/254 ร 1 = 4.5 in = 1143/254 ร 100/100 = 1143/254 ร 1 = 4.5 in
