1. Chapter 13 Class 10 Surface Areas and Volumes
  2. Serial order wise
  3. Ex 13.1

Ex 13.1, 1 - Class 10

Last updated at Dec. 8, 2016 by

Ex 13.1, 1 - 2 cubes of volume 64 cm3 are joined end to end - Surface Area - Added

  1. Chapter 13 Class 10 Surface Areas and Volumes
  2. Serial order wise
    3. Ex 13.1
    Ex 13.2 →
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Transcript

Ex 13.1, 1 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. It is given that Volume of 1 cube = 64 cm3 (side)3 = 64 cm3 a3 = 64 cm3 a3 = 43 a = 4 So, a = 4 cm Surface area of cuboid Length of cuboid = a + a = 4 cm + 4 cm = 8 cm Breadth = a = 4 cm Height = a = 4 cm Hence, l = 8 cm, b = 4 cm , h = 4 cm Area of cuboid = 2(lb + bh + hl) = 2(8 ×4+4×4+4×8) = 2(32 + 16 + 32) = 2(80) = 160 Hence, area of cuboid = 160 cm2

Chapter 13 Class 10 Surface Areas and Volumes
Serial order wise

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