Question

A solid cube whose edges are 4 cm each has been cut by a plane that bisects the three edges meeting at a corner and the smaller part is removed.

What is the surface area of remaining part ? (useLaTeX: \sqrt{3}=1.53 = 1.5)

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Answer 1

Given: edge of cube is 4 cm.

To find: What is the surface area of remaining part ?

Solution:

• Now we have given that a solid cube whose edges are 4 cm each has been cut by a plane that bisects the three edges meeting at a corner and the smaller part is removed.

• Now since the edge is bisecting, then remaining side will be 2 cm.

• A right angled triangle will be formed after cutting.

• So the hypotenuse will be:

                 h = √(2^2 + 2^2) = √8 = 2√2 cm

• So area of triangle to be excluded from 3 sides will be:

                 1/2 x b x h = 1/2 x 4 x 4 = 8 cm^2

• Area of inclined part will be:

                 √3 / 4 (side)^2 = √3 / 4 (2√2)^2

                 √3 / 4 (4x2) =  2√3

• Area of squares which are not cut is:

                 3 x (4 x 4) = 48

• So now adding all surface areas we get:

• So the total surface area will be:

                 48 + 3((4)^2 - 8 ) + 2√3

                 48 + 3(16 - 8) + 2√3

                 48 + 3(16 - 8) + 2√3

                 48 + 3(8) + 2√3

                 48 + 24 + 2√3

                 72 + 2√3

Answer:

          So the area is 72 + 2√3 cm^2.