Plese derive the formula for the diagonal of cube, cubiod and rectangle.

@Prakhar: Good attempt.! Here is the proper explanation.

For rectangle:

:

Consider a rectangle ABCD having length l and breadth b.

In right ΔBCD, 

diag. BD2 = BC2 + CD2  (By pythagoras theorem,as each angle is of 90°)

Thus, diagonal of a rectangle is given by , where l and b are length and breadth of  a rectangle respectively.

For cuboid:

Consider a cuboid ABCDWXYZ having length, breadth and height as l, b and h repectively and AX is the diagonal of the cuboid.

Consider face ABCD, having AC as the diagonal.

In ΔABC, by Pythagoras theorem,

AC2 = AB2+ BC2

⇒ AC2= l2 + b2    ... (1)

Consider right ΔACX,

diagonal AX2 = AC2 + CX2

⇒ AX2 = (l2+ b2) + h2  (Using (1))

⇒ AX2 = l2 + b2 + h2

Thus, diagonal of cuboid is given by , where l, b and h are the length, breadth and height of the cuboid respectively.

 

For cube:

We know diagonal of cuboid is given by .  (Proved above)

For cube l = b = h = a (say)

Thus, diagonal of cube =

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 Very conceptual

Diagonal of a cube

since all the three sides of a cube are equal therefore let us suppose each side as l

therefore

diagonal =sqrt(l^2+l^2+l^2)=sqrt(3)*l it is same as the diagonal of a cubiod but here all the sides are equal

same for the cubiod as the length breadth and height are l b h

and therefore diagonal= sqrt(l^2+b^2+h^2)

and rectangle it forms a right angled triangle with d diagonal as hypotenuse and l and b are length and breadth therfore diagonal equals sqrt(l^2+b^2)

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Thanks a lot, sir. But, I have a doubt. According to the diagram, angle ACX is not equal to 90 degrees. So, Pythagoras' theorem can't be applied in triangle ACX.

therefore, please tell another method.

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