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Properties of cuboids.

1. A cuboid is made up of six rectangles, each of the rectangles is called the face. In the figure above, ABFE, DAEH, DCGH, CBFG, ABCD and EFGH are the 6 faces of cuboid.

2. Base of cuboid – Any face of a cuboid may be called as the base of cuboid.

3. Edges – The edge of the cuboid is a line segment between any two adjacent vertices.

There are 12 edges. AB, AD, AE, HD, HE, HG, GF, GC, FE, FB, EF and CD

Opposite edges of a cuboid are equal.

4. Vertices – The point of intersection of the 3 edges of a cuboid is called the vertex of a cuboid.

A cuboid has 8 vertices

A, B, C, D, E, F, G, H

“All of a cuboid corners (vertices) are 90 degree angles”

5. Diagonal of cuboid – The length of diagonal of the cuboid of given by :

Diagonal of the cuboid $ = \sqrt {({l^2} + {b^2} + {h^2})} $

$HG = 11cm$

$FG = 4cm$

$BF = 8cm$

We have to find AG

In $\Delta EFG,$ angle $EFG = 90^\circ $

Because we know that “all of a cuboids”

And it is given that

$EF = 11cm$ (as opposite sides are equal)

$FG = 4cm$ (given)

So by using Pythagoras theorem we will find EG.

As EG is hypotenuse of triangle EFG so,

$E{G^2} = E{F^2} + F{G^2}$

$E{G^2} = {11^2} + {4^2}$

$E{G^2} = 137$

or

$EG = \sqrt {137} cm$ …..(1)

Now,

In $\Delta AEG,$ angle $AEG = 90^\circ $ (vertex of cuboid)

$AE = 8cm$ (given)

$EG = \sqrt {137} cm$ [from equation (1)]

So we can apply Pythagoras theorem in this triangle.

In $\Delta AEG$

$A{G^2} = A{E^2} + E{G^2}$

$A{G^2} = {8^2} + {(\sqrt {137} )^2}$

$A{G^2} = 64 + 137$

$A{G^2} = 201$

$AG = \sqrt {201} $

or

$AG = 14.17cm$

So, If ABCDEFGH is a cuboid. Then length of $AG = 14.17cm$

We know that in cuboid ABCDEFGH,

AG is the diagonal of cuboid.

So we can directly use the formula for cuboid diagonal.

Diagonal of the cuboid $ = \sqrt {({l^2} + {b^2} + {h^2})} $

Where

$l = length = HG = 11cm$

$b = breadth = FG = 4cm$

$h = height = BF = 8cm$

Substitute these values in above formula,

$AG = \sqrt {({{11}^2} + {4^2} + {8^2})} $

$AG = \sqrt {(121 + 16 + 64)} $

$AG = \sqrt {201} $

$AG = 14.17cm$