The side lengths of a cu be are congruent, so the edge length of the cu be is 65 mm
The length of diagonal A H is given as:
[tex]\mathbf{AH = 11.3cm}[/tex]
Considering triangle A FH, we have:
[tex]\mathbf{AH^2 = A\ F^2 + FH^2}[/tex]
So, we have:
[tex]\mathbf{11.3^2 = A\ F^2 + FH^2}[/tex]
Considering triangle HE F, we have:
[tex]\mathbf{FH^2 = EF^2 + EH^2}[/tex]
Substitute [tex]\mathbf{FH^2 = EF^2 + EH^2}[/tex] in [tex]\mathbf{11.3^2 = A\ F^2 + FH^2}[/tex]
[tex]\mathbf{11.3^2 = A\ F^2 + EF^2 + EH^2}[/tex]
The side lengths of a cu be are congruent.
This means that: A F = EF = EH
So, we have:
[tex]\mathbf{11.3^2 = A\ F^2 + A\ F^2 + A\ F^2}[/tex]
Evaluate like terms
[tex]\mathbf{11.3^2 = 3A\ F^2}[/tex]
Evaluate squares
[tex]\mathbf{127.69 = 3A\ F^2}[/tex]
Divide both sides by 3
[tex]\mathbf{42.563= A\ F^2}[/tex]
Take square roots
[tex]\mathbf{6.52= A\ F}[/tex]
Rewrite as:
[tex]\mathbf{A\ F = 6.52cm}[/tex]
Convert to m m
[tex]\mathbf{A\ F = 6.52 \times 10mm}[/tex]
[tex]\mathbf{A\ F = 65.2\ mm}[/tex]
Approximate
[tex]\mathbf{A\ F = 65\ mm}[/tex]
Hence, the edge length is 65 mm
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