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It is given that the surface area of cuboid is $758c{{m}^{2}}$ and length and breadth of the cuboid is equal to 14 cm and 11 cm respectively.

In the below diagram, we have drawn a cuboid having length, breadth and height as l, b and h respectively.

In the above diagram, we have shown length of AC by l, length of AE by b and length of AB by h.

We know that the formula of surface area of the cuboid is equal to:

$\text{Surface area}=2\left( lb+bh+hl \right)$……….. Eq. (1)

In the above formula, l, b and h are length, breadth and height of the cuboid respectively.

Substituting the value of l and b as 14 cm and 11 cm respectively and surface area as $758c{{m}^{2}}$ in eq. (1) we get,

$\begin{align}

& 758=2\left( 14\left( 11 \right)+\left( 11 \right)h+h\left( 14 \right) \right) \\

& \Rightarrow 758=2\left( 154+25h \right) \\

\end{align}$

Dividing 2 on both the sides of the above equation we get,

$\begin{align}

& \dfrac{758}{2}=154+25h \\

& \Rightarrow 379=154+25h \\

\end{align}$

Subtracting 154 on both the sides of the above equation we get,

$\begin{align}

& 379-154=25h \\

& 225=25h \\

\end{align}$

Dividing 5 on both the sides we get,

$\begin{align}

& \dfrac{225}{25}=h \\

& \Rightarrow 9cm=h \\

\end{align}$