91Ó°ÊÓ

The given$V−ABCD$ is a regular square pyramid.

Also,$AB=12$ and $BC=12$.

The edge length of square base is 12 and height of pyramid is 8.

The length of OMcan be calculated as:

$\begin{array}{c}OM=\frac{1}{2}AB\\ =\frac{1}{2}\left(12\right)\\ =6\end{array}$

In$\mathrm{Δ}VOM$ , apply Pythagoras Theorem to calculate the slant height.

$\begin{array}{c}{\left(VO\right)}^2+{\left(OM\right)}^2={\left(VM\right)}^2\\ 8^2+6^2=l{}^2\\ l^2=64+36\\ l^2=100\end{array}$

Further simplify.

$\begin{array}{c}l=\sqrt{100}\\ =Â\pm 10\end{array}$

The height cannot be negative. Neglect the negative value of l.

Therefore, the value of the slant height l is 10 unit.

In $\mathrm{Δ}VBC$, calculate the area.

$\begin{array}{c}\text{Area}\left(\mathrm{Δ}VBC\right)=\frac{1}{2}×\text{base}×\text{height}\\ \text{=}\frac{1}{2}×\left(12\right)×\left(10\right)\\ =6×10\\ =60{\text{unit}}^2\end{array}$

Now, calculate the lateral surface area.

$\begin{array}{c}\text{LateralSurfaceareaofpyramid}=4×\text{areaofone-triangularface}\\ \text{LateralSurfaceareaofpyramid}=4×60\\ =240{\text{unit}}^2\end{array}$

Therefore, the lateral surface area of pyramid shown in figur$240{\text{unit}}^2$e is