A right pyramid stands on a square base of diagonal 10√2 cm. If the height of the pyramid is 12 cm, the area (in cm²) of its slant surface is

$$\eqalign{ & {\text{Side of square}} = \frac{1}{{\sqrt 2 }} \times 10\sqrt 2 = 10{\text{ cm}} \cr & {\text{Slant height}} = \sqrt {{5^2} + {{12}^2}} = 13{\text{ cm}} \cr & {\text{Lateral surface area}} \cr & = \frac{1}{2} \times {\text{Perimeter of base}} \times {\text{Slant height}} \cr & = \frac{1}{2} \times 40 \times 30 \cr & = 260{\text{ c}}{{\text{m}}^2} \cr} $$