1. **Problem statement:** We have a square-based pyramid with a base side length of 6 cm and triangular faces with slant height 11 cm. We need to find the vertical height of the pyramid (the perpendicular distance from the apex to the base). 2. **Understanding the pyramid:** The base is a square with side length $6$ cm. The triangular faces are isosceles triangles with a slant height (the height of each triangular face) of $11$ cm. 3. **Key formula and approach:** The height $h$ of the pyramid can be found using the Pythagorean theorem in the right triangle formed by: - the height $h$ (vertical height from apex to base), - half the base diagonal (distance from center of base to midpoint of a side), - the slant height $l = 11$ cm (hypotenuse). 4. **Calculate half the base diagonal:** The diagonal $d$ of the square base is given by: $$d = 6\sqrt{2}$$ Half the diagonal is: $$\frac{d}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$$ 5. **Apply Pythagorean theorem:** $$h = \sqrt{l^2 - \left(\frac{d}{2}\right)^2} = \sqrt{11^2 - (3\sqrt{2})^2}$$ 6. **Calculate inside the square root:** $$11^2 = 121$$ $$(3\sqrt{2})^2 = 3^2 \times 2 = 9 \times 2 = 18$$ 7. **Calculate height:** $$h = \sqrt{121 - 18} = \sqrt{103} \approx 10.1489$$ 8. **Final answer:** The height of the pyramid is approximately $10.15$ cm (to 2 decimal places).