1. **State the problem:** We have a square-based pyramid with a slant edge of 13 cm and a base length of 10 cm. We need to find the slant height of the pyramid. 2. **Understand the terms:** - The base is a square with side length $10$ cm. - The slant edge is the edge from the apex to a vertex of the base, length $13$ cm. - The slant height is the height of a triangular face, measured from the apex perpendicular to the base edge (not the base vertex). 3. **Visualize the right triangle:** - The slant edge, slant height, and half the base diagonal form a right triangle. 4. **Calculate half the base diagonal:** - The diagonal of the square base is $d = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$ cm. - Half the diagonal is $\frac{d}{2} = 5\sqrt{2}$ cm. 5. **Apply the Pythagorean theorem:** - Let $l$ be the slant height. - Then, $$l^2 + \left(5\sqrt{2}\right)^2 = 13^2$$ - Simplify: $$l^2 + 25 \times 2 = 169$$ - $$l^2 + 50 = 169$$ - $$l^2 = 169 - 50 = 119$$ 6. **Find the slant height:** - $$l = \sqrt{119} \approx 10.91 \text{ cm}$$ **Final answer:** The slant height of the pyramid is approximately $10.91$ cm.