1. **State the problem:** We have a right square pyramid with a total surface area of $100 + 20\sqrt{146}$ square inches, including a base area of 100 square inches. We need to find the height of the pyramid. 2. **Recall the formulas:** - Surface area of a right square pyramid = Base area + Lateral area - Base area $= s^2$ where $s$ is the side length of the square base - Lateral area $= 2 s l$ where $l$ is the slant height - Height $h$, slant height $l$, and half the base side $\frac{s}{2}$ form a right triangle: $$l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}$$ 3. **Given:** - Base area $= 100$ so $s^2 = 100 \Rightarrow s = 10$ - Total surface area $= 100 + 20\sqrt{146}$ 4. **Find lateral area:** $$\text{Lateral area} = \text{Total surface area} - \text{Base area} = (100 + 20\sqrt{146}) - 100 = 20\sqrt{146}$$ 5. **Use lateral area formula:** $$20\sqrt{146} = 2 \times 10 \times l = 20l$$ Divide both sides by 20: $$\cancel{20} \sqrt{146} = \cancel{20} l \Rightarrow l = \sqrt{146}$$ 6. **Use right triangle relation to find height $h$:** $$l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}$$ Substitute $l = \sqrt{146}$ and $s = 10$: $$\sqrt{146} = \sqrt{h^2 + 5^2} = \sqrt{h^2 + 25}$$ Square both sides: $$146 = h^2 + 25$$ Solve for $h^2$: $$h^2 = 146 - 25 = 121$$ Take the positive root (height is positive): $$h = 11$$ **Final answer:** The height of the pyramid is $11$ inches.